Prasanna

Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Lesson 11.3 Answer Key Generating Equivalent Expressions.

Texas Go Math Grade 6 Lesson 11.3 Answer Key Generating Equivalent Expressions

Essential Question
How can you identify and write equivalent expressions?

Texas Go Math Grade 6 Lesson 11.3 Explore Activity Answer Key

Explore Activity 1
Identifying Equivalent Expressions
One way to test whether two expressions might be equivalent is to evaluate them for the same value of the variable.

Match the expressions in List A with their equivalent expressions in List B.

A. Evaluate each of the expressions in the lists for x = 3.

B. Which pair(s) of expressions have the same value for x = 3?

C. How could you further test whether the expressions in each pair are equivalent?

D. Do you think the expressions in each pair are equivalent? Why or why not?

Reflect

Go Math Grade 6 Lesson 11.3 Answer Key Question 1.
Error Analysis Lisa evaluated the expressions 2x and x² for x = 2 and found that both expressions were equal to 4. Lisa concluded that 2x and x2 are equivalent expressions. How could you show Lisa that she is incorrect?
Answer:
According to Lisa’s conclusion x² = 2x. If this was correct then this equation would have held true for all, values of x, but here this is not the case, as for x = 3, then 3² = 9 and 2(3) = 6 and 9 ≠ 6

Explore Activity 2
Modeling Equivalent Expressions
You can also use models to determine if two expressions are equivalent. Algebra tiles are one way to model expressions.

Determine if the expression 3(x + 2) is equivalent to 3x + 6.

A. Model each expression using algebra tiles.

B. The model for 3(x + 2) has ___ x tiles and ___ 1 tiles.
The model for 3x + 6 has ___ x tiles and ___ 1 tiles.
C. Is the expression 3(x + 2) equivalent to 3x + 6? Explain.

Reflect

Question 2.
Use algebra tiles to determine if 2(x – 3) is equivalent to 2x – 3. Explain your answer.
Answer:
Algebra tiles for the given expressions.

The parenthesis makes a difference in the given expressions. Parenthesis indicates grouping or multiplication which makes the expressions not equivalent
2 (x – 3) = 2x – 6 which is different from 2x – 3.
The expressions are not equivalent because the value of both expressions will not be similar.

Your Turn

For each expression, use a property to write an equivalent expression. Tell which property you used.

Question 3.
(ab)c = ____
Answer:
The operation in the expression is multiplication
You can use the Associative Property of Multiplication to write an equivalent expression:(ab)c = a(bc) (Final solution)
a(bc)

Go Math Lesson 11.3 6th Grade Answer Key Question 4.
3y + 4y = _____
Answer:
Given expression:
3y + 4y = 4y + 3y
According to the Commutative Property of Addition
3y + 4y = 4y + 3y. According to the Commutative Property of Addition.

Your Turn

Use the properties of operations to determine if the expressions are equivalent.

Question 5.
6x – 8; 2(3x – 5)
Answer:
Given expression:
2(3x – 5)
Apply distributive property to expand the parentheses
= 2(3x) + 2(-5)
Expand:
= 6x – 10x ≠ 6x – 8
6x – 10x ≠ 6x – 8

Question 6.
2 – 2 + 5x; 5x
Answer:
Determine if the expressions are equivalent
0 + 5x = 5x identity property of addition
5x = 5x
The expressions are equivalent because of the identity property of addition.

Question 7.
Jamal bought 2 packs of stickers and 8 individual stickers. Use x to represent the number of stickers in a pack of stickers and write an expression to represent the number of stickers Jamal bought. Is the expression equivalent to 2(4 + x)? Check your answer with algebra tile models.
Answer:
There are x stickers in each pack and 2 packs are bought so the total number of stickers bought is x × 2 = 2x. He also bought 8 more stickers, so the total number of stickers bought is 2x + 8.
Expand the expression: 2(4 + x) = 8 + 2x = 2x + 8. Yes, the 2 expressions are equivalent.

Your Turn

Combine like terms.

Question 8.
8y – 3y = ___________
Answer:
Combine like terms
8y – 3y 8y and 3y are like terms
8y – 3y = y(8 – 3) Distributive Property
= y(5) Subtract inside the parentheses
= 5y Commutative Property of Multiplication
5y Final Solution

Go Math Lesson 11.3 Independent Practice Answer Key Question 9.
6x² + 4(x² – 1) = ______
Answer:
Given expression:
6x² + 4(x² – 1)
Apply the distributive property to expand the parentheses:
= 6x² + 4x² – 4
Perform the indicated operation between like terms to combine like terms, therefore:
= 10x² – 4
6x² + 4(x² – 1)= 10x² – 4

Question 10.
4a⁵ – 2a⁵ + 4b + b = ___
Answer:
Given expression:
4a⁵ – 2a⁵ + 4b + b
Perform the indicated operation between like terms to combine like terms, therefore:
= 2a⁵ – 5b
4a⁵ – 2a⁵ + 4b + b = 2a⁵ + 5b

Question 11.
8m + 14 – 12 + 4n = ____
Answer:
Given expression:
8m + 14 – 12 + 4n
Perform the indicated operation between like terms to combine like terms, therefore:
= 8m – 2 – 4n
8m + 14 – 12 + 4n = 8m + 2 + 4n

Texas Go Math Grade 6 Lesson 11.3 Guided Practice Answer Key

Question 1.
Evaluate each of the expressions in the list for y 5. Then, draw lines to match the expressions in List A with their equivalent expressions in List B.

Answer:
Determine the equivalent expressions.


The equivalent expressions are:
a. 4 + 4y and 4 (y + 1) Distributive Property of Multiplication over Addition
b. 4 (y – 1) and 4y – 4 Distributive Property of Multiplication over Subtraction
c. 4y + 1 and 1 + 4y Commutative Property of Addition

Go Math Expressions Grade 6 Answer Key Algebra Lesson 11.3 Question 2.
Determine if the expressions are equivalent by comparing the models. (Explore Activity 2)

Answer:
The expressions are not equivalent based on the tiles shown. In x – 4, there are 1 tile for x and 4 tiles for -1 while in 2 (x – 2), there are 2 tiles for x and 4 tiles for -1.
x – 4 ≠ 2x – 4 by applying distributive property of multiplication over subtraction
The given expressions are not equivalent based on the models indicated.

For each expression, use a property to write an equivalent expression. Tell me which property you used.

Question 3.
ab = ______________________
Answer:
Given expression:
ab = ba
According to commutative property of multiplication
ab = ba according to commutative property of multiplication.

Question 4.
5(3x – 2) = ___
Answer:
The operation in the expression is Multiplication.
You can use the Distributive Property to write an equivalent expression: 5(3x – 2) = 15x – 10 (Final solution)
15x – 10

Use the properties of operations to determine if each pair of expressions is equivalent. (Example 2)

Question 5.
\(\frac{1}{2}\)(4 – 2x); 2 – 2x ___________
Answer:
Given expression:
\(\frac{1}{2}\)(4 – 2x)
Apply distributive property of multiplication to expand the parentheses:
= \(\frac{1}{2}\)(4) + \(\frac{1}{2}\)(-2x)
Expand:
= 2 – x ≠ 2 – 2x
\(\frac{1}{2}\)(4 – 2x); ≠ 2 – 2x

Go Math Answer Key Grade 6 Lesson 11.3 Question 6.
\(\frac{1}{2}\)(6x – 2); 3 – x _____
Answer:
Given expression:
\(\frac{1}{2}\)(6x – 2)
Apply the distributive property of multiplication to expand the parentheses:
= \(\frac{1}{2}\)(6x) + \(\frac{1}{2}\)(-2)
Expand:
= 3x – 1 ≠ 3 – x
\(\frac{1}{2}\)(6x – 2); 3 – x ≠ 3x – 1

Combine like terms. (Example 3)

Question 7.
32y + 12y _____________
Answer:
Combine like terms
32y + 12 32y and 12y are like terms.
32y + 12y = y(32 + 12) Distributive Property
= y(44) Add inside the parentheses.
= 44y Commutative Property of Multiplication
32y + 12y = 44y Final Solution
32y + 12y = 44y

Question 8.
12 + 3x – x – 12 = ___
Answer:
Combine like terms
12 + 3x – x – 12
12 + 3x – x – 12 = 2x Distributive Property
12 + 3x – x – 12 = 2x Final solution

Essential Question Check-In

Question 9.
Describe two ways to write equivalent algebraic expressions.
Answer:
The condensed form of an algebraic expression is by using parentheses, for example: 4(x + 3). Distributive property of multiplication can be applied here to expand it, therefore= 4x + 12. These 2 expressions are equal to each other.

Texas Go Math Grade 6 Lesson 11.3 Independent Practice Answer Key

For each expression, use a property to write an equivalent expression. Tell which property you used.

Question 10.
cd = ___
Answer:
Given expression:
cd = dc
According to commutative property of multiplication.
cd = dc according to commutative property of multiplication.

Equivalent Expressions Answer Key Go Math Lesson 11.3 Question 11.
x + 13 = ___
Answer:
The operation in the expression is Addition.
You can use the Commutative Property of Addition to write an equivalent expression: x + 13 = 13 + x (Final solution)
13 + x

Question 12.
4(2x – 3) = ___
Answer:
Given expression:
4(2x – 3) = 8x – 12
According to distributive property of multiplication
4(2x – 3) = 8x – 12 according to distributive property of multiplication.

Question 13.
2 + (a + b) = ___
Answer:
The operation in the expression is Addition.
You can use the Associative Property of Addition to write an equivalent expression: 2 + (a + b) = (2 + a) + b (Final solution)
(2 + a) + b

Question 14.
Draw algebra tile models to prove that 4 + 8x and 4(2x + 1) are equivalent.
Answer:
Algebra tile models for the given expression.

The expressions are equivalent as shown on the tiles. In 4 + 8, there are 4 tiles for +1 and 8 tiles for x which is similar to the tiles of 4(2x + 1).
The expressions are equivalent based on the diagram.

Combine like terms.

Question 15.
7x⁴ – 5x⁴ = ___________________
Answer:
7x⁴ – 5×4⁴ = x⁴(7 – 5) ← Distributive Property
= x⁴(2) ← Subtract inside the parentheses
= 2x⁴ ← Commutative Property of Multiplication
The final result is 2x⁴

Question 16.
32y + 5y = ________
Answer:
Given expression:
32y + 5y =
Perform the indicated operation between like terms to combine the like terms, therefore:
= 37y
32y + 5y = 37y

Question 17.
6b + 7b – 10 = ___
Answer:
Combine like terms
6b + 7b – 10 32y and 5y are like terms.
6b + 7b – 10 = 13b – 10 Distributive Property
13b – 10 Final Solution

Go Math 6th Grade Lesson 11.3 Answers Key Question 18.
2x + 3x + 4 = ___
Answer:
Combine like terms
2x + 3x + 4
2x + 3x + 4 = 5x + 4 Distributive Property
5x + 4 Final Solution

Question 19.
y + 4 + 3(y + 2) = ____
Answer:
Combine like terms
y + 4 + 3(y + 2)
y + 4 + 3(y + 2) = y + 4 + 3y + 6 Distributive Property
= y + 3y + 4 + 6 Commutative Property of Addition
= 4y + 10 Add
y + 4 + 3(y + 2) = 4y + 10
4y + 10 Final Solution

Question 20.
7a² – a² + 16 = ____
Answer:
Given expression:
7a² – a² + 16
Perform the indicated operation between like terms to combine the like terms, therefore:
= 6a² + 16
7a² – a² + 16 = 6a² + 16

Question 21.
3y² + 3(4y² – 2) = ____
Answer:
Given expression:
3y² + 3(4y² – 2)
Apply distributive property to expand the parentheses:
= 3y² + 12y² – 6
Perform the indicated operation between like terms to combine the like terms, therefore:
= 15y² – 6
3y² + 3(4y² – 2) = 15y² – 6

Question 22.
z² + z + 4z³ + 4z² = ___
Answer:
Combine like terms
z² + z + 4z³ + 4z²
z² + z + 4z³ + 4z² = 4z³ + 4z² + z² + z Comutative Property of Addition
= 4z³ + 5z² + z Distributive Property
z² + z + 4z³ + 4z² = 4z + 5z² + z
4z³ + 5z² + z Final Solution

Question 23.
0.5(x⁴ – 3) + 12 = ____
Answer:
Combine like terms
0.5(x⁴ – 3) + 12
0.5(x⁴ – 3) + 12 = 0.5x⁴ – 1.5 + 12 Distributive Property
= 0.5x⁴ + 10.5 Add
0.5(x⁴ – 3) + 12 = 0.5x⁴ + 10.5
0.5x⁴ + 10.5 Final Solution

Question 24.
\(\frac{1}{4}\)(16 + 4p) = ____
Answer:
Given expression:
\(\frac{1}{4}\)(16 + 4p)
Apply distributive property to expand the parentheses:
= \(\frac{1}{4}\)(16) + \(\frac{1}{4}\)(4p)
Expand:
= 4 + p
\(\frac{1}{4}\)(16 + 4p) = 4 + p

Question 25.
Justify Reasoning Is 3x + 12 – 2x equivalent to x + 12? Use two properties of operations to justify your answer.
Answer:
Determine if the expressions are equivalent.
= 3x – 2x + 12 commutative property of addition
= x + 12 combine like terms by subtracting the coefficient of the same variables
= 3 + 12 – 2x associative property of addition
= (3x – 2x) + 12 subtract the numbers inside the parenthesis
= x + 12

The given expressions are equivalent when the commutative or associative property of addition is applied.

Go Math Lesson 11.3 6th Grade Identifying Equivalent Expressions Question 26.
William earns $13 an hour working at a movie theater. Last week he worked three hours at the concession stand and three times as many hours at the ticket counter. Write and simplify an expression for the amount of money William earned last week.
Answer:
Last week he worked h hours at the concession stand and three times as many hours at the ticket counter This implies that he worked for 3 × h = 3h hours at the ticket counter The total number of hours worked is therefore: h + 3h = 4h.
His hourly rate at the theater is $13 so he earned $13 × 4h = $52h last week.

Question 27.
Multiple Representations Use the information in the table to write and simplify an expression to find the total weight of the medals won by the top medal-winning nations in the 2012 London Olympic Games. The three types of medals have different weights.

Answer:
Let g be the weight of a gold medal, s be the weight of a silver medal and b be the weight of a bronze medal.

The total weight of gold medals won by the 3 countries is g(46 + 38 + 29) = 113g. The total weight of silver medals won by the 3 countries is s(29 + 27 + 17) = 73s and the total weight of bronze medals won by the 3 countries is b(29 + 23 + 19) = 71b.
Their total sum is 113g + 73s + 71b.

Write an expression for the perimeters of each given figure. Simplify the expressions.

Question 28.
_____________

Answer:
Perimeter of a figure is the sum of its sides. Here 2 opposite sides are equal in length so the perimeter of the given figure is equal to 2(6) + 2(3x – 1):
Apply distributive property to expand the parentheses:
= 12 + 6x – 2
Perform the indicated operation between like terms to combine the like terms, therefore:
= 6x + 10
Perimeter of the figure shown is 6x + 10 millimeters.

Question 29.
___________

Answer:
Perimeter of a figure is the sum of its sides. Here 2 opposite sides are equal in Length and 4 other sides are also equal so the perimeter of the given figure is equal to 2(10.2) + 4(x + 4):

Apply distributive property to expand the parentheses:
= 20.4 + 4x + 16
Perform the indicated operation between like terms to combine the like terms, therefore:
= 4x + 36.4
Perimeter of the figure shown is 4x + 36.4 inches.

Texas Go Math Grade 6 Lesson 11.3 H.O.T. Focus On Higher Order Thinking Answer Key

Question 30.
Problem Solving Examine the algebra tile model.
a. Write two equivalent expressions for the model. ____

Answer:
The model shown represents 4 tiles of +1 and 6 tiles of -x. Therefore, the equivalent expressions for the model are 4 – 6x or 2 (2 – 3x).
The equivalent expressions are 4 – 6x and 2 (2 – 3)

b. What If? Suppose a third row of tiles identical to the ones above is added to the model. How does that change the two expressions?
Answer:
If a third row will be added, the expression will be 6 – 9x or 3 (2 – 3x). Since there will be 6 tiles of +1 and 9 tiles of -x.
The equivalent expressions are 6 – 9x and 3 (2 – 3x)

Question 31.
Communicate Mathematical Ideas Write an example of an expression that cannot be simplified, and explain how you know that it cannot be simplified.
Answer:
An example of such an expression is 8x + 100. It can be seen that the expression consists of 2 terms that are not like terms so the expression can no longer be simplified.

Question 32.
Problem-Solving Write an expression that is equivalent to 8(2y + 4) that can be simplified.
Answer:
Given expression:
8(2y + 4)
Apply the distributive property to expand the parentheses:
= 8(2y) + 8(4)
Simplify:
= 16y + 32
This expression can be broken down to form an expression with some like term that is equivalent to the given expression, therefore:
= 10y + 6y + 30 + 2
Note that there can be many answers to this question.

Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Lesson 11.2 Answer Key Evaluating Expressions.

Texas Go Math Grade 6 Lesson 11.2 Answer Key Evaluating Expressions

Essential Question
How can you use the order of operations to evaluate algebraic expressions?

Your Turn

Evaluate each expression for the given value of the variable.

Question 1.
4x; x = 8 ____
Answer:
Solution to this example is given below
4x; x = 8
4(8) Substitute 8 for x
32 Multiply
When x = 8, 4x = 32
32 Final solution

Question 2.
6.5 – n;n = 1.8 ___
Answer:
Given expression:
6.5 – n
Substitute n = 1.8 in the given expression:
= 6.5 – 1.8
Evaluate:
= 4.7
6.5 – n = 4.7

Practice and Homework Lesson 11.2 Answer Key 6th Grade Question 3.
\(\frac{m}{6}\);m = 18 ____
Answer:
Solution to this example is given below
\(\frac{m}{6}\); m = 18
\(\frac{18}{6}\) Substitute 18 for m
3 Divide
When m = 18, \(\frac{m}{6}\) = 3
3 Final solution

Your Turn

Evaluate each expression for n = 5.

Question 4.
3(n + 1) ____
Answer:
Solution to this example is given below
3(n + 1); n = 5
3(5 + 1) Substitute 5 for n
3(6) Add inside the parentheses
Multiply
When n = 5, 3(n + 1) = 18
18 Final solution

Question 5.
4(n – 4) + 14 ____
Answer:
Given expression:
4(n – 4) – 14
Substitute n = 5 in the given expression:
= 4(5 – 4) + 14
Simplify the expression in the parentheses:
= 4(1) + 14
Expand the parentheses:
= 4 + 14
Evaluate:
= 18
4(n – 4) + 14 = 18

Question 6.
6n + n² ____
Answer:
Solution to this example is given below
6n + n²; n = 5
6(5) + 5² Substitute 5 for n
6(5) + 25 Evaluate exponents
30 + 25 Multiply
55 Add
When n = 5, 6n + n² = 55
55 Final solution

Evaluate each expression for a = 3, b = 4, and c = -6.

Question 7.
ab – c ____
Answer:
Solution to this example is given below
ab – c; a = 3,b = 4, c = 6
3(4) – 6 Substitute 3 for a, 4 for b, and 6 for c
12 – 6 Multiply
6 Subtract
When a = 3, b = 4, and c = 6 ab – c = 6
6 Final solution

Question 8.
bc + 5a ___
Answer:
Solution to this example is given below
bc + 5a; a = 3, b = 4, c = 6
4(6) + 5(3) Substitute 3 for a, 4 for b, and 6 for c
24 + 15 Multiply
39 Subtract
When a = 3, b = 4, and c = 6 bc + 5a = 39
39 Final solution

Evaluating Expressions Lesson 11.2 Independent Practice Answer Key Question 9.
a² – (b + c) ____
Answer:
Evaluate the expression by substituting the given values.
= 3² – (4 + (-6)) substitute the values to the given expression
= 9 – (-2) evaluate the exponent and subtract the numbers inside the parenthesis
= 9 + 2 add the numbers
= 11 value of the expression
The value of the expression is 11.

Your Turn

Question 10.
The expression 6x² gives the surface area of a cube, and the expression x³ gives the volume of a cube, where x is the length of one side of the cube. Find the surface area and the volume of a cube with a side length of 2 m.
S = ____ m² ; V = ___ m³
Answer:
Given expression:
Surface Area = 6x²
Substitute x = 2 in the given expression:
Surface Area = 6(2)²
Evaluate:
Surface Area = 6(4) = 24
Surface area of the given cube is 24 square meters.
Given expression:
Volume = x³
Substitute x = 2 in the given expression:
Volume = 2³
Evaluate:
Volume = 8
Volume of the given cube is 8 cubic meters.

Question 11.
The expression 60m gives the number of seconds in m minutes. How many seconds are there in 7 minutes? ____ seconds
Answer:
Solution to this example is given below
60(m); m = 7
60(7) Substitute 7 for m
420 Multiply
When m = 7, 60(m) = 420
There are 420 seconds in 7 minutes.
420 Final solution

Texas Go Math Grade 6 Lesson 11.2 Guided Practice Answer Key

Evaluate each expression for the given value(s) of the variable(s). (Examples 1 and 2)

Question 1.
x – 7; x = 23 ________________
Answer:
Solution to this example is given below
x – 7; x = 23
23 – 7 Substitute 23 for x
16 Subtract
When x = 23, x – 7 = 16
16 Final solution

Question 2.
3a – b; a = 4, b = 6 _________
Answer:
Solution to this example is given below
3a – b; a = 4, b = 6
3(4) – 6 Substitute 4 for a, and 6 for b.
12 – 6 Multiply
6 Subtract
When a = 4, and b = 6, 3a – b = 6
6 Final solution

Question 3.
\(\frac{8}{t}\); t = 4 __________________
Answer:
Solution to this example is given below
\(\frac{8}{t}\); t = 4
\(\frac{8}{4}\) Substitute 4 for t
2 Divide
When t = 4, \(\frac{8}{t}\) = 2
2 Final solution

Question 4.
9 + m; m = 1.5 ___________
Answer:
Solution to this example is given below
9 + m; m = 1.5
9 + 1.5 Substitute 1.5 for m
10.5 Add
When m = 1.5, 9 + m = 10.5
10.5 Final solution

Lesson 11.2 Go Math 6th Grade Answer Key Question 5.
\(\frac{1}{2}\)\(\frac{1}{2}\)w + 2; w = \(\frac{1}{9}\) ___________________
Answer:
Solution to this example is given below
\(\frac{1}{2}\)w + 2; w = \(\frac{1}{9}\)
\(\frac{1}{2}\)(\(\frac{1}{2}\)) + 2 Substitute \(\frac{1}{9}\) for w
Substitute \(\frac{1}{18}\) + 2 Multiply
\(\frac{1+36}{18}\) Add
\(\frac{37}{18}\) Add
When w = \(\frac{1}{9}\), \(\frac{1}{2}\)w + 2 = \(\frac{37}{18}\)
\(\frac{37}{18}\) Final solution

Question 6.
5(6.2 + z); z = 3.8 _____________
Answer:
Given expression:
5(6.2 + z)
Substitute the value of z in the given expression:
= 5(6.2 – 3.8)
Simplify the parentheses:
= 5(10)
Expand the parentheses:
= 50
5(6.2 + z) = 50

Question 7.
The table shows the prices for games in Bella’s soccer league. Women’s Soccer league. Her parents and grandmother attended a soccer game. How much did they spend if they all went together in one car? Student tickets (Example 3)

a. Write an expression that represents the cost of one Parking carful of nonstudent soccer fans. Use x as the number of people who rode in the car and attended the game. __________ is an expression that represents the cost of one carful of nonstudent soccer fans.
Answer:
a. There are x Nonstudent people so the cost of their tickets will be 12 × x = 12x plus the parking fee, so the total cost is given by the expression: 12x + 5.

b. Since there are three attendees, evaluate the expression
12x + 5 for x = 3.
12(___) + 5 = _____ + 5 = _____
The family ___ spent to attend the game.
Answer:
Substitute x = 3 in the expression, therefore: 12(3) + 5 = 36 + 5 = $41.
The family spent $41 to attend the game.

Question 8.
Stan wants to add trim all around the edge of a rectangular tablecloth that measures 5 feet long by 7 feet wide. The perimeter of the rectangular tablecloth is twice the length added to twice the width. How much trim does Stan need to buy? (Example 3)

a. Write an expression that represents the perimeter of the rectangular tablecloth. Let l represent the length of the tablecloth and w represent its width. The expression would be ____.
Answer:
Here let length of the tablecloth be l and width be w so the expression for perimeter of the tablecloth becomes: 2l + 2w.

b. Evaluate the expression P = 2w + 2l for l = 5 and w = 7.
2(___) + 2(_____) = 14 + _____ = _____
Stan bought __________ of trim to sew onto the tablecloth.
Answer:
Here l = 5 and width be w = 7 so the value of perimeter of the tablecloth becomes: 2(5) + 2(7) = 10 + 14 = 24. The perimeter of the tablecloth is 24 feet. Stan bought 24 feet of trim to sew onto the tablecloth.

Question 9.
Essential Question Follow up How do you know the correct order in which to evaluate algebraic expressions?
Answer:
In any algebraic expression, the simplification of the parentheses has the highest priority followed by division, then multiplication, then addition and finally subtraction. These priorities are used to simplify algebraic expressions.

Texas Go Math Grade 6 Lesson 11.2 Independent Practice Answer Key

Question 10.
The table shows ticket prices at the Movie 16 theater. Let a represent the number of adult tickets, c the number of children’s tickets, and s the number of senior citizen tickets.

a. Write an expression for the total cost of tickets.
Answer:
The total cost of the three types of tickets for a adults, c children and s seniors is 8.75a + 6.5c + 6.5s.

b. The Andrews family bought 2 adult tickets, 3 children’s tickets, and 1 senior ticket. Evaluate your expression in part a to find the total cost of the tickets.
Answer:
Expression:
8.75a + 6.5c + 6.5s
Substitute a = 2, b = 3 and c = 1:
= 8.75(2) + 6.5(3) + 6.5(1)
Expand each parentheses:
= 17.5 + 19.5 + 6.5
Evaluate:
= 43.5
Total cost of tickets is $43.50

c. The Spencer family bought 4 adult tickets and 2 children’s tickets. Did they spend the same as the Andrews family? Explain.
Answer:
Expression:
Substitute a = 4, b = 2 and c = 0:
= 8.75(4) + 6.5(2) + 6.5(0)
Expand each parentheses:
= 35 + 13
Evaluate:
= 48
Total cost of tickets is $48. It can be seen that the 2 costs are not equal.

Question 11.
The area of a triangular sail is given by the expression \(\frac{1}{2}\)bh, where b is the length of the base and h is the height. What is the area of a triangular sail in a model sailboat when b = 12 inches and h = 7 inches?
A = ____ in.²
Answer:
Given expression:
\(\frac{1}{2}\)bh
Substitute b = 12 and h = 7:
= \(\frac{1}{2}\)(12)(7)
Simplify:
= 6(7)
Evaluate:
= 42
Area of the triangular sail in a model sailboat is 42 square inches.

Question 12.
Ramon wants to balance his checking account. He has $2,340 in the account. He writes a check for $140. He deposits a check for $268. How much does Ramon have left in his checking account? _________________
Answer:
The expression of his current balance is:
2340 – 140 + 268
The check is a withdrawal from the account so shown with a negative sign.
Evaluate:
= 2468
He will have $2468 in his account.

Go Math Lesson 11.2 Evaluate Expressions Answer Key Question 13.
Look for a Pattern Evaluate the expression 6x – x² for x = 0, 1, 2, 3,4, 5, and 6. Use your results to fill in the table and describe any pattern that you see.

Answer:
x 6x – x²
0 6(0) – 0² = 0
1 6(1) – 1² = 5
2 6(2) – 2² = 8
3 6(3) – 3² = 9
4 6(4) – 4² = 8
5 6(5) – 5² = 5
6 6(6) – 6² = o
The numbers increase from 0 to a maximum of 9 decrease in the same pattern and reach 0 again.

Question 14.
The kinetic energy (in joules) of a moving object can be calculated from the expression \(\frac{1}{2} m v^{2}\), where m is the mass of the object in kilograms and y is its speed in meters per second. Find the kinetic energy of a 0.145-kg baseball that is thrown at a speed of 40 meters per second.
E = __ joules
Answer:
Given expression:
\(\frac{1}{2}\)mv²
Substitute m = 0.145 and v = 40:
= \(\frac{1}{2}\)(0.145)(40)²
Simplify:
= 0.0725(1600)
Evaluate:
= 116
The kinetic energy of the baseball is 116 Joules.

Question 15.
The area of a square is given by x², where x is the length of one side. Mary’s original garden was in the shape of a square. She has decided to double the area of her garden. Write an expression that represents the area of Mary’s new garden. Evaluate the expression if the side length of Mary’s original garden was 8 feet.
Answer:
Given expression:
Area = s²
New length of the square garden is 2s²:
Area = 2s² = 2x²
Substitute x = 8:
Area = 2(8)²
Simplify:
Area = 2(64)
Evaluate:
Area = 128
Area of the new garden is 128 square feet

Question 16.
The volume of a pyramid with a square base is given by the expression \(\frac{1}{3} s^{2} h\), where s is the length of a side of the base and h is the height. Find the volume of a pyramid with a square base of side length 24 feet and a height of 30 feet.

Answer:
Solution to this example is given below
\(\frac{1}{3}\)s²h, s = 24, and h = 30
\(\frac{1}{3}\)(24)²(30) Substitute 24 for s, and 30 for h
(576)(30) Evaluate exponents
(576)(10) Divide
5760 Multiply
When s = 24, and h = 30, \(\frac{1}{3}\)s²h = 5760
Volume of the pyramid is 5760 cubic feet.
5760 Final solution

Texas Go Math Grade 6 Lesson 11.2 H.O.T. Focus On Higher Order Thinking Answer Key

Question 17.
Draw Conclusions Consider the expressions 3x(x – 2) + 2 and 2x² + 3x – 12.

a. Evaluate each expression for x = 2 and for x = 7. Based on your results, do you know whether the two expressions are equivalent? Explain.
Answer:
Check the value of the expression if they are equivalent when x = 2 and x = 7.
3(2)(2 – 2) + 2 = 2(2)² + 3(2) – 12 substitute for the value of x
6(0) + 2 = 2(4) + 6 – 12 perform the indicated operation inside the parentheses
0 + 2 = 8 + 6 – 12 simplify
2 = 2 value of the expression
The expressions are equivalent when the value of x is 2.
3(7)(7 – 2) +2 = 2(1)² + 3(7) – 12 substitute for the value of x
21(5) + 2= 2(49) + 21 – 12 perform the indicated operation inside the parentheses
105 + 2 = 98 + 21 – 12 simplify
107 = 107 value of the expression
The expressions are equivalent when the value of x is 7.

b. Evaluate each expression for x = 1. Based on your results, do you know whether the two expressions are equivalent? Explain.
Answer:
Check the value of the expression if they are equivalent when x = 1.
3(1)(1 – 2) + 2 = 2(1)² + 3(1) – 12 substitute for the value of x
3(-1) + 2 = 2 (1) + 3 – 12 perform the indicated operation inside the parentheses
-3 + 2 = 2 + 3 – 12 simplify
-1 ≠ -7 value of the expression
The expressions are not equivalent when the value of x is 1

Question 18.
Critique Reasoning Marjorie evaluated the expression 3x + 2 for x = 5 as shown:
3x + 2 = 35 + 2 = 37
What was Marjorie’s mistake? What is the correct value of 3x + 2 for x = 5?
Answer:
Given expression:
3x + 2
Substitute x = 5:
= 3(5) + 2
Simplify. Marjorie’s mistake was in the incorrect multiplication of 3 and 5:
= 15 + 2
Evaluate:
= 17
3x + 2 = 17

Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Module 10 Quiz Answer Key.

Texas Go Math Grade 6 Module 10 Quiz Answer Key

10.1 Exponents

Find the value of each power.

Question 1.
7³
Answer:
Solution to this example is given below 7³

Identify the base and the exponent.
The base is 7 and the exponent is 3.

Final Solution:
Evaluate: 7³ = 7 × 7 × 7 = 343

Question 2.
9²
Answer:
Solution to this example is given below 9²

Identify the base and the exponent.
The base is 9 and the exponent is 2.

Final Solution:
Evaluate: 9² = 9 × 9 = 81

Geometry Module 10 Test Answer Key Go Math Question 3.
\(\left(\frac{7}{9}\right)^{2}\)
Answer:
Solution to this example is given below \(\left(\frac{7}{9}\right)^{2}\)

Identify the base and the exponent.
The base is \(\frac{7}{9}\) and the exponent is 2.

Final Solution:
Evaluate: \(\left(\frac{7}{9}\right)^{2}\) = \(\frac{7}{9} \times \frac{7}{9}\) = \(\frac{79}{81}\)

Question 4.
\(\left(\frac{1}{2}\right)^{6}\)
Answer:
Solution to this example is given below \(\left(\frac{1}{2}\right)^{6}\)

Identify the base and the exponent.
The base is \(\frac{1}{2}\) and the exponent is 6.

Final Solution:
Evaluate: \(\left(\frac{1}{2}\right)^{6}\) = \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\) = \(\frac{1}{64}\)

Question 5.
\(\left(\frac{2}{3}\right)^{3}\)
Answer:
Solution to this example is given below \(\left(\frac{2}{3}\right)^{3}\)

Identify the base and the exponent.
The base is \(\frac{2}{3}\) and the exponent is 3.

Final Solution:
Evaluate: \(\left(\frac{2}{3}\right)^{3}\) = \(\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}\) = \(\frac{8}{27}\)

Question 6.
(- 3)⁵
Answer:
Simplify the given expression by identifying the base and exponent
The base is – 3 and the exponent is 5.
= (- 3) × (- 3) × (- 3) × (- 3) × (- 3) multiply the base by itself five times
= – 243
The simplified expression is – 243.

Question 7.
(- 2)⁴
Answer:
Simplify the given expression by identifying the base and exponent.
The base is – 2 and the exponent is 4.
= (- 2) × (- 2) × (- 2) × (- 2) multiply the base by itself four times
= 16
The simplified expression is 16.

Texas Go Math Grade 6 Module 10 Test Answers Question 8.
1.4²
Answer:
The solution to this example is given below 1.4²

Identify the base and the exponent.
The base is 1.4 and the exponent is 2.

Final Solution:
Evaluate: 1.4² = 1.4 × 1.4 = 1.96

10.2 Prime Factorization

Find the factors of each number.

Question 9.
96 _________________
Answer:
List the factors of 96

• 96 = 1 × 96
• 96 = 2 × 48
• 96 = 3 × 32
• 96 = 4 × 24
• 96 = 6 × 16
• 96 = 8 × 12

The factors of 96 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48 and 96.

Question 10.
120 ________________
Answer:
List the factors of 120

• 120 = 1 × 120
• 120 = 2 × 60
• 120 = 3 × 40
• 120 = 5 × 24
• 120 = 6 × 20
• 120 = 8 × 15
• 120 = 10 × 12

The factors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 60 and 120.

Find the prime factorization of each number.

Question 11.
58 ________________
Answer:
The prime factorization of a number is the number written as the product of its prime factors. For example, the prime factors of 58 are 2, and 29.

The prime factorization of 58 is 2 × 29

Question 12.
212 _______________
Answer:
The prime factorization of a number is the number written as the product of its prime factors. For example, the prime factors of 212 are 2, 2 and 53.

The prime factorization of 212 is 2 × 2 × 53 or 2² × 53

Question 13.
2,800 ______________
Answer:
The prime factorization of a number is the number written as the product of its prime factors. For example, the prime factors of 2800 are 2, 2, 2, 2, 5, 5 and 7.

The prime factorization of 2800 is 2 × 2 × 2 × 2 × 5 × 5 × 7 or 2⁴ × 5² × 7

Question 14.
900 ______________
Answer:
The prime factorization of a number is the number written as the product of its prime factors. For example, the prime factors of 900 are 2, 2, 3, 3, 5 and 5.

The prime factorization of 2800 is 2 × 2 × 3 × 3 × 5 × 5 or 2² × 3² × 5²

10.3 Order of Operations

Simplify each expression using the order of operations.

Question 15.
(21 – 3) ÷ 3²
Answer:
Solution to this example is given below
(21 – 3) ÷ 3²

(21 – 3) ÷ 3² = (21 – 3) ÷ 9 Evaluate 3²
= 18 ÷ 9 Perform operations inside parentheses.
= 2 Divide.
= 2

Question 16.
7² × (6 ÷ 3)
Answer:
Solution to this example is given below
7² × (6 ÷ 3)

7² × (6 ÷ 3)> = 49 × (6 ÷ 3) Evaluate 7²
= 49 × 2 Perform operations inside parentheses.
= 98 Multiply.
= 98

Question 17.
17 + 15 ÷ 3 – 2⁴
Answer:
Solution to this example is given below
17 + 15 ÷ 3 – 2⁴

17 + 15 ÷ 3 – 2⁴ = 17 + 15 ÷ 3 – 16 Evaluate 2⁴.
= 17 + 5 – 16 Divide.
= 22 – 16 Add.
= 6 Subtract.
= 6

Question 18.
(8 + 56) ÷ 4 – 3²
Answer:
The solution to this example is given below
(8 + 56) ÷ 4 – 3²

(8 + 56) ÷ 4 – 3² = (8 + 56) ÷ 4 – 9 Evaluate 3².
= 64 ÷ 4 – 9 Perform operations inside parentheses.
= 16 – 9 Divide.
= 7 Subtract.
= 7

Module 10 Answer Key Texas Go Math Grade 6 Question 19.
The nature park has a pride of 7 adult lions and 4 cubs. The adults eat 6 pounds of meat each day and the cubs eat 3 pounds. Simplify 7 × 6 + 4 × 3 to find the amount of meat consumed each day by the lions.
Answer:
Solution to this example ¡s given below 7 × 6 + 4 × 3

7 × 6 + 4 × 3 = 7 × 6 + 12 Multiply.
= 42 + 12 Multiply.
= 54 Add.
= 1
Each day by the lions is equal to 54 pounds.

Essential Question

Question 20.
How do you use numerical expressions to solve real-world problems?
Answer:
Most real world problems can be converted to algebraic expressions and equations to solve them. For example if a person earns a fixed monthly salary of $x and has expenses a, b and c, then the expenses can be added to evaluate the total expenses of the month and this sum can be subtracted front the amount of monthly salary to evaluate monthly savings and so on.

Texas Go Math Grade 6 Module 10 Mixed Review Texas Test Prep Answer Key

Selected Response

Question 1.
Which expression has a value that is less than the base of that expression?
(A) 2³
(B) \(\left(\frac{5}{6}\right)^{2}\)
(C) 3²
(D) 4⁴
Answer:
(B) \(\left(\frac{5}{6}\right)^{2}\)

Explaination:
Identify the expression which has a value of less than the basa

(A) The base is 2 and the exponent is 3.
= 2 × 2 × 2 multiply the base three times
= 8
8 is not less than 2

(B) The base is 2 and the exponent is 3.
= \(\frac{5}{6} \times \frac{5}{6}\) multiply the base twice
= \(\frac{25}{36}\)
\(\frac{25}{36}\) is less than \(\frac{5}{6}\)

(C) The base is 3 and the exponent is 2.
= 3 × 3 multiply the base twice
= 9
9 is not less than 3

(D) The base is 2 and the exponent is 3.
= 4 × 4 × 4 × 4 multiply the base tour times
= 256
256 is not less than 4

The expression with a value less than the base is B. \(\left(\frac{5}{6}\right)^{2}\) because the base is a fraction.

Question 2.
After the game the coach bought 9 chicken meals for $5 each and 15 burger meals for $6 each. What percent of the total amount the coach spent was used for the chicken meals?
(A) 33\(\frac{1}{3}\)%
(B) 45%
(C) 66\(\frac{2}{3}\)%
(D) 90%
Answer:
(A) 33\(\frac{1}{3}\)%

Explaination:
Solution to this example is given below
9 × 5 + 15 × 6
9 × 5 + 15 × 6 = 9 × 5 + 90 Multiply.
= 45 + 90 Multiply.
= 135 Add.

We now calculate the percentage
Percent = \(\frac{45}{135}\) × 100%
= \(\frac{4500}{135}\)
= 33\(\frac{1}{3}\)%

Question 3.
Which operation should you perform first when you simplify 75 – (8 + 45 ÷ 3) × 7?
(A) addition
(B) division
(C) multiplication
(D) subtraction
Answer:
(B) division

Explaination:
Solution to this example is given below
175 – (8 + 45 ÷ 3) × 7
175 – (8 + 45 ÷ 3) × 7 = 175 – (8 + 15) × 7 [Divide.]
= 175 – 23 × 7 [Perform operations inside parentheses.]
= 175 – 161 [Multiply.]
= 14 [Subtract.]

B is the correct option
Division is first operation

Question 4.
At Tanika’s school, three people are chosen in the first round. Each of those people chooses 3 people in the second round, and so on. How many people are chosen in the sixth round?
(A) 18
(B) 216
(C) 243
(D) 729
Answer:
(D) 729

Explaination:
First round = 3¹ = 3
Second round = 3² = 9
Third round = 3³ = 27
Fourth round = 3⁴ = 81
Fifth round = 3⁵ = 243
Sixth round = 3⁶ = 729
There are 729 people chosen in the sixth round.

Question 5.
Which expression shows the prime factorization of 100?
(A) 2² × 5²
(B) 10 × 10
(C) 10¹⁰
(D) 2 × 5 × 10
Answer:
(A) 2² × 5²

Explaination:
The prime factorization of a number is the number written as the product of its prime factors. For example, the prime factors of 100 are 2, 2, 5 and 5

The prime factorization of 100 is 2 × 2 × 5 × 5 or 2² × 5²

Question 6.
Which number has only two factors?
(A) 21
(B) 23
(C) 25
(D) 27
Answer:
(B) 23

Explaination:
Let’s check which number has two factors

(A) List the factors of 21
21 = 1 × 21
21 = 3 × 7
Option A is not the correct answer

(B) List the factor of 23
23 = 1 × 23
Option B is the correct answer

(C) List the factors of 25
25 = 1 × 25
25 = 5 × 5
Option C is not the correct answer

(D) List the factors of 27
27 = 1 × 27
27 = 3 × 9
Option D is not the correct answer

Grade 6 Go Math Module 10 Answer Key Pdf Question 7.
Which expression is equivalent to 3.6 × 3.6 × 3.6 × 3.6?
(A) 3.6 × 4
(B) 36³
(C) 3⁴ × 6⁴
(D) 3.6⁴
Answer:
(D) 3.6⁴

Explaination:
Solution to this example is given below
3.6 × 3.6 × 3.6 × 3.6
Find the base, or the numbers being multiplied. The base is 3.6

Find the exponent by counting the number of the 3.6s being multiplied.
The exponent is 4

Question 8.
Which expression gives the prime factorization of 80?
(A) 2⁴ × 10
(B) 2 × 5 × 8
(C) 2³ × 5
(D) 2⁴ × 5
Answer:
(D) 2⁴ × 5

Explaination:
The prime factorization of a number is the number written as the product of its prime factors. For example, the prime factors of 80 are 2, 2, 2, 2 and 5.

The prime factorization of 80 is 2 × 2 × 2 × 2 × 5 or 2⁴ × 5

Gridded Response

Question 9.
Alison raised 10 to the 5th power. Then she divided this value by 100. What was the quotient?

Answer:
Evaluate the expression.
= 10⁵ ÷ 100 [evaluate the exponent]
= 100,000 ÷ 100 [divide the numbers]
= 1,000
The answer on the grid must be 1,000.

Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Lesson 10.3 Answer Key Order of Operations.

Texas Go Math Grade 6 Lesson 10.3 Answer Key Order of Operations

Reflect

Question 1.
In C, why does it makes sense to write the values as powers? What is the pattern for the number of e-mails in each wave for Amy?
Answer:
It is easier to write the values as powers to identify the pattern that can be observed in the given situation.
Pattern for the diagram of Amy:
1st wave = 3¹ = 3
2nd wave = 3² = 9
The pattern is 3¹ and 3².

Your Turn

Simplify each expression using the order of operations.

Question 2.
(3 – 1)⁴ + 3 _______________
Answer:
Evaluate the expression.
= 2⁴ + 3 subtract the numbers inside the parenthesis and evaluate the exponent
= 16 + 3 add the numbers
= 19
The simplified expression is 19.

Go Math 6th Grade Lesson 10.3 Answer Key Question 3.
24 ÷ (3 × 2²) _______________
Answer:
Evaluate the expression.
= 24 ÷ (3 × 4) evaluate the exponent inside the parenthesis and multiply
= 24 ÷ 12 divide the numbers
= 2
The simplified expression is 2.

Simplify each expression using the order of operations.

Question 4.
– 7 × (- 4) ÷ 14 – 2²
Answer:
Evaluate the expression.
= – 7 × (- 4) ÷ 14 – 4 evaluate the exponent
= 28 ÷ 14 – 4 multiply – 7 and – 4 then divide the answer by 14
= 2 – 4 subtract the numbers
= – 2
The simplified expression is – 2.

Question 5.
– 5(- 3 + 1)³ – 3
Answer:
Evaluate the expression.
= – 5 (- 2)^{3 – 3} subtract the numbers inside the parenthesis and evaluate the exponent
= – 5(- 8) – 3 multiply – 5 and – 8
= 40 – 3 subtract the numbers
= 37

Texas Go Math Grade 6 Lesson 10.3 Guided Practice Answer Key

Question 1.
In a video game, a guppy that escapes a net turns into three goldfish. Each goldfish can turn into two betta fish. Each betta fish can turn into two angelfish. Complete the diagram and write the number of fish at each stage. Write and evaluate an expression for the number of angelfish that can be formed from one guppy.

Answer:
1 guppy fish
3 goldfish
3 × 2 betta fish
3 × 2² = 12 angel fish

Complete to simplify each expression.

Question 2.
4 + (10 – 7)² ÷ 3 = 4 + ( ____________ )² ÷ 3
= 4 + ____________ ÷ 3
= 4 + ____________
= ____________
Answer:
Evaluate the expression.
= 4 + 3² ÷ 3 subtract the numbers inside the parenthesis and evaluate the exponent
= 4 + 9 ÷ 3 divide 9 by 3
= 4 + 3 add the number
= 7
The simplified expression is 7.

Go Math Lesson 10.3 6th Grade Answer Key Question 3.
36 ÷ 2² – 4 × 2 = 36 ÷ ____________ – 4 × 2
= ____________ – 4 × 2
= ____________ – 8
= ____________
Answer:
Evaluate the expression.
= 36 ÷ 4 – 4 × 2 evaluate the exponent
= 9 – 4 × 2 divide 36 by 4 then multiply 4 by 2
= 9 – 8 subtract the numbers
= 1
The simplified expression is 1.

Question 4.
2 + (- 24 ÷ 2³) – 9 = – 2 + (- 24 ÷ _________ ) – 9
= 2 + ____________ – 9
=____________ – 9
= ____________
Answer:
Evaluate the expression
= 2 + (- 24 ÷ 8) – 9 evaluate the exponent and divide the numbers inside the parenthesis
= 2 + (- 3) – 9 subtract the numbers
= – 1 – 9 add the numbers
= – 10
The simplified expression is – 10

Question 5.
– 4² × (- 3 × 2 + 8) = – 4² × (_______ + 8)
= – 4² × _________
= _________ × _________
= _________
Answer:
Evaluate the expression
= – 4² × (-6 + 8) multiply – 3 by 2 inside the parenthesis and subtract
= – 4² × 2 evaluate the exponent
= 16 × 2 multiply the numbers
= 32
The simplified expression is 32.

Essential Question Check – In

Question 6.
How do you use the order of operations to simplify expressions with exponents?
Answer:
The order of operations to simplify expressions with exponents starts with simplifying the bracket (if any), solving the numerical value of power first, followed by division, multiplication, addition, and then subtraction.

Simplify each expression using the order of operations.

Go Math Grade 6 Lesson 10.3 Answer Key Question 7.
5 × 2 + 3² __________
Answer:
Solution to this example is given below = 5 × 2 + 3²

5 × 2 + 3² = 5 × 2 + 9 Evaluate 3².
= 10 + 9 Multiply.
= 19 Add.
= 19

Question 8.
15 – 7 × 2 + 2³ _____________
Answer:
Solution to this example is given below
15 – 7 × 2 + 2³
15 – 7 × 2 + 2³ = 15 – 7 × 2 + 8 Evaluate 2.
= 15 – 14 – 8 Multiply.
= 1 + 8 Subtract.
= 9 Add.
= 9

Question 9.
(11 – 8)² – 2 × 6 _____________
Answer:
Evaluate the expression.
= 3² – 2 × 6 subtract the numbers inside the parenthesis and evaluate the exponent
= 9 – 2 × 6 multiply – 2 by 6
= 9 – 12 subtract the numbers
= – 3
The simplified expression is – 3.

Question 10.
6 + 3(13 – 2) – 5² _____________
Answer:
Solution to this example is given below
6 + 3(13 – 2) – 5²

6 + 3(13 – 2) – 5² = 6 + 3(13 – 2) – 25 Evaluate 5².
= 6 + 3 × 11 – 25 Perform operations inside parentheses.
= 6 + 33 – 25 Multiply.
= 6 + 8 Subtract.
= 14 Add.
= 14

Question 11.
12 + \(\frac{9^{2}}{3}\) _____________
Answer:
Solution to this example is given below
12 + \(\frac{9^{2}}{3}\)

12 + \(\frac{9^{2}}{3}\) = 12 + \(\frac{81}{3}\) Evaluate 9².
= 12 + 27 Divide.
= 39 Add.
= 39

Lesson 10.3 Answer Key Go Math Grade 6 Question 12.
\(\frac{8+6^{2}}{11}\) + 7 × 2 _____________
Answer:
The solution to this example is given below
\(\frac{8+6^{2}}{11}\) + 7 × 2

\(\frac{8+6^{2}}{11}\) + 7 × 2 = \(\frac{8+6^{2}}{11}\) + 7 × 2 Evaluate 6².
= \(\frac{44}{11}\) + 7 × 2 Add.
= 4 + 7 × 2 Divide.
= 4 + 14 Multiply.
= 18 Add.

Question 13.
Explain the Error Jay simplified the expression – 3 × (3 + 12 ÷ 3) 4. For his first step, he added 3 + 12 to get 15. What was Jay’s error? Find the correct answer.
Answer:
Evaluate the expression
= – 3 × (3 + 4) – 4 divide 12 by 3 then add to 3
= – 3 × 7 – 4 multiply – 3 by 7
= – 21 – 4 add the numbers
= – 25
Jay made an error on his first step. In the order of operations, inside a parenthesis, division must be done first than addition, Therefore, the simplified expression is – 25.

Question 14.
Multistep A clothing store has the sign shown in the shop window. Pani sees the sign and wants to buy 3 shirts and 2 pairs of jeans. The S cost of each shirt before the discount is $12, and the cost of each pair of jeans is $19 before the discount.

a. Write and simplify an expression to find the amount Pani pays if a $3 discount is applied to her total.
Answer:
Evaluate the total cost when only 1 $3 off coupon is applied. Therefore, Cost = 3(12) + 2(19) – 3 = $71.

b. Pani says she should get a $3 discount on the price of each shirt and a $3 discount on the price of each pair of jeans. Write and simplify an expression to find the amount she would pay if this is true.
Answer:
Evaluate the total cost when only 1 $3 off coupon is applied. on every piece of clothing. Therefore, Cost = 3(12 – 3) + 2(19 – 3) = $59.

c. Analyze Relationships Why are the amounts Pani pays in a and b different?
Answer:
In a, the coupon is applied only once, while in b the coupon is applied for each piece of clothing, so here 5 times. This is why the result of b is $12 less than that of a.

d. If you were the shop owner, how would you change the sign? Explain.
Answer:
It would be better to write that $3 off the total purchase.

Go Math Grade 6 Lesson 10.3 Practice Answer Key Question 15.
Ellen is playing a video game in which she captures butterflies. There are 3 butterflies on screen, but the number of butterflies doubles every minute. After 4 minutes, she was able to capture 7 of the butterflies.

a. Look for a Pattern Write an expression for the number of butterflies after 4 minutes. Use a power of 2 in your answer.
Answer:
In the 1st minute there are 3 butterfLies on the screen and they double every minute, so here the repeating number is 2, so the expression for butterflies on the screen after n minutes is given by 3 × 2ⁿ.

b. Write an expression for the number of butterflies remaining after Ellen captured the 7 butterflies. Simplify the expression.
Answer:
Use the expression obtained to soLve for n = 4 and subtract 7 from it, therefore: 3 × 2⁴ – 7 = 41. There were 41 butterflies on the screen after 4 minutes and after Ellen had caught 7.

Question 16.
Show how to write, evaluate and simplify an expression to represent and solve this problem: Jeff and his friend each text four classmates about a concert. Each classmate then texts four students from another school about the concert. If no one receives the message more than once, how many students from the other school receive a text about the concert?
Answer:
Jeff and his friend, so 2 people are texting 4 people each, so 2 × 4 = 8 people are being notified about the concert. Then each 8 of them texts to 4 more, so the total number of people of the other school to get notified about the concert are 8 × 4 = 32.

H.O.T. Focus on Higher Order Thinking

Question 17.
Geometry The figure shown is a rectangle. The green shape in the figure is a square. The blue and white shapes are rectangles, and the area of the blue rectangle is 24 square inches.

a. Write an expression for the area of the entire figure that includes an exponent. Then find the area.
Answer:
Area of the blue rectangle is 24 square inches, area of the white rectangle is 2 × 6 = 12 square inches, and area of the square is 6² = 36 square inches. Therefore the total area of the figure is 24 + 12 + 36 = 72 square inches.

b. Find the dimensions of the entire figure.
Answer:
The side length of the square is 6 inches and then there is an additional 2 units below it so it can be said that the left and right side length of the given figure is 6 + 2 = 8 inches. The area is 72 square inches so this can be divided by the obtained side length to evaluate the width of the rectangle, therefore: \(\frac{72}{8}\) = 9 inches. The figure is a rectangle with a length of 8 inches and a width of 9 inches.

Lesson 10.3 Answer Key 6th Grade Go Math Question 18.
Analyze Relationships Roberto’s teacher writes the following statement on the board: The cube of a number plus one more than the square of the number is equal to the opposite of the number. Show that the number is – 1.
Answer:
Evaluate the expression.
(- 1)³ + 1 + (- 1)² = 1 translating the problem
– 1 + 1 + 1 = 1 evaluate the exponents
– 1 + 2 = 1 subtract the numbers
1 = 1
The expression is (- 1)³ + 1 + (- 1)² = 1.

Question 19.
Persevere in Problem Solving Use parentheses to make this statement true: 8 × 4 – 2 × 3 + 8 ÷ 2 = 25
Answer:
Solution to this example is given below
8 × 4 – (2 × 3 + 8) ÷ 2 (we use parentheses like this)

8 × 4 – (2 × 3 + 8) ÷ 2 = 8 × 4 – (6 + 8) ÷ 2 Perform operations inside parentheses.
= 8 × 4 – 14 ÷ 2 Perform operations inside parentheses
= 8 × 4 – 7 Divide.
= 32 – 7 Multiply.
= 25 Subtract.
= 25

Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Lesson 10.2 Answer Key Prime Factorization.

Texas Go Math Grade 6 Lesson 10.2 Answer Key Prime Factorization

Your Turn

List all the factors of each number.

Question 1.
21 ______________
Answer:
List the factors of 21
21 = 1 × 21
21 = 3 × 7
The factors of 21 are 1, 3, 7, 21.

Question 2.
37 ______________
Answer:
List the factors of 37
37 = 1 × 37
The factors of 37 are 1, 37,

Lesson 1.2 Prime Factorization Answers 6th Grade Question 3.
42 ______________
Answer:
List the factors of 42

• 42 = 1 × 42
• 42 = 2 × 21
• 42 = 3 × 14
• 42 = 6 × 7

The factors of 42 are 1, 2, 3, 6, 7, 14, 21, 42.

Question 4.
30 ______________
Answer:
List the factors of 30

• 30 = 1 × 30
• 30 = 2 × 15
• 30 = 3 × 10
• 30 = 5 × 6

The factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30

Reflect

Question 5.
What If? What will the factor tree for 240 look like if you start the tree with a different factor pair? Check your prediction by creating another factor tree for 240 that starts with a different factor pair.

Answer:
Prime factorization of 240 using factor tree.

Prime factorization of 240 using factor tree.

The prime factors of 240 are 5 ∙ 3 ∙ 2 ∙ 2 ∙ 2 ∙ 2 or 5 ∙ 3 ∙ 2⁴

Reflect

Grade 6 Go Math Lesson 10.2 Practice Answer Key Question 6.
Complete a factor tree and a ladder diagram to find the prime factorization of 54.

Answer:
Prime factorization of 54 using factor tree.

Prime factorization of 54 using a ladder diagram.

The prime factors of 54 are 2 ∙ 3 ∙ 3 ∙ 3 or 2 ∙ 3³

Question 7.
Communicate Mathematical Ideas If one person uses a ladder diagram and another uses a factor tree to write a prime factorization, will they get the same result? Explain.
Answer:
If one person uses a ladder diagram and another uses a factor tree to write a prime factorization, they will get the same result because the prime factors of a number are identical an independent of the method used to evaluate them.

Texas Go Math Grade 6 Lesson 10.2 Guided Practice Answer Key

Use a diagram to list the factor pairs of each number.

Question 1.
18
Answer:
Given number: 18

Write the given number as a product of its factors, therefore:

• 18 = 2 × 9
• 18 = 3 × 6

The factors of 18 are 1, 2, 3, 6, 9 and 18.

Question 2.
52
Answer:
Given number: 52

Write the given number as a product of its factors, therefore:
52 = 2 × 26
52 = 4 × 13
The factors are of 52 are 1, 2, 4, 13, 26 and 52.

Question 3.
Karl needs to build a stage that has an area of 72 square feet. The length of the stage should be longer than the width. What are the possible whole number measurements for the length and width of the stage?
Complete the table with possible measurements of the stage.

Answer:
Possible measurements of the stage.

List the possible measurements of the stage based on the factors of 72.

Use a factor tree to find the prime factorization of each number.

Question 4.
402

Answer:
Prime factorization of 402 using a factor tree.
First line of factor tree 6 × 67 = 402
Factor out 6 and you will have 3 × 2.
The complete prime factors of 402 is 3 ∙ 2 ∙ 67

Another way to show the prime factorization of 402 using a factor tree.
First line of factor tree 2 × 201 = 402
Factor out 201 and you will have 3 × 67.
The complete prime factors of 402 is 2 ∙ 3 ∙ 67

Prime factors of 402 are 3 ∙ 2 67.

Grade 6 Go Math Lesson 10.2 Prime Factorization Answer Key Question 5.
36
Answer:
Prime factorization of 36 using a factor tree.

Prime factors of 36 are 3 ∙ 2 ∙ 3 ∙ 2 or 3² ∙ 2²

Use a ladder diagram to find the prime factorization of each number.

Question 6.
32
Answer:
Determine the prime factors of 32 using continuous division.

Prime factors of 32 are 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 or 2⁵

Question 7.
27
Answer:
Determine the prime factors of 27 using continuous division.

The prime factors of 27 are 3 ∙ 3 ∙ 3 or 3³

Essential Question Check-In

Question 8.
Tell how you know when you have found the prime factorization of a number.
Answer:
The prime factorization of a number is verified by studying its factors. They must all be prime numbers.

Question 9.
Multiple Representations Use the grid to draw three different rectangles so that each has an area of 12 square units and they all have different widths. What are the dimensions of the rectangles?

Answer:
The dimensions of the rectangle are:
a. 1 by 12
b. 2 by 6
c. 3 by 4

Diagram of the rectangle

Dimensions are: 1 by 12, 2 by 6, 3 by 4.

Question 10.
Brandon has 32 stamps. He wants to display the stamps in rows, with the same number of stamps in each row. How many different ways can he display the stamps? Explain.
Answer:
Given Number = 32

Write the given number as a product of its factors, therefore:
32 = 2 × 16
32 = 4 × 8
This implies that he can arrange the stamps in 4 different ways. He can make 2 columns each containing 16 stamps or 2 rows each containing 16 stamps Or he can make 4 columns each containing 8 stamps or 4 rows each containing 8 stamps.

He can arrange the stamps in 4 different ways. He can make 2 columns each containing 16 stamps or 2 rows each containing 16 stamps. Or he can make 4 columns each containing 8 stamps or 4 rows each containing 8 stamps.

Question 11.
Communicate Mathematical Ideas How is finding the factors of a number different from finding the prime factorization of a number?
Answer:
Finding the factors of a number indicates all the factors that may be used for the number which can be a combination of prime numbers and composite numbers. Finding the prime factorization of a number breaks a composite number into product of its prime factors.

Example:
12 – factors are 1 and 12, 2 and 6, 3 and 4
12 – prime factors are 2 ∙ 2 ∙ 3 or 2² ∙ 3

Factors are the numbers to be multiplied.
Prime factorization results to product of prime factors of the number.

Find the prime factorization of each number.

Question 12.
891 ______________
Answer:
The prime factorization of a number is the number written as the product of its prime factors. For example, the prime factors of 891 are 3, 3, 3, 3 and 11.

The prime factorization of 891 is:
891 = 3 × 3 × 3 × 3 × 11 or 3⁴ × 11
= 3⁴ × 11

Question 13.
5.4 ______________
Answer:
The prime factorization of a number is the number written as the product of its prime factors. For example, the prime factors of 504 are 2, 2, 2, 3, 3 and 7.

The prime factorization of 504 is:
504 = 2 × 2 × 2 × 3 × 3 × 7 or 2³ × 3² × 7
= 2³ × 3² × 7

Question 14.
23 ______________
Answer:
The prime factorization of a number is the number written as the product of its prime factors. For example, the prime factors of 23 are 1 and 23.

The prime factorization of 23 is:
23 = 1 × 23

Question 15.
230 ______________
Answer:
The prime factorization of a number is the number written as the product of its prime factors. For example, the prime factors of 230 are 2, 5 and 23.

The prime factorization of 230 is:
230 = 2 × 5 × 23

Question 16.
The number 2 is chosen to begin a ladder diagram to find the prime factorization of 66. What other numbers could have been used to start the ladder diagram for 66? How does starting with a different number change the diagram?
Answer:
Prime factorization of 66 using ladder diagram.

It will only change the order or arrangement of the prime factors but the prime factors are still the same.
The number 3 can be used to start the diagram however the prime factors will still be the same.

Question 17.
Critical Thinking List five numbers that have 3, 5, and 7 as prime factors.
Answer:

• 105 = 3 ∙ 5 ∙ 7
• 315 = 3 ∙ 3 ∙ 5 ∙ 7
• 525 = 3 ∙ 5 ∙ 5 ∙ 7
• 735 = 3 ∙ 5 ∙ 7 ∙ 7
• 1,575 = 3 ∙ 3 ∙ 5 ∙ 5 ∙ 7

The numbers are 105, 315, 525, 735, and 1,575.

Question 18.
In a game, you draw a card with three consecutive numbers on it. You can choose one of the numbers and find the sum of its prime factors. Then you can move that many spaces. You draw a card with the numbers 25, 26, 27. Which number should you choose if you want to move as many spaces as possible? Explain.
Answer:
Given number:
25, 26, 27

Write the given numbers as a product of their prime factors, therefore:

• 25 = 5 × 5
• 26 = 2 × 13
• 27 = 3 × 3 × 3

Evaluate the sum of the prime factors, therefore:

• 5 + 5 = 10
• 2 + 13 = 15
• 3 × 3 × 3 = 9

It can be seen that 15 > 10 > 9 therefore the choice of the number 26 will result in the maximum number of moves.

Prime Factorization Grade 6 Pdf Lesson 10.2 Question 19.
Explain the Error When asked to write the prime factorization of the number 27, a student wrote 9 3. Explain the error and write the correct answer.
Answer:
Given number = 27
Write the given number as a product of its prime factors, therefore:
27 = 3 × 3 × 3
Prime factors of the given number is 3 . The student’s solution is incorrect because the factor 9 ¡s not a prime factor.
The student’s solution is incorrect because the factor 9 is not a prime factor.

H.O.T. Focus On Higher Order Thinking

Question 20.
Communicate Mathematical Ideas Explain why it is possible to draw more than two different rectangles with an area of 36 square units, but it is not possible to draw more than two different rectangles with an area of 15 square units. The sides of the rectangles are whole numbers.
Answer:
Given number = 36
Write the given number as a product of its factors, therefore:

• 36 = 2 × 18 = Rectang1e 1
• 36 = 3 × 12 = Rectangle 2
• 36 = 4 × 9 = Rectangle 3
• 36 = 36 × 1 = Rectangle 4
• 36 = 6 × 6 = Square

It can be seen that 4 rectangles with different dimensions can be made each having an area of 36 square units

Given number = 15
Write the given number as a product of its factors, therefore:

• 15 = 3 × 5 = Rectangle 1
• 15 = 15 × 1 = Rectangle 2

It can be seen that only 2 rectangles can be made here because the given area is a product of prime factors or the number itself.

Question 21.
Critique Reasoning Alice wants to find all the prime factors of the number you get when you multiply 17 ∙ 11 ∙ 13 ∙ 7. She thinks she has to use a calculator to perform all the multiplications and then find the prime factorization of the resulting number. Do you agree? Why or why not?
Answer:
It can be seen that the number itself has factors that are prime numbers, so no calculator is required and the prime factors: 17, 11, 13 and 7 can directly be deduced.

The prime factors are 7, 11, 13 and 17.

Question 22.
Look for a Pattern Ryan wrote the prime factorizations shown below, If he continues this pattern, what prime factorization will he show for the number one million? What prime factorization will he show for one billion?
10 = 5 ∙ 2
100 = 5² ∙ 2²
1,000 = 5³ ∙ 2³
Answer:
If the pattern continues, the number of zeros will determine the exponent of the base 10.
1 million = 1,000,000 = 5⁶ . 2⁶
1 billion = 1,000,000,000 = 5⁹ . 2⁹

The prime factorization for 1 million is 5⁶ . 2⁶
The prime factorization for 1 billion is 5⁹ . 2⁹

Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Module 10 Answer Key Generating Equivalent Numerical Expressions.

Texas Go Math Grade 6 Module 10 Answer Key Generating Equivalent Numerical Expressions

Texas Go Math Grade 6 Module 10 Are You Ready? Answer Key

Find the product.

Question 1.
992 × 16
Answer:
Given expression = 992 × 16
Rewrite the given expression as a product of a number and a difference, therefore = 16(1000 – 8)
Apply distributive property to expand the parenthesis = 16000 – 128

Evaluate = 15872
9920 × 16 = 15872

Go Math Grade 6 Module 10 Answer Key Question 2.
578 × 27
Answer:
Solution to this example is given below

Final Solution = 15,606

Question 3.
839 × 65
Answer:
Given expression = 839 × 65
Rewrite the given expression as a product of a number and a difference, therefore = 65(800 + 39)
Apply distributive property to expand the parenthesis = 52000 + 2535

Evaluate = 54535
839 × 65 = 54535

Question 4.
367 × 23
Answer:
Solution to this example is given below

Final Solution = 8,441

Find the product.

Question 5.
7 × 7 × 7
Answer:
Solution to this example is given below

Go Math Expressions Grade 6 Answer Key Module 10 Test Answers Question 6.
3 × 3 × 3 × 3
Answer:
Solution to this example is given below

Question 7.
6 × 6 × 6 × 6 × 6
Answer:
Solution to this example is given below

Question 8.
2 × 2 × 2 × 2 × 2 × 2 × 2
Answer:
Solution to this example is given below

Divide.

Question 9.
20 ÷ 4
Answer:
Solution to this example is given below
20 ÷ 4 = ?
Think: 4 times what number equals 20?
⇒ 4 × 5 = 20
⇒ 20 ÷ 4 = 5
So, 20 ÷ 4 = 5.

Grade 6 Go Math Module 10 Answer Key Question 10.
21 ÷ 7
Answer:
Solution to this example is given below
21 ÷ 7 = ?
Think: 7 times what number equals 20?
⇒ 7 × 3 = 21
⇒ 21 ÷ 7 = 3
So, 21 ÷ 7 = 3.

Question 11.
42 ÷ 7
Answer:
Solution to this example is given below
42 ÷ 7 = ?
Think: 7 times what number equals 20?
⇒ 7 × 6 = 42
⇒ 42 ÷ 7 = 6
So, 42 ÷ 7 = 6.

Question 12.
56 ÷ 8
Answer:
Solution to this example is given below
56 ÷ 8 = ?
Think: 8 times what number equals 20?
⇒ 8 × 7 = 56
⇒ 56 ÷ 8 = 7
So, 56 ÷ 8 = 7.

Texas Go Math Grade 6 Module 9 Reading Start-Up Answer Key

Visualize Vocabulary
Use the words to complete the graphic. You may put more than one word in each box.

Understand Vocabulary

Complete the sentences using the preview words.

Question 1.
A number that is formed by repeated multiplication by the same factor is a _________________.
Answer:
Power

Go Math Answer Key Grade 6 Generating Equivalent Expressions Question 2.
A rule for simplifying expressions is _____________________
Answer:
Order of Operations.
It can consist of multiple operations but follow a certain order.

Question 3.
The _________________ is a number that is multiplied. The number that indicates how many times this number is used as a factor is the ____________ .
Answer:
base and exponent.

Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Lesson 13.3 Answer Key Multiplication and Division Inequalities with Positive Numbers.

Texas Go Math Grade 6 Lesson 13.3 Answer Key Multiplication and Division Inequalities with Positive Numbers

Texas Go Math Grade 6 Lesson 13.3 Explore Activity Answer Key

Modeling One-Step Inequalities

You can use algebra tiles to solve inequalities that involve multiplying positive numbers.

Dominic is buying school supplies. He buys 3 binders and spends more than $9. How much did he spend on each binder?
A. Let x represent the cost of one binder. Write an inequality.

B. The model shows the inequality from A
There are _________ x-tiles, so draw circles to separate the tiles into ________ equal groups. _____________
How many units are in each group? _________

C. What values make the inequality you wrote in A true? Graph the solution of the inequality. _________

Reflect

Question 1.
Analyze Relationships Is 3.25 a solution of the inequality you wrote in A ? If so, does that solution make sense for the situation?
Answer:
Inequality for A is the following:
3 ∙ x > 9
We can substitute 3.2 for x and check if inequality is true, so we have the following:
3 ∙ 3.25 > 9
9 ∙ 75 > 9
We got that inequality is true, so, 3.25 can be the solution. And yes, this solution makes sense for the situation.

Go Math Lesson 13.3 Answer Key Grade 6 Multiplication Question 2.
Represent Real-World Problems Rewrite the situation in A to represent the inequality 3x < 9.
Answer:
The modified situation that represents the given inequality could be:
Dominic is buying school supplies. He buys 3 binders and spends less than 9$. How much did he spend on each binder?

Your Turn

Solve each inequality. Graph and check the solution.

Question 3.
5x ≥ 100

Answer:
The first step in solving a given inequality is dividing both sides by 5 and getting:
\(\frac{5 x}{5} \geq \frac{100}{5}\)
x ≥ 20
Now, we will graph the solution:

We will check the solution by substituting a solution from the shaded part of the graph into the original inequality.
We will substitute 30 for x into the original inequality and get:
5 ∙ 30 ≥ 100
150 ≥ 100
So, the inequality is true
x ≥ 20

Question 4.
\(\frac{z}{4}\) < 11

Answer:
First step in solving given inequality is multiplying both sides by 4 and get:
4(\(\frac{z}{4}\)) < 11(4)
x < 44
Now, we will graph the solution:

We will check the solution by substituting a solution from the shaded part of the graph into the original inequality.
We will substitute 40 for x into the original inequality and get:
\(\frac{40}{4}\) < 11
10 < 11
So, the inequality is true.
z < 44

Reflect

Question 5.
Represent Real-World Problems Write and solve a real-world problem for the inequality 4x ≤ 60.
Answer:
Justin buys snacks. He bought 4 packs and spent $60 or less than $60. How much costs one pack of snacks?
Let x represent a price of one pack of snacks. According to given information.
we have the following inequality:
4x ≤ 60
In order to solve it. We will divide it by 4 and get:
\(\frac{4 x}{4} \leq \frac{60}{4}\)
x ≤ 15
So, one pack of snacks costs $15 or less than $15.

Your Turn

Question 6.
A paperweight must weigh less than 4 ounces. Brittany wants to make 6 paperweights using sand. Write and solve an inequality to find the possible weight of the sand she needs.
Answer:
Let x represent the possible weight of the sand Brittany needs. According to all informations, we have the following inequality:
6x < 4
We will, divide by 6 given inequality in order to solve previous inequality and get:
\(\frac{6x}{6}\) < \(\frac{4}{6}\)
x < \(\frac{2}{3}\)
So, the weight of the sand must be less than \(\frac{2}{3}\) ounce.
We will check solution substituting \(\frac{1}{3}\) for x in original inequality and get:
6 ∙ \(\frac{1}{3}\) < 4
2 < 4
So, the inequality is true.
x < \(\frac{2}{3}\)

Texas Go Math Grade 6 Lesson 13.3 Guided Practice Answer Key

Question 1.
Write the inequality shown on the model. Circle groups of tiles to show the solution. Then write the solution. (Explore Activity)

Inequality: ___________
Solution: ___________
Answer:

Solve each inequality. Graph and check the solution.  (Example 1)

Question 2.
8y < 320 _________

Answer:
First step ¡s multiplying both sides by 8 and get:
\(\frac{8y}{8}\) < \(\frac{320}{80}\)
y < 40
Now, we will graph the solution:

We will now check the solution by substituting a solution from the shaded part of the graph into the original inequality.
We will substitute 30 for y into the original inequality and get:
8 ∙ 30 < 320
240 < 320
So, the inequality is true.
y < 40

Go Math Grade 6 Lesson 13.3 Answer Key Question 3.
\(\frac{r}{3}\) ≥ 11 __________

Answer:
First we will multiply both sides by 3 and get:
3(\(\frac{r}{3}\)) ≥ 11 ∙ 3
r ≥ 33
Now, we will graph the solution:

We will now check the solution by substituting a solution from the shaded part of graph into the original inequality.
We will substitute 39 for r into original inequality and get:
\(\frac{39}{3}\) ≥ 11
13 ≥ 11
So, the inequality is true.
r ≥ 33

Question 4.
Karen divided her books and put them on 6 shelves. There were at least 14 books on each shelf. How many books did she have? Write and solve an inequality to represent this situation. (Example 2)
Answer:
Let x represent number of books Karen had. According to previous informations, we need to solve following inequality:
\(\frac{x}{6}\) > 14
First, we will multiply by 6 both sides and get:
6(\(\frac{x}{6}\)) > 14.6
x > 84
Next thing we will do is to check the solution by substituting a value in the solution set in the original inequality.
We will substitute 90 for x and get:
\(\frac{90}{6}\) > 14
15 > 14
So, the inequality is true.
Conclusion is that Karen had at least 84 book.
x > 84

Essential Question Check-In

Question 5.
Explain how to solve and check the solution to 5x < 40 using properties of inequalities.
Answer:
In order to solve this inequality, we need to divide both sides by 5 and get:
\(\frac{5x}{5}\) < \(\frac{40}{5}\)
x < 8
So, we got that set of solutions is x < 8. Now we will check the solution by substituting a value in the solution set in the original inequality. We will substitute 5 for x and get:
5 ∙ 5 < 40
25 < 40
So, the inequality is true.

Texas Go Math Grade 6 Lesson 13.3 Independent Practice Answer Key

Write and solve an inequality for each problem.

Question 6.
Geometry The perimeter of a regular hexagon is at most 42 inches. Find the possible side lengths of the hexagon.

Answer:
We can notice that hexagon has 6 sides, so, let x be length of one side of hexagon. Because this is a regular hexagon, all sides have equal lengths. The perimeter is calculated by multiplying side length by 6, so, according to given informations in this task, we get the following inequality:
6x ≤ 42
In order to solve this inequality, we will divide both sides by 6 and get:
\(\frac{6x}{6}\) ≤ \(\frac{42}{6}\)
x ≤ 7
So, the possible length of side of the hexagon ¡s at most 7.

Question 7.
Tamar needs to make at least $84 at work on Tuesday to afford dinner and a movie on Wednesday night. She makes $14 an hour at her job. How many hours does she need to work on Tuesday?
Answer:
Let x represent a number of hours Tamara needs to work on Tuesday.
According to all the information in the task, we have the following inequality:
14x ≥ 84
In order to solve this inequality, we need to divide both sides by 14 and get:
\(\frac{14x}{14}\) ≥ \(\frac{84}{14}\)
x ≥ 6
So, Tamara needs to work at least 6 hours on Tuesday

Lesson 13.3 Answer Key Multiplication and Division Inequalities Question 8.
In a litter of 7 kittens, each kitten weighs more than 3.5 ounces. Find the possible total weight of the litter.
Answer:
Let x represent the possible total. weight of the litter.
According to all information in the task, we have the following inequality:
\(\frac{x}{7}\) > 3.5
We will multiply both sides by 7 and get:
7(\(\frac{x}{7}\)) > 3.5 ∙ 7
x > 24.5
So, the possible total weight of the litter is more than 24.5 ounces.

Question 9.
To cover his rectangular backyard, Will needs at least 1 70.5 square feet of sod. The length of Will’s yard is 15.5 feet. What are the possible widths of Will’s yard?
Answer:
Let x represent width of Will’s yard. In order to calculate it, we need to solve the following inequality:
15.5x ≥ 170.5
We wiLl divide both sides by 15.5 and get:
\(\frac{15.5x}{15.5}\) ≥ \(\frac{170.5}{15.5}\)
x ≥ 11
So, the width of Will’s yard need to be at least 11 feet.

Solve each inequality. Graph and check the solution.

Question 10.
10x ≤ 60

Answer:
First step is to divide both sides by 10, so, we get:
\(\frac{10x}{10}\) ≤ \(\frac{60}{10}\)
x ≤ 6
Now, we wilt graph the solution:

We will, now check the solution by substituting a solution from the shaded part of graph into the original inequality.
We will, substitute 3 for z into original inequality and get:
10 ∙ 3 ≤ 60
30 ≤ 60
So, the inequality is true.
x ≤ 6

Question 11.
\(\frac{t}{2}\) > 0

Answer:
First step is to multiply both sides by 2, so, we get:
2(\(\frac{t}{2}\)) > 0 ∙ 2
t > 0
Now, we will graph the solution:

We will now check the solution by substituting a solution from the shaded part of graph into the original inequality.
We will substitute 2 for t into original inequality and get:
\(\frac{2}{2}\) > 0
1 < 0 So, the inequality is true t > o

Question 12.
Steve pays less than $32 per day to rent his apartment. August has 31 days. What are the possible amounts Steve could pay for rent in August?
Answer:
Let x represent the possible amount Steve could pay for rent in August According to all informations, we have the following inequality we need to solve:
\(\frac{x}{31}\) < 32
We will multiply both sides by 31 and get:
31(\(\frac{x}{31}\)) < 32 ∙ 31
x < 992
So, Steve could pay for amount in August less than 992$.

Question 13.
If you were to graph the solution for exercise 12, would all points on the graph make sense for the situation? Explain.
Answer:
No, points which make sense for the situation will be between 992 and 0. Points with negative values do not make sense for this a real-world situation.

Question 14.
Multistep Lina bought 4 smoothies at a health food store. The bill was less than $16.
a. Write and solve an inequality to represent the cost of each smoothie.
Answer:
Let x represent the cost of each smoothie. We need to soLve the following inequaLity in order to caLculate x.
4x < 16
We need to divide both sides by 4 and get:
\(\frac{4x}{4}\) < \(\frac{16}{4}\)
x < 4
So, the price of each smoothie in less than 4$.

b. What values make sense for this situation? Explain.
Answer:
Values that makes sense for this situation are between 4 and 0.
Because, no one sells nothing for price 0.
So only previous values make sense for this situation.

c. Graph the values that make sense for this situation on the number line.
Answer:
According to part (b), we will graph the values that make sense for this situation, it would be:

Solve each inequality.

Question 15.
\(\frac{p}{13}\) ≤ 30
Answer:
First step is to multiply both sides by 13:
13 (\(\frac{p}{13}\)) ≤ 30 ∙ 13
p ≤ 390
Now, we will graph the solution:

Next step is to check the solution by substituting a solution from the shaded part of the graph into the original inequality.
We will substitute 360 for p and get:
\(\frac{360}{13}\) ≤ 30
27.69 ≤ 30
So, the inequality is true
p ≤ 390

Question 16.
2t > 324
Answer:
First step is to divide both sides by 2, so, we have the following:
\(\frac{2t}{2}\) > \(\frac{324}{2}\)
t > 162
Now, we will graph the solution:

Next step is to check the solution by substituting a soLution from the shaded part of the graph into the original inequality.
We will substitute 170 for t and get:
2 ∙ 170 > 324
340 > 324
So, the inequality is true.
t > 162

Multiplication Questions for Grade 6 Go Math Question 17.
12y ≥ 1
Answer:

The next step is to check the solution by substituting a solution from the shaded part of the graph into the original inequality.
We will substitute 10 for y and get:
12 ∙ 10 ≥ 1
120 ≥ 1
So, the inequality is true.
y ≥ \(\frac{1}{12}\)

Question 18.
\(\frac{x}{9.5}\) < 11
Answer:
We will multiply both sides by 9.5 and get:
9.5 (\(\frac{x}{9.5}\)) < 11.9.5
x < 104.5
Now, we will graph the solution:

Next step is to check the solution by substituting a solution from the shaded part of the graph into the original inequality.
We will substitute 90 for x and get:
\(\frac{90}{9.5}\) < 11
9.47 < 11
So, the inequality is true.
x < 104.5

The sign shows some prices at a produce stand.

Question 19.
Tom has $10. What is the greatest amount of spinach he can buy?
Answer:
Let x represent the amount of spinach Tom can buy. According to all informations in the task, we have the following inequality:
3x ≤ 10
In order to solve this inequaLity, we need to divide both sides by 3 and get:
\(\frac{3x}{3}\) ≤ \(\frac{10}{3}\)
x ≤ 3.33
So, conclusion is that Tom can buy at most 3.33 pounds of spinach.

Question 20.
Gary has enough money to buy at most 5.5 pounds of potatoes. How much money does Gary have?
Answer:
Let x represent sum of money Gary have. According to data from the table and informations from task, we have the following inequality:
\(\frac{x}{0.50}\) ≤ 0.50
In order to solve this inequality, we need to multiply both sides by 0.50 and get:
0.50(\(\frac{x}{0.50}\)) ≤ 5.5 ∙ 0.50
x ≤ 2.75
So, conclusion is that Gary has at most 2.75.

Question 21.
Florence wants to spend no more than $3 on onions. Will she be able to buy 2.5 pounds of onions? Explain.
Answer:
Let x represent the amount of onions Florence can buy.
According to the table and informations, we get following inequality:
1.25x ≤ 3
In order to soLve this inequality, we need to divide both sides by 1.25 and get:
\(\frac{1.25 x}{1.25} \leq \frac{3}{1.25}\)
x ≤ 2.4
So, she can buy at most 2.4 pounds of onions. Conclusion is she can not buy 2.5 pounds of onions.

Question 22.
The produce buyer for a local restaurant wants to buy more than 30 lb of onions. The produce buyer at a local hotel buys exactly 12 pounds of spinach. Who spends more at the produce stand? Explain.
Answer:
Let x represent the sum of money the produce buyer for a local restaurant will spend for buying onions. So, we have the following inequality :
\(\frac{x}{1.25}\) > 30
In order to solve this inequality, we need to multiply both sides by 1.25 and get:
1.25(\(\frac{x}{1.25}\)) > 30 ∙ 1.25
x > 37.5
So, the produce buyer will spend more that $ 37.5 to by more than 30 lb of onions.
Now, let y represent sum of money the produce buyer for a local hotel will spend for buying exactly 12 pounds of spinach. We have the following equation we need to solve:
y ÷ 3 = 12
y = 12.3
y = 36
So, we can notice that the produce buyer will spend exactly $36 to buy 12 pounds of spinach.
Conclusion is that the produce buyer for a local restaurant will spend more money than the produce buyer for a local hotel.

H.O.T. Focus on Higher Order Thinking

Question 23.
Critique Reasoning A student solves \(\frac{r}{5}\) ≤ \(\frac{2}{5}\) and gets r ≤ \(\frac{2}{25}\) What is the correct solution? What mistake might the student have made?
Answer:
The mistake which student might made is that he divided by 5, not multiplied by 5
So, the right way to solve this inequality is to multiply by 5 and get:
5(\(\frac{r}{5}\)) ≤ \(\frac{2}{5}\) ∙ 5
r ≤ 2
So, the solution is r ≤ 2.

Question 24.
Represent Real-World Problems Write and solve a word problem that can be represented with 240 ≤ 2x.
Answer:
Mary want’s to spend $240 or more on buying chocolate as presents for guests at lier party. One pack costs $2, so, how many packs of chocolate she can buy?
Let x represent the number of packs of chocolate Mary can buy. According to all informations. we get the following inequality:
2x ≥ 240
We need to divide both sides by 2 and get:
\(\frac{2 x}{2} \geq \frac{240}{2}\)
x ≥ 120
So, she can at least buy 120 packs of chocolate or more.

Question 25.
Persevere in Problem Solving A rectangular prism has a length of 13 inches and a width of \(\frac{1}{2}\) inch. The volume of the prism is at most 65 cubic inches. Find all possible heights of the prism. Show your work.
Answer:
First, we will calculate the area of basis of this rectangular prism multiplying the length by the width and get:
B = 13 ∙ \(\frac{1}{2}\) = \(\frac{13}{2}\)
So, the area of bassis of rectangular prism is \(\frac{13}{2}\) square inches. Formula for calculating the volume is:
V = B ∙ H
Where V is the volume, B is the area of basis and H is its height.
According to all informations in the task, we have the following inequality we need to solve to find all possible heights of prism, where x represent height:
\(\frac{13}{2}\)x ≤ 65
We need to multiply both sides by \(\frac{2}{13}\) and get:
\(\frac{2}{13} \cdot \frac{13}{2} x \leq \frac{65}{1} \cdot \frac{2}{13}\)
x ≤ 10
So, the height of the prism can be at most 10 inches.

Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Lesson 12.2 Answer Key Addition and Subtraction Equations.

Texas Go Math Grade 6 Lesson 12.2 Answer Key Addition and Subtraction Equations

Texas Go Math Grade 6 Lesson 12.2 Explore Activity Answer Key

Modeling Equations

A puppy weighed 6 ounces at birth. After two weeks, the puppy weighed 14 ounces. How much weight did the puppy gain?
Let x represent the number of ounces gained.

To answer this question, you can solve the equation 6 + x = 14.
Algebra tiles can model some equations. An equation mat represents the two sides of an equation. To solve the equation, remove the same number of tiles from both sides of the mat until the x tile is by itself on one side.
A. Model 6 + x = 14.

B. How many 1 tiles must you remove on the left side so that the x tile is by itself? ___________ Cross out these tiles on the equation mat.
C. Whenever you remove tiles from one side of the mat, you must remove the same number of tiles from the other side of the mat. Cross out the tiles that should be removed on the right side of the mat.
D. How many tiles remain on the right side of the mat? ____________ This is the solution of the equation.
The puppy gained ____________ ounces.

Reflect

Question 1.
Communicate Mathematical Ideas How do you know when the model shows the final solution? How do you read the solution?
Answer:
The model shows final solution when only the variable is left in one side of the mat and the the same number of tiles have been removed. Thus, the remaining tiles on the right side of the mat are 12 which indicates that the puppy gained 12 ounces.

Example 1

Solve the equation a + 15 = 26. Graph the solution on a number line.

Reflect

Go Math Grade 6 Answer Key Lesson 12.2 Question 2.
Communicate Mathematical Ideas How do you decide which number to subtract from both sides?
Answer:
The reason for solving an equation is to determine the value of the variable on which the equation holds true. For this purpose, the variable is isolated on 1 side of the equation. Therefore, the number associated with the variable on the variable side of the equation is subtracted (or added as required) from both sides of the equation.

Your Turn

Question 3.
Solve the equation 5 = w + 1.5.

Graph the solution on a number line.
w = ____________
Answer:
Solution to this example is given below
5 = w + 1.5
5 = w + 1.5 Notice that 15 is added tow
5 – 1.5 = w + 1.5 – 1.5 Subtract both sides of the equation by 15
w = 3.5 Simplify
Check; 5 = w + 1.5
5 = 3.5 + 1.5 Substitute 3.5 for w
5 = 5 Add on the right side.
w = 3.5 Final solution
Graph the solution on a number line.

Example 2

Solve the equation y – 21 = 18. Graph the solution on a number line.

Reflect

Question 4.
Communicate Mathematical Ideas How do you know whether to add on both sides or subtract on both sides when solving an equation?
Answer:
The reason for solving an equation is to determine the value of the variable on which the equation holds true. For this purpose, the variable is isolated on 1 side of the equation. Therefore, the number associated with the variable on the variable side of the equation is subtracted if its sign is positive or added if its sign is negative from both sides of the equation.

Your Turn

Go Math Grade 6 Answers Pdf Lesson 12.2 Question 5.
Solve the equation h – \(\frac{1}{2}\) = \(\frac{3}{4}\).
Graph the solution on a number line.

h = ___________
Answer:
Solution to this example is given below.

Graph the solution on a number line.

Example 3

Find the measure of the unknown angle.

Your Turn

Question 6.
Write and solve an equation to find the measure of the unknown angle.

Answer:
It can be seen that the 2 angles are complementary angles so the equation becomes:
x + 65° = 90°
Add 65° on both sides of the equation to isolate the variable on 1 side of the equation:
x + 65° – 65° = 90° – 65°
Evaluate the variable:
x = 25°

Go Math Answer Key Grade 6 Lesson 12.2 Practice Geometry Question 7.
Write and solve an equation to find the measure of a complement of an angle that measures 42°.
Answer:
The sum of an unknown angle and a 42° angle is 90°. What is the measure of the unknown angle?
Write an equation
x + 42° = 90°
x + 42° – 42° = 90° – 42° Subtract 42 from both sides
x = 48° Simplify
x = 48° Final solution
The unknown angle measures 48°

Example 4

Write a real-world problem for the equation 21.79 + x = 25. Then solve the equation.
21.79 + x = 25

Reflect

Question 8.
What If? How might the real-world problem change if the equation were x – 21.79 = 25 and Joshua still spent $21.79 on roses?
Answer:
In the context of the example, r w the amount to be spent on the card. Here it represents tile total budget. as when this equation is solved: r = 25 + 21.79 = $46.79 and the situation can be to evaluate the total amount such that there should be $25 left to spend on a gift after spending $21.79 on roses.

Your Turn

Question 9.
Write a real-world problem for the equation x – 100 = 40. Then solve the equation.
Answer:
A real world situation to model the given equation can be the total amount required to he saved, if 100 are in savings and still $40 more are required.
Solve the equation: x = 100 + 40 = 140.
x = 140

Texas Go Math Grade 6 Lesson 12.2 Guided Practice Answer Key

Question 1.
A total of 14 guests attended a birthday party. Three guests stayed after the party to help clean up. How many guests left when the party ended? (Explore Activity)
a. Let x represent the _____________
Answer:
The variable x represents the number of guests who left the party when it ended.

b.

Answer:

c. Draw algebra tiles to model the equation. _________ friends left when the party ended.

Answer:
There are 11 friends left when the party ended.

Solve each equation. Graph the solution on a number line. (Examples 1 and 2)

Go Math Grade 6 Pdf Solving Addition and Subtraction Equations Answers Question 2.
2 = x – 3 x = ____________

Answer:
Solution to this example is given below
2 = x – 3
2 = x – 3 Notice that 3 is subtracted from x
2 + 3 = x – 3 + 3 Add both sides of the equation by 3
x = 5 Switch sides
Check; 2 = x – 3
2 = 5 – 3 Substitute 5 for x
2 = 2 Subtract on the right side.
x = 5 Final solution
Graph the solution on a number line.

Question 3.
s + 12.5 = 14 s = ___________

Answer:
Solution to this example is given below
s + 12.5 = 14
s + 12.5 = 14 Notice that 12.5 is added to s
s + 12.5 – 12.5 = 14 – 12.5 Subtract both sides of the equation by 125
s = 1.5 Simplify
Check; s + 12.5 = 14
1.5 + 12.5 = 14 Substitute 1.5 for s
14 = 14 Add on the left side.
s = 1.5 Final solution
Graph the solution on a number line.

Question 4.
h + 6.9 = 11.4
h = _________
Answer:
Solution to this example is given below
h + 6.9 = 11.4
h + 6.9 = 11.4 Notice that 6.9 is added to h
h + 6.9 6.9 = 11.4 – 6.9 Subtract both sides of the equation by 6.9
h = 4.5 Simplify
Check; h + 6.9 = 11.4
4.5 + 6.9 = 11.4 Substitute 4.5 for h
11.4 = 11.4 Add on the Left side.
h = 4.5 Final solution
h = 4.5

Question 5.
82 + p = 122
p = ____________
Answer:
Given equation:
82 + p = 122
Add 82 on both sides of the equation to isolate the variable on 1 side of the equation:
82 + p – 82 = 122 – 82
Evaluate the variable:
p = 40

Go Math Grade 6 Answer Key Pdf Lesson 12.2 Question 6.
n + \(\frac{1}{2}\) = \(\frac{7}{4}\)
n = ___________
Answer:
Given equation:
n + \(\frac{1}{2}\) = \(\frac{7}{4}\)
Add –\(\frac{1}{2}\) on both sides of the equation to isolate the variable on 1 side of the equation:
n + \(\frac{1}{2}\) – \(\frac{1}{2}\) = \(\frac{7}{4}\) – \(\frac{1}{2}\)
Evaluate the variable using Least common denominator technique:
n = \(\frac{7-1 \times 2}{4}\)
Simplify:
n = \(\frac{7-2}{4}=\frac{5}{4}\)
n = \(\frac{5}{4}\)

Question 7.
Write and solve an equation to find the measure of the unknown angle. (Example 3)

Answer:
It can be seen that the 2 angles are supplementary angles so the equation becomes:
x + 45° = 180°
Add -45° on both sides of the equation to isolate the variable on 1 side of the equation:
x + 45° – 45° = 180° – 45°
Evaluate the variable:
x = 135°

Question 8.
Write a real-world problem for the equation x – 75 = 200. Then solve the equation. (Example 4)
Answer:
John needs to save a certain amount of money for a new bike. He has already saved $75. If he requires $200 more. how much does he need to save?
Equation: x – 75 = 200, therefore x = 200 + 75 = 275

Essential Question Check-In

Question 9.
How do you solve equations that contain addition or subtraction?
Answer:
Case 1. If the sign of the constant associated with the variable on the variable side of the equation is positive, then the constant is subtracted from both sides of the equation, to isolate the variable on 1 side of the equation. For example: x + 5 = 12 is soLved by subtracting 5 from both sides of the equation, therefore the intermediate step becomes: x + 5 – 5 = 12 – 5 and the solution is x = 7.

Case 2. If the sign of the constant associated with the variable on the variable side of the equation is negative, then the constant is added to both sides of the equation, to isolate the variable on 1 side of the equation. For example: x – 5 = 12 is soLved by adding 5 to both sides of the equation, therefore the intermediate step becomes: x – 5 + 5 = 12 + 5 and the solution is x = 17.

Texas Go Math Grade 6 Lesson 12.2 Independent Practice Answer Key

Write and solve an equation to answer each question.

Question 10.
A wildlife reserve had 8 elephant calves born during the summer and now has 31 total elephants. How many elephants were in the reserve before summer began?
Answer:
Solution to this example is given below
x + 8 = 31
x + 8 = 31 Notice that 8 ¡s added to x
x + 8 – 8 = 31 – 8 Subtract both sides of the equation by 8
x = 23 Simplify
Check; x + 8 = 31
23 + 8 = 31 Substitute 23 for x
31 = 31 Add on the left side.
x = 23 Final solution
There were 23 elephants in the reserve before summer began.

Question 11.
My sister is 14 years old. My brother says that his age minus twelve is equal to my sister’s age. How old is my brother?
Answer:
Let the brother’s age be x, then the equation of this age is:
x – 12 = 14
Add 12 on both sides of the equation to isolate the variable on 1 side of the equation:
x – 12 + 12 = 14 + 12
Evaluate the variable:
x = 26
The brother is 26 years old.

Question 12.
Kim bought a poster that cost $8.95 and some colored pencils. The total cost was $21.35. How much did the colored pencils cost?
Answer:
Solution to this example is given below
x + 8.95 = 21.35
x + 8.95 = 21.35 Notice that 895 is added to x
x + 8.95 8.95 = 21.35 – 8.95 Subtract both sides of the equation by 8.95
x = 12.4 Simplify
Check; x + 8.95 = 21.35
12.4 + 8.95 = 21.35 Substitute 12.4 for x
21.35 = 21.35 Add on the left side.
x = 12.4 Final solution
The colored pencils cost 12.4 dollars.

Go Math Sixth Grade Answer Key Practice and Homework Lesson 12.2 Question 13.
The Acme Car Company sold 37 vehicles in June. How many compact cars were sold in June?

Answer:
Let the number of compact cars sold in Junes be x, then the equation of total cars sold is:
x + 8 = 37
Add -8 on both sides of the equation to isolate the variable on 1 side of the equation:
x + 8 – 8 = 37 – 8
Evaluate the variable:
x = 29
29 compact cars were sold in June.

Question 14.
Sandra wants to buy a new MP3 player that is on sale for $95. She has saved $73. How much more money does she need?
Answer:
Solution to this example is given below
x + 73 = 95
x + 73 = 95 Notice that 73 is added to x
X + 73 – 73 = 95 – 73 Subtract both sides of the equation by 73
x = 22 Simplify
Check; x + 73 = 95
22 + 73 = 95 Substitute 22 for x
95 = 95 Add on the left side
x = 22 Final solution
22 more are required.

Question 15.
Ronald spent $123.45 on school clothes. He counted his money and discovered that he had $36.55 left. How much money did he originally have?
Answer:
Let the total amount of money be x, then the equation of total cost is:
x = 123.45 + 36.55
Evaluate the variable:
x = 160
He originally had $160.

Question 16.
Brita withdrew $225 from her bank account. After her withdrawal, there was $548 left in Brita’s account. How much money did Brita have in her account before the withdrawal?
Answer:
Let the total amount of money be x, then the equation of money after withdrawal is:
x – 225 = 548
Add 225 on both side of the equation to isolate the variable on 1 side of the equation:
x – 225 + 225 = 548 + 225
Evaluate the variable:
x = 773
There were $773 in the account before the withdrawal.

Question 17.
Represent Real-World Problems Write a real-world situation that can be represented by 15 + c = 17.50. Then solve the equation and describe what your answer represents for the problem situation.
Answer:
Solution to this example is given below
15 + c = 17.50
15 + c = 17.50   Notice that 15 is added to c
15 + c – 15 = 17.50 – 15 Subtract both sides of the equation by 15
c = 2.5 Simplify
Check; 15 + c = 17.50
15 + 2.5 = 17.50  Substitute 25 for c
17.50 = 17.50   Add on the left side
c = 2.5 Final solution
Here c = 2.5 is the amount in dollars that can be spent on the card.

Question 18.
Critique Reasoning Paula solved the equation 7 + x = 10 and got 17, but she is not certain if she got the correct answer. How could you explain Paula’s mistake to her?
Answer:
Given equation:
7 + x = 10
Add -7 on both sides of the equation to isolate the variable on 1 side of the equation:
7 + x – 7 = 10 – 7
Evaluate the variable:
x = 3
Paula added 7 to both sides of the equation instead of adding -7.

H.O.T. Focus on Higher Order Thinking

Question 19.
Multistep Handy Dandy Grocery is having a sale this week. If you buy a 5-pound bag of apples for the regular price, you can get another bag for $1.49. If you buy a 5-pound bag of oranges at the regular price, you can get another bag for $2.49.

a. Write an equation to find the discount for each situation using a for apples and r for oranges.
Answer:
The equation to find the discount for each situation using a for the amount of the discount for apples is a + 1.49 = 2.99 and r for the amount of the discount for oranges is r + 2.49 = 3.99
Solve equation 1: a = 2.99 – 1.49 = 1.5
Solve equation 2: r = 3.99 – 2.49 = 1.5

b. Which fruit has a greater discount? Explain.
Answer:
It can be seen that the discount in both cases is the same and equal to $1.50.

Lesson 12.2 Practice Problems Answer Key Grade 6 Question 20.
Critical Thinking An orchestra has twice as many woodwind instruments as brass instruments. There are a total of 150 brass and woodwind instruments.
a. Write two different addition equations that describe this situation. Use w for woodwinds and b for brass.
Answer:
Let w be the number of woodwind instruments and b for brass instruments, then according to the first situation given: w = 2b. The second equation is w + b = 150
Substitute the value of w = 2b:
2b + b = 150
Simplify:
3b = 150
Divide both sides of the equation with 3 to evaluate the variable:
b = \(\frac{150}{3}\) = 50

b. How many woodwinds and how many brass instruments satisfy the given information?
Answer:
There are 50 brass instruments and 2(50) = 100 woodwinds instruments.

Question 21.
Look for a Pattern Assume the following: a + 1 = 2, b + 10 = 20, c + 100 = 200, d + 1,000 = 2,000,…
a. Solve each equation for each variable.
Answer:
a + 1 – 1 = 2 – 1 subtraction property of equality
a = 1 value of a
b + 10 – 10 = 20 – 10 subtraction property of equality
b = 10 value of b
c + 100 – 100 = 200 – 100 subtraction property of equality
c = 100 value of c
d + 1,000 – 1, 000 = 2,000 – 1,000 subtraction property of equality
d = 1,000 value of d

b. What pattern do you notice between the variables?
Answer:
The pattern between variables is increasing by multiples of 10.

c. What would be the value of g if the pattern continues?
Answer:
Following the same pattern of the variables:
a = 1
b = 10
c = 100
d = 1,000
e = 10,000
f = 100,000
g = 1,000,000

Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Module 12 Quiz Answer Key.

Texas Go Math Grade 6 Module 12 Quiz Answer Key

Texas Go Math Grade 6 Module 12 Ready to Go On? Answer Key

12.1 Writing Equations to Represent Situations

Determine whether the given value is a solution of the equation.

Question 1.
p – 6 = 19; p = 13
Answer:
Solution or not
13 – 6 = 19 substitute for the value of p
7 ≠ 19    13 is not a solution
The given value of 13 is not a solution to the equation.

Grade 6 Go Math Module 12 Answer Key with Solution Question 2.
62 + j = 74; j = 12.
Answer:
Solution or not
62 – 12 = 74 substitute for the value of j
74 = 74     12 is a solution
The given value of 12 is a solution for the equation.

Question 3.
\(\frac{b}{12}\) = 5; b = 60.
Answer:
Solution to this example is given below
\(\frac{b}{12}\) = 5; b = 60
\(\frac{60}{12}\) = 5 Substitute 60 for b
5 = 5 Divide.
60 is a solution of \(\frac{b}{12}\)
b = 60

Question 4.
7w = 87; w = 12
Answer:
Solution to this example is given below
7w = 87; w = 12
17(12) = 87 Substitute 12 for w
84 = 87 Multiply.
12 is a not of solution of the equation 7w = 87
w ≠ 12

Question 5.
18 – h = 13; h = -5.
Answer:
Solution or not
18 – (-5) = 13 substitute for the value of h
18 + 5 = 13 add the numbers
23 ≠ 13 -5 is not a solution
The given value of -5 is not a solution for the equation.

6th Grade Math Equations Module 12 Review Quiz Question 6.
6g = -86; g = -16
Answer:
Solution or not
6 (-16) = -86 substitute for the value of g
-96 ≠ 86 -16 is not a solution
The given value of 16 is not a solution for the equation.

Write an equation to represent the situation.

Question 7.
The number of eggs in the refrigerator e decreased by 5 equals 18.
Answer:
The total eggs were e and they decreased by 5 to reduced to 18, so the equation of the situation becomes: e – 5 = 18

Question 8.
The number of new photos p added to the 17 old photos equals 29.
Answer:
Equation:
p + 17 = 29
where:
p is the number of new photos
17 is the number of old photos
29 is the total number of photos

12.2 Addition and Subtraction Equations

Solve each equation.

Question 9.
r – 38 = 9
Answer:
SoLution to this example is given below
r – 38 = 9
r – 38 + 38 = 9 + 38 Add 38 to both sides
r = 47 Simplify
r = 47

Question 10.
h + 17 = 40
Answer:
Solution to this example is given below
h + 17 = 40
h + 17 – 17 = 40 – 17 Subtract 17 to both sides
h = 23 Simplify
h = 23

Question 11.
n + 75 = 155
Answer:
Given equation:
n + 75 = 155
Add -75 on both sides of the given equation to isolate the variable on 1 side of the equation:
n + 75 – 75 = 155 – 75
Simplify to evaluate the variable:
n = 80

Module 12 Quiz Answer Key Go Math 6th Grade Question 12.
q – 17 = 18
Answer:
Solution to this example is given below
q – 17 = 18
q – 17 + 17 = 18 + 17 Add 17 to both sides
q = 35 Simplify
q = 35

12.3 Multiplication and Division Equations

Solve each equation.

Question 13.
8z = 112
Answer:
Given equation:
8z = 112
Divide both sides of the given equation with 8 to isoLate the variable on 1 side of the equation:
\(\frac{8z}{8}\) = \(\frac{112}{8}\)
Simplify to evaluate the variable:
z = 14

Question 14.
\(\frac{d}{14}\) = 7
Answer:
Solution to this example is given below

Go Math Quiz for Grade 6 Module 12 Test Answers Question 15.
\(\frac{f}{28}\) = 24
Answer:
Solution to this example is given below

Question 16.
3a = 57
Answer:
Solution to this example is given below

Essential Question

Question 17.
How can you solve problems involving equations that contain addition, subtraction, multiplication, or division?
Answer:
For explaining this question, lets consider an equation: 2x + 9 = 15 The first step to solve this equation will be to shift all the contents to the non variable side of the equation. This is done by adding 9 to both sides of the equation so the equation transforms to 2x = 15 – 9 = 6. The next step to change the coefficient of x to 1. This is done by dividing both sides of the equation with 2, so the equation transforms to x = \(\frac{6}{2}\) = 3. This is the solution of the equation.

Texas Go Math Grade 6 Module 12 Mixed Review Texas Test Prep Answer Key

Selected Response

Question 1.
Kate has gone up to the chalkboard to do math problems 5 more times than Andre. Kate has gone up 11 times. Which equation represents this situation?
(A) a – 11 = 5
(B) 5a = 11
(C) a – 5 = 11
(D) a + 5 = 11
Answer:
(D) a + 5 = 11

Explanation:
Let Andre’s turns be a and Kate has had 5 more turns so the equation to represent the number of Kates’s turns becomes a + 5 = 11.

6th Grade Math Quiz with Answers Module 12 Question 2.
For which equation is y = 7 a solution?
(A) 7y = 1
(B) y – 26 = -19
(C) y + 7 = 0
(D) \(\frac{y}{2}\) = 14
Answer:
(B) y – 26 = -19

Explanation:
(A) 7 – (7) = 1 substitute for the value of y
49 ≠ 17 is not a solution

(B) 7 – 26 = – 9 substitute for the value of y
-19 = -19 7 is a solution

(C) 7 + 7 = 0 substitute for the value of y
14 ≠ 0 7 is not a solution

(D) \(\frac{7}{2}\) = 14 substitute for the value of y
3.5 ≠ 14 7 is not a solution

The equation (B) y – 26 = -19 shows that the value of 7 is a solution.

Question 3.
Which is an equation?
(A) 17 + x
(B) 45 ÷ x
(C) 20x = 200
(D) 90 – x
Answer:
(C) 20x = 200

Explanation:
An equation is presented by 2 different expressions equated with each other by an equal to sign, therefore option C shows an equation.

Grade 6 Math Quiz Module 12 Answer Key Question 4.
The number line below represents which equation?

(A) -4 + 7 = 3
(B) -4 – 7 = 3
(C) 3 + 7 = -4
(D) 3 – 7 = -4
Answer:
(D) 3 – 7 = -4

Explanation:
The arrow started at 3 and moved to the left 7 gaps and ended at -4. Therefore the equation for the number line is 3 – 7 = -4.

Question 5.
Becca hit 7 more home runs than Beverly. Becca hit 21 home runs. How many home runs did Beverly hit?
(A) 3
(B) 14
(C) 21
(D) 28
Answer:
(B) 14

Explanation:
Let Beverly’s runs be a and Becca has hit 7 more runs so the equation to represent Becca’s runs is: x + 7 = 21. Solve this equation for x, therefore: x = 21 – 7 = 14.

Grade 6 Go Math Answer Key Module 12 Quiz Question 6.
Jeordie spreads out a rectangular picnic blanket with an area of 42 square feet. Its width is 6 feet. Which equation could you use to find its length?
(A) 6x = 42
(B) 42 – x = 6
(C) \(\frac{6}{x}\) = 42
(D) 6 + x = 42
Answer:
(A) 6x = 42

Explanation:
The area of a rectangular is the product of its length and width Therefore, 6x = 42 is the equation to be used to evaluate x, the length of the rectangular blanket.

Question 7.
What is a solution to the equation 6t = 114?
(A) t = 19
(B) t = 108
(C) t = 120
(D) t = 684
Answer:
(A) t = 19

Explanation:
Examine each part of the equation.
t is the unknown value you want to find.
6 is multiplied by t.
= 114 means that after multiplying 6 and t, the result is 114.
Use the equation to solve the problem.
6t = 114
\(\frac{6 t}{6}=\frac{114}{6}\) Divide both sides by 6
t = 19 Simplify
A is the option is correct answer.

Question 8.
The area of a rectangular deck is 680 square feet. The deck’s width is 17 feet. What is its length?
(A) 17 feet
(B) 20 feet
(C) 40 feet
(D) 51 feet
Answer:
(C) 40 feet

Explanation:
A = l ∙ w formula for the area of a rectangle
680 square feet = l ∙ 17 feet substitute for the given values
\(\frac{680}{17}=\frac{l \cdot 17}{17}\) divide both sides of the equation by 17
40 feet = l length of the rectangle
The length of the rectangular deck is 40 feet.

Gridded Response

Module 12 Answer Key Go Math Quiz 6th Grade Question 9.
Sylvia earns $7 per hour at her after-school job. One week she worked several hours and received a paycheck for $91. Write and solve an equation to find the number of hours in which Sylvia would earn $91.

Answer:
Equation:
7x = 91
where:
7 is the amount she earns per hour
x is the number of hours she worked
91 is the total. amount she earned
\(\frac{7 x}{7}=\frac{91}{7}\) divide both sides of the equation by 7
x = 13 number of hours she worked in a week
The answer in the grid is 13.00.