Texas Go Math

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 12.3 Answer Key Multiplication and Division Equations.

Texas Go Math Grade 6 Lesson 12.3 Answer Key Multiplication and Division Equations

Essential Question
How do you solve equations that contain multiplication or division?

Texas Go Math Grade 6 Lesson 12.3 Explore Activity Answer Key

Explore Activity
Modeling Equations
Deanna has a cookie recipe that requires 12 eggs to make 3 batches of cookies. How many eggs are needed per batch of cookies?
Let x represent the number of eggs needed per batch.


To answer this question, you can use algebra tiles to solve 3x = 12.
A. Model 3x = 12

B. There are 3 x-tiles, so draw circles to separate the tiles into 3 equal groups. One group has been circled for you.

C. How many -hi-tiles are in each group? ____
This is the solution of the equation.
___ eggs are needed per batch of cookies.

Reflect

Question 1.
Look for a Pattern Why does it make sense to arrange the 12 tiles in 3 rows of 4 instead of any other arrangement of 12 + 1 -tiles, such as 2 rows of 6?
Answer:
The arrangement of tiles must be 3 rows because it is indicated in the problem that the eggs will be used in 3 batches of cookies.
The solution must indicate 3 batches of cookies which represent the 3 rows.

Your Turn

Solve the equation 3x = – 21. Graph the solution on a number line.

Go Math Grade 6 Lesson 12.3 Answer Key Question 2.
x = ______

Answer:
Determine the value of x
\(\frac{3 x}{3}\) = \(\frac{-21}{3}\) division property of equality
x = -7 value of x
Graph of the solution.

The value of the variable is -7.

Your Turn

Solve the equation \(\frac{y}{9}\) = -1. Graph the solution on a number line.

Question 3.
\(\frac{y}{9}\) = -1
y = ___

Answer:
Determine the vaLue of y.
9 • \(\frac{y}{9}\) = -1 • 9 multiply both sides of the equation by 9
y = -9 value of the variable y
Graph of the solution.

The value of the variable is -9.

Your Turn

Question 4.
Roberto is dividing his baseball cards equally among himself, his brother, and his 3 friends. Roberto was left with 9 cards. How many cards did Roberto give away? Write and solve an equation to solve the problem.
Answer:
Let the total number of cards be x and each person receives 9 so the equation becomes:
\(\frac{x}{5}\) = 9
Multiply both sides of the equation with 5 to isolate the variable on 1 side of the equation:
\(\frac{x}{5}\) × 5 = 9 × 5
Simplify to evaluate the variable:
x = 45
There were a total of 45 cards and he was left with 9, so he gave away 45 – 9 = 36.
he gave away 36 cards.

Your Turn

Go Math Practice and Homework Lesson 12.3 Answer Key Question 5.
Write a real-world problem for the equation 11x = 385. Then solve the equation.
Answer:
Examine each part of the equation.
x is the unknown value you want to find.
11 is multiplied by x.
= 385 means that after multiplying 11 and x, the result is 385.
Use the equation to solve the problem.
11x = 385
\(\frac{11 x}{11}\) = \(\frac{385}{11}\) Divide both sides by 11
x = 35 Final solution

Texas Go Math Grade 6 Lesson 12.3 Guided Practice Answer Key

Question 1.
Caroline ran 15 miles in 5 days. She ran the same distance each day. Write and solve an equation to determine the number of miles she ran each day. (Explore Activity)
a. Let x represent the _____
Answer:
Variable
The variable x represents the number of miles ran each day.

b.
Answer:
Equation
Representation

c. Draw algebra tiles to model the equation.

Caroline ran ___ miles each day.
Answer:
Algebra tiles
Based from the diagram, Caroline ran 3 miles each day.

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

Question 2.
x ÷ 3 = 3

Answer:
Determine the value of x.
3 . \(\frac{x}{3}\) = 3 . 3 multiply both sides of the equation by 3
x = 9 value of the variable x
Graph of the solution.

The value of the variable is 9.

6th Grade Math Multiplication Lesson 12.3 Answer Key Question 3.
4x = -32

Answer:
Determine the value of x
\(\frac{4 x}{4}\) = \(\frac{-32}{4}\) divide both sides of the equation by 4
x = -8 value of the variable x
Graph of the solution.

The value of the variable is 8.

Question 4.
The area of the rectangle shown is 24 square inches. How much longer is its length than its width? (Example 3)
Answer:
Area of a rectangle is the product of its length and width so the equation of area becomes:
6 × w = 24

Divide both sides of the equation with 6 to isolate the variable on 1 side of the equation:
\(\frac{6 w}{6}\) = \(\frac{24}{6}\)
Simplify to evaluate the variable:
w = 4
Therefore, the length is 6 – 4 = 2 inches longer than the width.
The length of this rectangle is 2 inches longer than its width.

Essential Question Check-in

Question 5.
How do you solve equations that contain multiplication or division?
Answer:
Case 1. If the coefficient associated with the variable on the variable side of the equation is being multiplied with it, then this coefficient is converted to 1 by dividing both sides of the equation with the same constant, to isolate the variable on 1 side of the equation. For example: 5x = 60 is solved by dividing both sides of the equation with 5, therefore the intermediate step becomes: \(\frac{5 x}{5}\) = \(\frac{60}{5}\) and the solution is x = 12.

Case 2. If the coefficient associated with the variable on the variable side of the equation is dividing it, then this coefficient is converted to 1 by multiplying both sides of the equation with the same constant, to isolate the
variable on 1 side of the equation. For example: \(\frac{x}{5}\) = 10 is solved by multiplying both sides of the equation with 5, therefore the intermediate step becomes: \(\frac{x}{5}\) × 5 = 10 × 5 and the solution is x = 50.

Texas Go Math Grade 6 Lesson 12.3 Independent Practice Answer Key

Write and solve an equation to answer each question.

Question 6.
Jorge baked cookies for his math class’s end-of-year party. There are 28 people in Jorge’s math class including Jorge and his teacher. Jorge baked enough cookies for everyone to get 3 cookies each. How many cookies did Jorge bake?
Answer:
Solution to this example is given below
\(\frac{x}{28}\) = 3 Notice that 28 is divided from x
\(\frac{x}{28}\) × 28 = 3 × 28 × 9 Multiply both sides of the equation by 28
x = 84 Simplify
Check; \(\frac{x}{28}\) = 3
\(\frac{84}{28}\) = 3 Substitute 84 for x
3 = 3 Divide on the left side.
x = 84 Final solution
Jorge baked 84 cookies

Question 7.
Sam divided a rectangle into 8 congruent rectangles that each have the area shown. What is the area of the rectangle before Sam divided it?

Answer:
Solution to this example is given below
\(\frac{x}{8}\) = 5 Notice that 8 is divided from x
\(\frac{x}{28}\) × 8 = 5 × 8 × 9 Multiply both sides of the equation by 8
x = 40 Simplify
Check; \(\frac{x}{8}\) = 5
\(\frac{40}{8}\) = 5 Substitute 40 for x
5 = 5 Divide on the Left side.
x = 40 Final solution
The area of the rectangle was 40 square centimeters.
x = 40

6th Grade Long Division Lesson 12.3 Go Math Question 8.
Carmen participated in a read-a-thon. Mr. Cole pledged $4.00 per book and gave Carmen $44. How many books did Carmen read?
Answer:
Examine each part of the equation.
x is the unknown value you want to find.
4 is multiplied by x.
44 means that after multiplying 4 and x, the result is 44.
Use the equation to solve the problem.
4x = 44
\(\frac{4 x}{4}\) = \(\frac{44}{4}\) Divide both sides by 4
x = 11 Final solution
Carmen read 11 books
x = 11

Question 9.
Lee drove 420 miles and used 15 gallons of gasoline. How many miles did Lee’s car travel per gallon of gasoline?
Answer:
Let the mpg of the car be x, then the equation of the total distance traveled is:
15x = 420
Step 2
Divide both sides of the equation with 15 to isolate the variable on 1 side of the equation:
\(\frac{15 x}{15}\) = \(\frac{420}{15}\)
Simplify to evaluate the variable:
x = 28
Lee’s car travelled 28 mites per gallon of gasoline.

Question 10.
On some days, Me’vin commutes 3.5 hours per day to the city for business meetings. Last week he commuted for a total of 14 hours. How many days did he commute to the city?
Answer:
Examine each part of the equation.
x is the unknown value you want to find.
3.5 is multiplied by x.
= 14 means that after multiplying 3.5 and x, the result is 14.
Use the equation to solve the problem.
3.5x = 14
\(\frac{3.5 x}{3.5}\) = \(\frac{14}{3.5}\) Divide both sides by 3.5
x = 4 Final solution
He commuted to the city on 4 days
x = 4

Question 11.
Dharmesh has a square garden with a perimeter of 132 feet. Is the area of the garden greater than 1,000 square feet?

Answer:
Perimeter of a square is the product of 4 and its side length s. Here the equation becomes:
4s = 132
Divide both sides of the equation with 4 to isolate the variable on 1 side of the equation:
\(\frac{4 s}{4}\) = \(\frac{132}{4}\)
Simplify to evaluate the variable:
s = 33
Area of a square is the square of its side length, therefore the equation becomes:
4 = 33² = 33 × 33
Evaluate:
A = 1089 > 1000
The area of the given square is greater than 1000 square feet.

Question 12.
Ingrid walked her dog and washed her car. The time she spent walking her dog was one-fourth the time it took her to wash her car. It took Ingrid 14 minutes to walk the dog. How long did it take Ingrid to wash her car?
Answer:
Let the total time spent in washing the car be x, then the equation for the time spent in walking the dog 5:
\(\frac{1}{4}\) × x = 14

Multiply both sides of the equation with 4 to isolate the variable on 1 side of the equation:
\(\frac{1}{4}\) × x × 4 = 14 × 4
Simplify to evaluate the variable:
x = 56
She took 56 minutes to wash her car.

Question 13.
Representing Real-World Problems Write and solve a problem involving money that can be solved with a multiplication equation.
Answer:
5 friends go out for a movie anti end up spending $50 on tickets. How much did each ticket cost?
Here the equation will he 5x = 50 and x = \(\frac{50}{5}\) = 10. Kach ticket costed $10.

Question 14.
Representing Real-World Problems Write and solve a problem involving money that can be solved with a division equation and has a solution of 1,350.
Answer:
Evaluate the total amount of funds raised a group of 5 friends if each one of them raised $270.
Let the total profit be x, then the equation for the profit of one person be is:
\(\frac{x}{5}\) = 270
Multiply both sides of the equation with 5 to isolate the variable on 1 side of the equation:
\(\frac{x}{5}\) × 5 = 5 × 270
Simplify to evaluate the variable:
x = 1350

Texas Go Math Grade 6 Lesson 12.3 H.O.T. Focus On Higher Order Thinking Answer Key

Question 15.
Communicate Mathematical Ideas Explain why 7 • \(\frac{x}{7}\) = x. How does this relate to solving division equations?
Answer:
If the given equation is multiplied, it will be \(\frac{7 x}{7}\), and in the simplified form will become x. It is related to solving division equations in order to isolate the variable to get its value on the other side of the equation.
It makes the variable isolated on one side of the equation.

Go Math 6th Grade Answer Key Multiplication and Division Worksheets Pdf Question 16.
Critical Thinking A number tripled and tripled again is 729. What is the number?
Answer:
The solution to this example ¡s given below
3(3x) = 729
3(3x) = 729 Notice that 3x is multiplied to x
9x = 729 Simplify
\(\frac{9 x}{9}\) = \(\frac{729}{9}\) Divide both sides of the equation by 9
x = 81 Simplify
Check; 3(3x) = 729
3(3 × 81) = 729 Substitute 81 for x
729 = 729 Multiply on the left side.
x = 81 Final solution
The number is 81.
x = 81

Question 17.
Multistep Andre has 4 times as many model cars as Peter, and Peter has one-third as many model cars as Jade. Andre has 36 model cars.
a. Write and solve an equation to find how many model cars Peter has.
Answer:
Let the number of Peter’s cars be x, then the equation for the number of Andre’s cars 5:
4x = 36
Divide both sides of the equation with 4 to isolate the variable on 1 side of the equation:
\(\frac{4 x}{4}\) = \(\frac{36}{4}\)
Simplify to evaluate the variable:
x = 9
Peter has 9 cars.

b. Using your answer from part a, write and solve an equation to find how many model cars Jade has.
Answer:
Peter has 9 cars, then the equation for the number of Jades cars y 5:
\(\frac{y}{3}\) = 9
Multiply both sides of the equation with 3 to isolate the variable on 1 side of the equation:
\(\frac{y}{3}\) × 3 = 9 × 3
Simplify to evaluate the variable:
y = 27
Jade has 27 cars.

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 Answer Key Equations and Relationships.

Texas Go Math Grade 6 Module 12 Answer Key Equations and Relationships

Texas Go Math Grade 6 Module 12 Are You Ready? Answer Key

Evaluate the expression.

Question 1.
4(5 + 6) – 15 ___________
Answer:
Solution to this example is given below
4(5 + 6) – 15
4(5 + 6) 15 Perform operations inside parentheses.
= 4(11) – 15 Multiply.
= 44 – 15 Subtract.
= 29

Question 2.
8(2 + 4) + 16 ___________
Answer:
Solution to this example is given below
8(2 + 4) + 16
8(2 + 4) + 16 Perform operations inside parentheses.
= 8(6) + 16 Multiply.
= 48 + 16 Add.
= 64

Texas Go Math Grade 6 Answer Key Pdf Module 12 Question 3.
3(14 – 7) – 16 ___________
Answer:
Given expression:
3(14 – 7) – 16
Simplify the parentheses:
= 3(7) – 16
Expand the parentheses:
= 21 – 16
Evaluate:
= 5
3(14 – 7) – 16 = 5

Question 4.
6(8 – 3) + 3(7 – 4) ___________
Answer:
Solution to this example is given below
6(8 – 3) + 3(7 – 4)
6(8 – 3) + 3(7 – 4) = 6(5) + 3(3) Perform operations inside parentheses
= 30 + 9 Multiply
= 39 Add.
= 39

Question 5.
10(6 – 5) – 3(9 – 6) ___________
Answer:
Solution to this example is given below
10(6 – 5) – 3(9 – 6)
10(6 – 5) – 3(9 – 6) = 10(1) – 3(3) Perform operations inside parentheses.
= 10 – 9 Multiply
= 1 Subtract.
= 1

Question 6.
7(4 + 5 + 2) – 6(3 + 5) ___________
Answer:
Solution to this example ¡s given below
7(4 + 5 + 2) – 6(3 + 5)
7(4 + 5 + 2) – 6(3 + 5) = 7(11) – 6(8) Perform operations inside parentheses.
= 77 – 48 Multiply.
= 29 Subtract.
= 29

Question 7.
2(8 + 3) + 4² ____________
Answer:
Given expression:
2(8 + 3) + 4²
Simplify the parentheses and power:
= 2(11) + 16
Expand the parentheses:
= 22 + 16
Evaluate:
= 38
2(8 + 3) + 4² = 38

Question 8.
7(14 – 8) – 6² ___________
Answer:
Given expression:
7(14 – 8) – 6²
Simplify the parentheses and power:
= 7(6) – 36
Expand the parentheses:
= 42 – 36
Evaluate:
= 6
7(14 – 8) – 6² = 6

Module 12 Answer Key Texas Go Math Grade 6 Question 9.
8(2 + 1)² – 4² ___________
Answer:
Given expression:
8(2 + 1)² – 4²
Simplify the parentheses and power:
= 8(3)² – 16
Simplify:
= 8(9) – 16
Expand the parentheses:
= 72 – 16
Evaluate:
= 56
8(2 + 1)² – 4² = 56

Write an algebraic equation for the word sentence.

Question 10.
A number increased by 7.9 is 8.3 ____________
Answer:
Let the number be x, then the expression for the given statement is: x + 7.9 = 8.3

Question 11.
17 is the sum of a number and 6. ____________
Answer:
The sum of a number and 6 ¡s 17.
The sum of x and 6 is 17 Represent the unknown with a variable.
x + 6 is 17 Determine the operation
x + 6 = 17 Determine the placement of the equal sign
x + 6 = 17 Final solution

Question 12.
The quotient of a number and 8 is 4. ____________
Answer:
Let the number be x, then the expression for the given statement is: \(\frac{x}{8}\) = 4

Question 13.
81 is three times a number. ____________
Answer:
The product of a number and 3 is 81
The product of x and 3 is 81 Represent the unknown with a variable
3 × x is 81 Determine the operation
3 × x = 81 Determine the placement of the equal sign.
3 × x = 81 Final solution

Go Math Grade 6 Module 12 Answer Key Question 14.
The difference between 31 and a number is 7. ____________
Answer:
The difference between 31 and a number is 7.
The difference between 31 and x is 7 Represent the unknown with a variable.
31 – x is 7 Determine the operation
31 – x = 7 Determine the placement of the equal sign.
31 – x = 7 Final solution

Question 15.
Eight less than the number is 19. ____________
Answer:
Let the number be x, then the expression for the given statement is: x – 8 = 19

Texas Go Math Grade 6 Module 12  Reading Start-Up Answer Key

Visualize Vocabulary

Use the ✓ words to complete the graphic.

Understand Vocabulary

Match the term on the left to the correct expression on the right.

Answer:
1 – C. Algebraic expression is a mathematical statement that includes one or more variables. It also consists of terms and constant.
2 – A. An equation is a mathematical sentence that two expressions are equal.
3 – B. A solution of the equation is a value of the variable that makes the equation true.

Active Reading
Booklet Before beginning the module, create a booklet to help you learn the concepts in this module. Write the main idea of each lesson on each page of the booklet. As you study each lesson, write important details that support the main idea, such as vocabulary and formulas. Refer to your finished booklet as you work on assignments and study for tests.

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 12.1 Answer Key Writing Equations to Represent Situations.

Texas Go Math Grade 6 Lesson 12.1 Answer Key Writing Equations to Represent Situations

Your Turn

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

Question 1.
11 = n + 6; n = 5
Answer:
Given equation:
11 = n + 6
Substitute the value of n in the given equation:
11 = 5 + 6
Simplify:
11 = 11
Since the left and right hand side of the equation are equal, this implies that the given value of the variable is the solution of the given equation.
n = 5 is the solution of the given equation.

Go Math Grade 6 Lesson 12.1 Answer Key Question 2.
y – 6 = 24; y = 18
Answer:
Given equation:
y – 6 = 24
Substitute the value of y in the given equation:
18 – 6 = 24
Simplify:
12 = 24
But:
12 ≠ 24
Since the left and right hand side of the equation are not equal this implies that the given value of the variable is not the solution of the given equation.
y = 18 is not the solution of the given equation.

Question 3.
\(\frac{36}{x}\) = 9; x = 4
Answer:
Substitute the value to evaluate the equation.
\(\frac{36}{4}\) = 9 substitute for the value of x
9 = 9 the value of x is a solution to the equation
The given value of 4 is a solution to the equation.

Write an equation to represent each situation.

Question 4.
Marilyn has a fish tank that contains 38 fish. There are 9 goldfish and f other fish.
Answer:
Solution to this example is given below

38 = 9 + f

Question 5.
Juanita has 102 beads to make n necklaces. Each necklace will have 17 beads.
Answer:
There are a total of 102 beads to make n necklaces and each must contain 17 beads, so the equation representing this situation becomes: \(\frac{102}{n}\) = 17.

Go Math Grade 6 Answer Key Pdf Lesson 12.1 Question 6.
Craig is c years old. His 12-year-old sister Becky is 3 years younger than Craig.
Answer:
Craig is 3 years older than his 12-year-old sister, so his age is c = 12 + 3.

Question 7.
Sonia rented ice skates for h hours. The rental fee was $2 per hour and she paid a total of $8.
Answer:
Solution to this example is given below

2 × h = 8

Example

Sarah used a gift card to buy $47 worth of groceries. Now she has $18 left on her gift card. Write an equation to determine whether Sarah had $65 or $59 on the gift card before buying groceries.

Reflect

Question 8.
What expressions are represented in the equation x – 47 = 18? How does the relationship represented in the equation help you determine if the equation is true?
Answer:
The expression represented in the equation is 47 less than the amount of the card. The relationship represented in the equation became true when the value of the card was substituted. Based from the substitution done, 65 is a solution to the equation and not 59.

Your Turn

Question 9.
On Saturday morning, Owen earned $24 raking leaves. By the end of the afternoon he had earned a total of $62. Write an equation to determine whether Owen earned $38 or $31 on Saturday afternoon.
Answer:
Let the unknown amount earned be x, then the equation of total earning becomes:
x + 24 = 62
Solve for x:
x = 62 – 24 = 38
$38 were earned on Saturday afternoon.

Texas Go Math Grade 6 Lesson 12.1 Guided Practice Answer Key

Determine whether the given value is a solution of the equation. (Example 1)

Question 1.
23 = x – 9; x = 14 __________

Answer:
Solution to this example is given below
23 = x – 9; x = 14
23 = 14 – 9 Substitute 14 for x
23 = 5 Subtract.
14 is a not of solution of the equation 23 = x – 9
x ≠ 14

Go Math Grade 6 Lesson 12.1 Independent Practice Answer Key Question 2.
\(\frac{n}{13}\) = 4; n = 52 ___________

Answer:
Solution to this example is given below
\(\frac{n}{13}\) = 4; n = 52
\(\frac{52}{13}\) = 4 Substitute 52 for n
4 = 4 Divide
52 is a solution of \(\frac{n}{13}\) = 4
n = 52

Question 3.
14 + x = 46; x = 32 ___________
Answer:
Solution to this example is given below
14 + x = 46; x = 32
14 + 32 = 46 Substitute 32 for x
46 = 46 Add
32 is a solution of 14 + x = 46
x = 32

Question 4.
17y = 85; y = 5 ___________
Answer:
Solution to this example is given below
17y = 85; y = 5
17(5) = 85 Substitute 5 for y
85 = 85 Multiply.
5 is a solution of 17y = 85
y = 5

Texas Go Math Grade 6 Lesson 12.1 Practice Geometry Answers Question 5.
25 = \(\frac{k}{5}\); k = 5 ___________
Answer:
Given equation:
25 = \(\frac{k}{5}\)
Substitute the given value of k in the given equation:
25 = \(\frac{5}{5}\)
Simplify:
25 = 1
Compare:
25 ≠ 1
Since the left and right hand side of the equation are not equal, this implies that the given value of the variable is not a solution of the given equation.
k = 5 is not a solution of the given equation.

Question 6.
2.5n = 45; n = 18 ____________
Answer:
Given equation:
2.5n = 45
Substitute the given value of n in the given equation:
2.5(18) = 45
Simplify:
45 = 45
Since the left and right hand side of the equation are equal, this implies that the given value of the variable is a solution of the given equation.
n = 18 is a solution of the given equation.

Question 7.
21 = m + 9; m = 11 ___________
Answer:
Solution to this example is given below
21 = m + 9; m = 11
21 = 11 + 9 Substitute 11 for m
21 = 20 Add
11 is a not of solution of the equation 21 = m + 9
m ≠ 11

Question 8.
21 – h = 15; h = 6 ____________
Answer:
Solution to this example is given below
21 – h = 15; h = 6
21 – 6 = 15 Substitute 6 for h
15 = 15 Subtract.
6 is a solution of 21 – h = 15
h = 6

Question 9.
d – 4 = 19; d = 15 _____________
Answer:
Given equation:
d – 4 = 19
Substitute the given value of d in the given equation:
15 – 4 = 19
Simplify:
11 = 19
Compare:
11 ≠ 19
Since the left and right hand side of the equation are not equal, this implies that the given value of the variable is not a solution of the given equation.
d = 15 is not a solution of the given equation.

6th Grade Math Problems with Answers Lesson 12.1 Question 10.
5 + x = 47; x = 52 ____________
Answer:
Solution to this example is given below
5 + x = 47; x = 52
5 + 52 = 47 Substitute 52 for x
57 = 47 Add
52 is a not of solution of the equation 5 + x = 47
x ≠ 52

Question 11.
w- 9 = 0; w = 9 ____________
Answer:
Solution to this example is given below
w – 9 = 0; w = 9
9 – 9 = 0 Substitute 9 for w
0 = 0 Subtract.
9 is a solution of w – 9 = 0
w = 9

Question 12.
5q = 31; q = 13 _____________
Answer:
Given equation:
5q = 31
Substitute the given value of q in the given equation:
5(13) = 31
Simplify:
65 = 31
Compare:
65 ≠ 31
Since the left and right hand side of the equation are not equal, this implies that the given value of the variable is not a solution of the given equation.
q = 13 is not a solution of the given equation.

Question 13.
Each floor of a hotel has r rooms. On 8 floors, there are a total of 256 rooms. Write an equation to represent this situation. (Example 2)

Answer:
Solution to this example is given below

8 × r = 256

6th Grade Math Answer Key Hands on Equations Lesson 12.1 Question 14.
In the school band, there are 5 trumpet players and five flute players. There are twice as many flute players as there are trumpet players. Write an equation to represent this situation. (Example 3)
Answer:
There are twice as many flute players as there are trumpet players, so f = 2 × 5

Question 15.
Pedro bought 8 tickets to a basketball game. He paid a total of $208. Write an equation to determine whether each ticket costs $26 or $28. (Example 3)
Answer:
Let the cost of 1 ticket = be x then the equation of total cost is:
8 × x = 208
Solve for x:
x = \(\frac{208}{8}\)
x = 26
Each ticket costs $26.

Question 16.
The high temperature was 92°F. This was 24°F higher than the overnight low temperature. Write an equation to determine whether the low temperature was 62°F or 68°F. (Example 3)
Answer:
Let the low temperature be x then the equation for high temperature is:
x + 24 = 92
Solve for x:
x = 92 – 24
Evaluate:
x = 68
The low temperature was 68°F.

Essential Question Check-In

Question 17.
Tell how you can determine whether a number is a solution of an equation.
Answer:
A number is checked if it is a solution of an equation or not by substituting its value in the equation and simplifying it. If the left and right hand side of the equation are equal then the number is a solution of the equation otherwise not.

Texas Go Math Grade 6 Lesson 12.1 Independent Practice Answer Key

Question 18.
Andy is one-fourth as old as his grandfather, who is 76 years old. Write an equation to determine whether Andy is 19 or 22 years old.
Answer:
Let Andy’s age be x, then the equation becomes:
x = 76 × \(\frac{1}{2}\)
Solve for x:
x = 19
Andy is 19 years old.

Question 19.
A sleeping bag weighs 8 pounds. Your backpack and sleeping bag together weigh 31 pounds. Write an equation to determine whether the backpack without the sleeping bag weighs 25 or 23 pounds.
Answer:
Let the weight of the backpack be x, then the equation becomes:
x + 8 = 31
Solve for x:
x =31 – 8
Evaluate:
x = 23
The sleeping bag weighs 23 pounds.

Expressions and Equations 6th Grade Answer Key Question 20.
Halfway through a bus route, 23 students have been dropped off and 48 students remain on the bus. Write an equation to determine whether there are 61 or 71 students on the bus at the beginning of the route.
Answer:
Let the total number of students on the bus be x, then the equation becomes:
x = 23 + 48
Solve for x:
x = 71
There are 71 students on the bus at the beginning of the route.

Question 21.
The table shows the distance between Greenville and nearby towns. The distance between Artaville and Greenville is 13 miles less than the distance between Greenville and Jonesborough.

a. Write two equations that state the relationship of the distances between Greenville, Artaville, and Jonesborough.
Answer:
Equations
x = 29 – 13 and 29 = 13 + x
x is the distance between Greenville and Artaville
29 is the distance between Greenville and Jonesborough
13 Less than the distance between Greenville and Jonesborough

b. Describe what your variable represents.
Answer:
The variable x represents the distance between Greenville and Artaville.

Question 22.
Write an equation that involves multiplication, contains a variable, and has a solution of 5. Can you write another equation that has the same solution and includes the same variable and numbers but uses division? If not, explain. If possible, write the equation.
Answer:
Possible equations are:
15x = 75 and x = \(\frac{75}{15}\) which both have a solution of 5

Question 23.
How are expressions and equations different? Explain using a numerical example.
Answer:
An expression represents a single value can be evaluated ¡f there ¡s an indicated value of the variable while an equation shows an equality of two expressions which may be true for some values.
Expression: 29 – 3x or 7x + 8y
Equation: 29 – 3x = 35 or 7x + 8 = 1
The expression indicates a single value while the equation shows equality between expressions.

Question 24.
Explain the Error The problem states that Ursula earns $9 per hour. To write an expression that tells how much money Ursula earns for h hours, Joshua wrote \(\frac{9}{h}\). Sarah wrote 9h. Whose expression is correct and why?
Answer:
Here the rate is given and the number of hours are given, so the total earning is a product of these 2, that is $9 × h = $9h. so Sarah is correct.

6th Grade Math Equations and Answers Lesson 12.1 Question 25.
Communicate Mathematical Ideas A dog weighs 44 pounds and the veterinarian thinks it needs to lose 7 pounds. Mikala wrote the equation x + 7 = 44 to represent the situation. Kirk wrote the equation 44 – x = 7. Which equation is correct? Can you write another equation that represents the situation?
Answer:
Kirk’s equation 44 – x = 7 is correct because the final weight x of the dog must be less than 44 pounds by 7 pounds.
This situation can also be represented by the equation: x = 44 – 7.
Kirk is correct.

Question 26.
Multiple Representations The table shows the ages of Cindy and her dad.

a. Write an equation that relates Cindy’s age to her dad’s age when Cindy is 18.
Answer:
Equation
y = x + 26
y is the age of the dad of Cindy
x is the age of Cindy
26 is the difference in their ages

b. Determine if 42 is a solution to the equation. Show your work.
Answer:
y = x + 26 substitute for the value of x and y
42 = 18 + 26 add the numbers
42 ≠ 44
42 is not a solution to the equation

c. Explain the meaning of your answer in part b.
Answer:
Based from the previous solution, it shows that 42 is not a solution to the equation It indicates that the dad of Cindy is not 42 years old when Cindy is 18 years old.
The age of the dad of Cindy is not 42 when she is 18 years old. Her dad is 44 years old.

H.O.T. Focus on Higher Order Thinking

Question 27.
Critical Thinking In the school band, there are 4 trumpet players and f flute players. The total number of trumpet and flute players is 12. Are there twice as many flute players as trumpet players? Explain.
Answer:
Form the equation for total number of band members. Therefore: f + 4 = 12. Solve this equation for f, therefore: f = 12 – 4 = 8.
This implies that there are 8 flute players in the band. It can also be seen that 8 = 4 × 2 so it can be said that yes, there are twice as many flute players as trumpet players.

Question 28.
Problem Solving Ronald paid $162 for 6 tickets to a basketball game. During the game he noticed that his friend paid $ 130 for 5 tickets. The price of each ticket was $26. Was Ronald overcharged? Justify your answer.
Answer:
If each tickets costs $26. then the cost of 6 tickets should he 6 × $26 = $156 < $162. Therefore, it can be said that Ronald was overcharged by $162 – $156 = $6.

Question 29.
Communicate Mathematical Ideas Tariq said you can write an equation by setting an expression equal to itself. Would an equation like this be true? Explain.
Answer:
If an expression is equated with itself, then all, terms of the equation will cancel out each other and the equation will reduce to 0 = 0, which does not have any numerical significance. For example: x + 2 = x + 2, when solving for x, the equation is rearranged: x – x = 2 – 2 which reduces to 0 = 0.

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 13 Answer Key Inequalities and Relationships.

Texas Go Math Grade 6 Module 13 Answer Key Inequalities and Relationships

Texas Go Math Grade 6 Module 13 Are You Ready? Answer Key

Write an integer to represent each situation.

Question 1.
a loss of $75
Answer:
Because here we have loss, we will have -, so, the required integer is 75.

Question 2.
a football player’s gain of 9 yards
Answer:
Here, because we have gain, we will have +. so, the required integer is 9.

Question 3.
spending $1,200 on a flat-screen TV __________
Answer:
Because here we have spent, we will have -, so the required integer is -1.200.

Question 4.
a climb of 2,400 feet _________
Answer:
Because here we have climbed, we will have +, so the required integer is 2.400.

Find the product or quotient.

Question 5.
6 × 9 __________
Answer:
Here, the signs of the integers are the same, so, the product will be positive. We have that result is 54.

Question 6.
15 ÷ (-5) ________
Answer:
Here, the signs of integers are different, so the quotient will be negative.
We have that result is -3.

Go Math Grade 6 Module 13 Answer Key Question 7.
-8 × 6 _______
Answer:
Here, the signs of integers are different, so the product will be negative.
So, the result is -48.

Question 8.
-100 ÷ (10) _________
Answer:
Here, the signs of integers are different, so the quotient will be negative.
We have that result is -10.

Question 9.
3 × (-7) ________
Answer:
Here, the signs of integers are different, so the product will be negative.
So, the result is -21.

Question 10.
-64 ÷ 8 ________
Answer:
Here, the signs of integers are different, so the quotient will be negative.
We have that result is -8.

Question 11.
-8 × (-2) __________
Answer:
Here, the signs of the integers are the same, so, the product will be positive We have that result is 16.

Question 12.
32 ÷ 2 ___________
Answer:
Here, the signs of the integers are the same, so, the quotient will be positive. We have that result is 16.

Solve.

Question 13.
9p = 108 __________
Answer:
Here we have that p is multiplied by 9, so, we will multiply both sides by the reciprocal, \(\frac{1}{9}\), in order to isolate variable. After that we will simplify and get the following:
\(\frac{1}{9}\) ∙ 9p = 108 ∙ \(\frac{1}{9}\)
p = \(\frac{108 \cdot 1}{9}\)
p = 12

Question 14.
\(\frac{3}{5}\)n = 21 _________
Answer:
3 5
Here we have that n is multiplied by \(\frac{3}{5}\), so, we will multiply both sides by the reciprocal, \(\frac{3}{5}\), in order to isolate the variable.
After that we will simplify and get the following:
\(\frac{5}{3}\) ∙ \(\frac{3}{5}\)n = 21 ∙ \(\frac{5}{3}\)
n = \(\frac{21 \cdot 5}{3}\)
n = 35

Question 15.
\(\frac{4}{7}\)k = 84 __________
Answer:
\(\frac{7}{4}\) ∙ \(\frac{4}{7}\)k = 84 ∙ \(\frac{7}{4}\)
k = \(\frac{84 \cdot 7}{4}\)
k = 147

Module 13 Go Math Grade 6 Answer Key Question 16.
\(\frac{3}{20}\) e = 24 ___________
Answer:
\(\frac{20}{3}\) ∙ \(\frac{3}{20}\)e = 24 ∙ \(\frac{20}{3}\)
e = \(\frac{24 \cdot 20}{3}\)
e = 160

Texas Go Math Grade 6 Module 13 Reading Start-Up Answer Key

Visualize Vocabulary

Use the ✓ words to complete the graphic.

Understand Vocabulary

Match the term on the left to the correct expression on the right.

Answer:
1 – A. Solution of an inequality is a value or values that make the inequality true. ConcLusion ¡s that A matches to this term.
2 – C. Coefficient is the number that is multiplied by the variable in an algebraic expression. Conclusion is that C matches to this term.
3 – B. Constant is a specific number whose value does not change. Conclusion is that B matches this term.

Active Reading
Two-Panel Flip Chart Create a two-panel flip chart to help you understand the concepts in this module. Label one flap “Adding and Subtracting Inequalities.” Label the other flap “Multiplying and Dividing InequalitiesTM As you study each lesson, write important ideas under the appropriate flap.

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 11 Quiz Answer Key.

Texas Go Math Grade 6 Module 11 Quiz Answer Key

Texas Go Math Grade 6 Module 11 Ready to Go On? Answer Key

11.1 Modeling Equivalent Expressions

Write each phrase as an algebraic expression.

Question 1.
p divided by 6 _________________
Answer:
Solution to this example is given below
p divided by 6. The operation is division. (Final solution)
The algebraic expression is \(\frac{p}{6}\)
\(\frac{p}{6}\)

Question 2.
65 less than j _________
Answer:
Solution to this example is given below
65 less than j. The operation is subtraction. (Final solution)
The algebraic expression is j – 65
j – 65

Module Quiz Ready To Go On Answers 6th Grade Question 3.
the sum of 185 and h _________
Answer:
The algebraic expression for the given statement: the sum of 185 and h is h + 185.

Question 4.
the product of 16 and g _____
Answer:
The algebraic expression for the given statement: the product of 16 and g is 16 × g = 16g,
16 × g = 16g

Question 5.
Let x represent the number of television show episodes that are taped in a season. Write an expression for the number of episodes taped in 4 seasons. _____________________
Answer:
The number of television show episodes that are taped in a season are x so in 4 seasons, there will be x × 4 = 4x episodes. There will be 4x episodes in 4 seasons.

11.2 Evaluating Expressions

Evaluate each expression for the given value of the variable.

Question 6.
8p; p = 9 _________________
Answer:
Solution to this example is given below
8p; p = 9
8(9) Substitute 9 for p
72 Multiply
When p = 9, 8p = 72
72 Final solution

Module 11 Quiz Answers Go Math Expressions Grade 6 Answer Key Question 7.
11 + r; r = 7 _____
Answer:
Solution to this example is given below
11 + r; r = 7
11 + 7 Substitute 7 for r
18 Add
When r = 7, 11 + r = 18
18 Final solution

Question 8.
4(d + 7); d = -2 ___________
Answer:
Simplify the expression
= 4 (-2 + 7) substitute for the value of d
= 4(5) perform the operation inside the parenthesis then multiply
= 20
The value of the expression is 20.

Question 9.
\(\frac{-60}{m}\); m = 5 __________
Answer:
Solution to this example is given below
\(\frac{-60}{m}\); m = 5
\(\frac{-60}{5}\) Substitute 5 for m
-12 Divide
When m = 5, \(\frac{-60}{m}\) = -12
12 Final solution

Grade 6 Math Test with Answers Module 11 Quiz Question 10.
To find the area of a triangle, you can use the expression b × h ÷ 2, where b is the base of the triangle and h is its height. What is the area of a triangle with a base of 6 and a height of 8? _______________________
Answer:
Solution to this example is given below
b × h ÷ 2; b = 6, h = 8
6 × 8 ÷ 2 Substitute 6 for b, and 8 for h
48 ÷ 2 Multiply
24 Divide
When b = 6, h = 8 b × h ÷ 2 = 24
Area of the triangle is 24 square units
24 Final solution

11.3 Generating Equivalent Expressions

Question 11.
Draw lines to match the expressions in List A with their equivalent expressions in List B.

Answer:
a. of List B: Given expression:
7(1 + x)
Apply distributive property of multiplication to expand the parentheses:
= 7 + 7x
This is equal to b of List A.

c. of List B: Given expression:
7(x + 2)
Apply distributive property of multiplication to expand the parentheses:
= 7x + 14
This is equal to a of List A.

c. of List A: Given expression:
7(x – 1)
Apply distributive property of multiplication to expand the parentheses:
= 7x – 7
This is equal to b of List B

Essential Question

Question 12.
How can you determine if two algebraic expressions are equivalent?
Answer:
Problems involving equivalent expressions are solved using the rules of algebra. If 2 expressions are equivalent then they are identical in their simplified form.

Texas Go Math Grade 6 Module 11 Mixed Review Texas Test Prep Answer Key

Selected Response

Question 1.
Which expression represents the product of 83 and x?
(A) 83 + x
(B) 83 ÷ x
(C) 83x
(C) 83 – x
Answer:
(C) 83x

Explanation:
Product implies multiplication so the product of 83 and x is equal to 83 × x = 83x
Option C.

Module 11 Quiz for Grade 6 with Answers Question 2.
Which phrase describes the algebraic expression \(\frac{r}{9}\)?
(A) the product of r and 9
(B) the quotient of r and 9
(C) 9 less than r
(D) r more than 9
Answer:
(B) the quotient of r and 9

Explanation:
\(\frac{r}{9}\) is a fraction which implies that r is divided by 9 or the quotient of r and 9
Option B

Question 3.
Rhonda was organizing photos in a photo album. She took 60 photos and divided them evenly among p pages. Which algebraic expression represents the number of photos on each page?
(A) P – 60
(B) 60 – p
(C) \(\frac{p}{60}\)
(D) \(\frac{60}{p}\)
Answer:
(D) \(\frac{60}{p}\)

Explanation:
A total of 60 photos are to be divided equal on p pages imply a problem of division where each page has \(\frac{60}{p}\) photos.
Option D

Question 4.
Using the algebraic expression 4n + 6, what is the greatest whole-number value of n that will give you a result less than 100?
(A) 22
(B) 23
(C) 24
(D) 25
Answer:
(B) 23

Explanation:
Given expression:
4n + 6
According to the given condition, the inequality becomes:
4n + 6 < 100
Solve for n:
4n < 100 – 6
Or
n < \(\frac{94}{4}\)
Simplify:
n < 23.5
Therefore, for n = 23, the value of the expression will be the greatest and less than 100.
Option B

Question 5.
Evaluate 7w – 14 for w = 9.
(A) 2
(B) 18
(C) 49
(D) 77
Answer:
(C) 49

Explanation:
Given expression: 7w – 14
Substitute the value of w:
7(9) – 14
Expand:
= 63 – 14
Evaluate:
= 49
Option (C)

Question 6.
Katie has read 32% of a book. If she has read 80 pages, how many more pages does Katie have left to read?
(A) 40
(B) 170
(C) 200
(D) 250
Answer:
(B) 170

Explanation:
Data:
Portion = 80
Total = x
Percent = 32

Write equation of percentage:

There are 250 pages in the book, 80 of them are read so 250 – 80 = 170 remain to be read.
Option B.

Grade 6 Math Answer Key Module 11 Test Answers Question 7.
The expression 12(x + 4) represents the total cost of CDs Mel bought in April and May at $12 each. Which property is applied to write the equivalent expression 12x + 48?
(A) Associative Property of Addition
(B) Associative Property of Multiplication
(C) Commutative Property of Multiplication
(D) Distributive Property
Answer:
(D) Distributive Property

Explanation:
The operation in the expression is Multiplication
You can use the Distributive Property of Multiplication to write an equivalent expression: 12(x + 4) = 12x + 48 (This option is the correct answer)
D

Gridded Response

Question 8.
When traveling in Europe, Bailey converts the temperature given in degrees Celsius to a Fahrenheit temperature by using the expression 9x ÷ 5 + 32, where x is the Celsius temperature. Find the temperature in degrees Fahrenheit when it is 15 °C.

Answer:
Determine the temperature in degrees Fahrenheit
= 9(15) ÷ 5 + 32 substitute for the value of x then multiply
= 135 ÷ 5 + 32 divide 135 by 5
= 27 + 32 add the numbers
= 59 temperature in degrees Fahrenheit
The answer in the grid must be 59.00.

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 13 Quiz Answer Key.

Texas Go Math Grade 6 Module 13 Quiz Answer Key

Texas Go Math Grade 6 Module 13 Ready to Go On? Answer Key

13.1 Writing Inequalities

Write an inequality to represent each situation, then graph the solutions.

Question 1.
There are fewer than 8 gallons of gas in the tank. __________________

Answer:
Draw a solid circle at 8 to show that 8 is not a solution.
Shade the number line to the left of 8 to show that numbers less than 8 are solutions.
Check your answer.
Substitute 7 for g.
7 < 8; 7 is less than 8, so 7 is a solution
Graph the solution on a number line.

Texas Go Math Grade 6 Module 13 Test Math Answers Question 2.
There are at least 3 pieces of gum left in the pack. __________________

Answer:
Because here we have “at least 3″, the required inequality would be graphed on the following picture.

The inequality that describes the given situation would be x ≥ 3.

Question 3.
The valley was at least 4 feet below sea level. ______________________

Answer:
Because here we have “at least 4″, the required inequality would be graphed on the following picture.

The inequality which describes given situation would be x ≥ 4.

13.2 Addition and Subtraction Inequalities

Solve each inequality. Graph the solution.

Question 4.
c – 28 > -32

Answer:
We will add 28 to both sides in order to solve this inequality and get:
c > -4
Now, we will graph the solution:

Now we will check our solution by substituting some value from the set of solutions we got in original inequation. We will substitute 0 for c and get:
0 – 28 > -32
-28 > -32
So, the inequality is true.
c <> -4

Texas Go Math Grade 6 Module 13 Quiz Answers Question 5.
v + 17 ≤ 20.

Answer:
In order to solve given inequality, we need to subtract 17 from both sides and get:
v ≤ 3
Now, we will graph the solution:

Question 6.
Today’s high temperature of 80°F is at least 16° warmer than yesterday’s high temperature. What was yesterday’s high temperature?
Answer:
Let x represent yesterday’s high temperature. According to all informations, we get the following inequality:
80 – x ≥ 16
We need to subtract 80 from both sides and get:
-x ≥ -64
Next step is to multiply both sides by -1. -1 is negative number, so, we need to reverse the inequality symbol for the statement to still be true.
(-x) (-1) ≤ (-64) (-1)
x ≤ 64
So, yesterday’s high temperature was at most 64°F.

13.3, 13.4 Multiplication and Division inequalities

Solve each inequality. Graph the solution.

Question 7.
7f ≤ 35

Answer:
First step is to divide both sides by 7.
\(\frac{7f}{7}\) ≤ \(\frac{35}{7}\)
f ≤ 5
Now, we will graph the solution:

Now we will check the solution by substituting a soLution from the shaded part of the graph into the original inequality.
We will substitute 2 for f and get:
7 ∙ 2 ≤ 35
14 ≤ 35
So, the inequality is true.
f ≤ 5

Texas Go Math Grade 6 Inequality Quiz Answer Key Question 8.
\(\frac{a}{2}\) < 4

Answer:
First step is to multiply by 2.
2(\(\frac{a}{2}\)) < 4(2)
a < 8
Now, we will graph the solution:

Now, we will check the solution by substituting a solution from the shaded part of the graph into the original inequality. We will substitute 4 for a and get:
\(\frac{4}{2}\) < 4
2 < 4
So, the inequality is true.
a < 8

Question 9.
-25g ≥ 150

Answer:
First step is to divide both sides by -25. -25 is negative number, so we need to reverse the inequality symbol for the statement to still be true.
\(\frac{-25g}{-25}\) ≤ \(\frac{150}{-25}\)
g ≤ -6
Now, we will graph the solution:

Now we will check the solution by substituting a solution from the shaded part of the graph into the original inequality.
We will substitute -8 for g and get:
-25(-8) ≥ 150
200 ≥ 150
So, the inequality is true.
g ≤ -6

Question 10.
\(\frac{k}{-3}\) < 3

Answer:
First step is to multiply by -3. -3 is negative number, so we need to reverse the inequality symbol for the statement to still be true.
\(\frac{k}{-3}\)(-3) > 3(-3)
k > -9
Now, we will graph the solution:

Now we will check the solution by substituting a solution from the shaded part of the graph into the original inequality.
So, we will substitute 0 for k and get:
\(\frac{0}{-3}\) < 3
0 < 3 So, the inequality is true. k > -9

Texas Go Math Grade 6 Module 13 Mixed Review Texas Test Prep Answer Key

Selected Response

Question 1.
Em saves at least 20% of what she earns each week. If she earns $140 each week for 4 weeks, which inequality describes the total amount she saves?
(A) t > 112
(B) t ≥ 112
(C) t < 28 (D) t ≤ 28
Answer:
(D) t ≤ 28

Explanation:
First we will calculate how much is 20% of 140. So we will multiply 140 by \(\frac{20}{100}\) = \(\frac{1}{5}\).
140 x \(\frac{1}{5}\) = \(\frac{140 \times 1}{1 \times 5}\) = 28
So, if t represent Em’s savings, the inequality which describes given situation is: t ≤ 28

Question 2.
Which number line represents the inequality r > 6?

Answer:
(B)

Explanation:
This inequality represents numbers on the right side of 6 and because there is symbol >, the circle at 6 is not full. So, conclusion is that number line which represents the inequality r > 6 is B.

Question 3.
For which inequality below is z = 3 a solution?
(A) z + 5 ≥ 9
(B) z + 5 > 9
(C) z + 5 ≤ 8
(D) z + 5 < 8
Answer:
(C) z + 5 ≤ 8

Explanation:
Solution of inequality at C is the following:
z ≤ 3
This solution includes 3, so, for this inequality, z = 3 is a solution.

Question 4.
What is the solution to the inequality -6x < -18?
(A) x > 3
(B) x < 3
(C) x ≥ 3
(D) x ≤ 3
Answer:
(A) x > 3

Explanation:
In order to solve this inequality, we need to divide both sides by -6. -6 is negative number, so, we need to reverse the inequality symbol for the statement to still be true.
\(\frac{-6x}{-6}\) > \(\frac{-18}{-6}\)
x > 3

Question 5.
The number line below represents the solution to which inequality?

(A) \(\frac{m}{4}\) > 2.2
(B) 2m < 17.6 (C) \(\frac{m}{3}\) > 2.5
(D) 5m > 40
Answer:
(A) \(\frac{m}{4}\) > 2.2

Explanation:
On the given number line there are numbers right from 8.8 and 9 is not included. That is actually m > 8.8.
We can notice that this is the solution of inequality at A.
4(\(\frac{m}{4}\)) > (2.2) (4)
m > 8.8

Module 13 Test Answers Texas Go Math Grade 6 Question 6.
Which number line shows the solution to w – 2 ≤ 8?

Answer:
(B)

Explanation:
In order to solve given inequality, we need to add 2 to both sides:
w ≤ 10
Solution we got appropriate to B.

Gridded Response

Question 7.
Hank needs to save at least $150 to ride the bus to his grandparent’s home, If he saves $12 a week, what is the least number of weeks he needs to save?

Answer:
Let x represent the number of weeks which are needed to Hank to save required money. Using all information in this task, we got the following inequality:
12x ≥ 150
In order to soLve it, we need to divide both sides by 12.
\(\frac{12x}{12}\) ≥ \(\frac{150}{12}\)
x ≥ 12.5
So, Hank will need at least 12.5 weeks to save required money.

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 11.1 Answer Key Modeling Equivalent Expressions.

Texas Go Math Grade 6 Lesson 11.1 Answer Key Modeling Equivalent Expressions

Essential Question
How can you write algebraic expressions and use models to decide if expressions are equivalent?

Texas Go Math Grade 6 Lesson 11.1 Explore Activity Answer Key

Explore Activity
Modeling Equivalent Expressions
Equivalent expressions are expressions that have the same value.

The scale shown to the right is balanced.
A. Write an expression to represent the circles on the left side of the balance. ______
B. The value of the expression on the left side is ____.
C. Write an expression to represent the circles on the right side of the balance. ____
D. The value of the expression on the right side is ____.
E. Since the expressions have the same value, the expressions are ____.
F. What will happen if you remove a circle from the right side of the balance?
___________
G. If you add a circle to the left side of the balance, what can you do to the right side to keep the scale in balance?

Reflect

Question 1.
What If? Suppose there were 2 + 5 circles on the right side of the balance and 3 on the left side of the balance. What can you do to balance the scale? Explain how the scale models equivalent expressions.
Answer:
left side = right side to make the scale balance
3 + n = 2 – 5 subtract 3 to both sides of the equation
3 + n – 3 = (2 + 5) – 3 perform the operation inside the parenthesis
n = 7 – 3 subtract the numbers
n = 4 number of circles to be added on the left side
4 circles must be added on the left side to balance the scale.

Your Turn

Write each phrase as an algebraic expression.

Question 2.
n times 7 ____
Answer:
The algebraic expression for the phrase is in.
The expression is 7n.

Go Math Answer Key Grade 6 Lesson 11.1 Question 3.
4 minus y ____
Answer:
The solution to this example is given below
The difference of 4 and y The operation is subtraction.
The algebraic expression is 4 – y (Final solution)
4 – y

Question 4.
13 added to x ____
Answer:
Solution to this example is given below
The sum of 13 and x. The operation is addition. (Final solution)
The algebraic expression is 13 + x
13 + x

Write a phrase for each expression.

Question 5.
\(\frac{x}{12}\) ______
Answer:
Solution to this example is given below
\(\frac{x}{12}\) The operation is Division.
The quotient of x and 12 (Final solution)

Question 6.
10y ____
Answer:
Solution to this example is given below
10y The operation is Multiplication. (Final solution)
The product of 10 and y

Your Turn

Draw a bar model to represent each expression.

Question 7.
t – 2

Answer:
Bar model of the expression: t – 2:

Lesson 11.1 Answer Key Equivalent Expressions Question 8.
4y

Answer:
Bar model of the expression: 4y:

Your Turn

Question 9.
On a math quiz, Tina scored 3 points more than Julia. Juan scored 2 points more than Julia and earned 2 points in extra credit. Write an expression and draw a bar model to represent Tina’s score and Juan’s score. Did Tina and Juan make the same grade on the quiz? Explain.

Answer:
Let Julia’s score be x, then Tina’s score is x + 3 and Juan’s score is x + 2 + 2 = x + 4. It can be seen that Juan’s score is not same as Tin a’s but more by 1 point.

Texas Go Math Grade 6 Lesson 11.1 Guided Practice Answer Key

Question 1.
Write an expression in the right side of the scale that will keep the scale balanced. (Explore Activity)

Answer:
The following expressions will make the scale balance:
a. 5 + 5
b 1 + 9
c. 2 + 8
d. 4 + 6

Expressions to balance the scale: 5 + 5, 1 + 9, 2 + 8, 4 + 6

Write each phrase as an algebraic expression. (Example 1)

Question 2.
3 less than y _____________________
Answer:
Solution to this example is given below
The difference of y and 3. The operation is subtraction. (Final solution)
The algebraic expression is y – 3
y – 3

Question 3.
The product of 2 and p ___________
Answer:
Solution to this example is given below
The product of 2 and The operation is multiplication (Final solution)
The algebraic expression is 2 × p

Write a phrase for each algebraic expression. (Example 1)

Question 4.
y + 12 ______
Answer:
Solution to this example is given below
y + 12 The operation is Addition. (Final solution)
The sum of y and 12

Go Math Grade 6 Lesson 11.1 Answer Key Question 5.
\(\frac{p}{10}\) ____
Answer:
Solution to this example is given below
\(\frac{p}{10}\) The operation is Division. (Final solution)
The quotient of p and 10

Question 6.
Draw a bar model to represent the expression m ÷ 4. (Example 2)

Answer:
Bar model where is the big black box and each blue box is equal to \(\frac{m}{4}\) = m ÷ 4 = m/4

At 6 p.m., the temperature in Phoenix, AZ, t, is the same as the temperature in Tucson, AZ. By 9 p.m., the temperature in Phoenix has dropped 2 degrees and in Tucson it has dropped 4 degrees. By 11 p.m., the temperature in Phoenix has dropped another 3 degrees. (Example 3)

Question 7.
Represent the temperature In each city with an algebraic expression and a bar model.

Answer:
Representation of the problem and algebraic expression.
Phoenix = t – 2 – 3
Tucson = t – 4

The algebraic expression for Phoenix is t – 2 – 3 and Tucson is t – 4.

Question 8.
Are the expressions that represent the temperatures in the two cities equivalent? Justify your answer.
Answer:
From the solution of question 6, it can be seen that the 2 temperatures are not equivalent as the temperature in Phoenix has dropped by a total of 5 degrees while that in Tucson, only dropped by 4 degrees.

Essential Question Check-In

Question 9.
How can you use expressions and models to determine if expressions are equivalent?
Answer:
Algebraic expressions and models can easily determine that expressions are equivalent. Models are used to represent the word problems by graphical representation while algebraic expressions are used by translating the word problem into variables and constant

Models are graphical representations of the word problem while algebraic expression is a translation using variables and constants.

Texas Go Math Grade 6 Lesson 11.1 Independent Practice Answer Key

Question 10.
Write an algebraic expression with the constant 7 and the variable y.
Answer:
Solution to this example is given below
The product of 7 and y. The operation is multiplication (Final solution)
The algebraic expression is 7 × y
7 × y

Question 11.
Write an algebraic expression with two variables and one constant.
Answer:
The algebraic expression is m(n + 5).
Expression is m(n + 5).

Go Math Expressions Answer Key Lesson 11.1 Question 12.
What are the variables in the expression x + 8 – y?
Answer:
The variable is part of an algebraic expression that is represented by a single letter. Therefore, the variables ¡n the given expression are x and y.

The variables are x and y.

Question 13.
Identify the parts of the algebraic expression x + 15.
Constant(s) ________
Variable(s) _______
Answer:
The parts of an algebraic expression from the given are:

• Constant: 15
• Variable: x

The constant is 15 and the variable is x.

Write each phrase as an algebraic expression.

Question 14.
n divided by 8 ______
Answer:
The word divided indicates division and so the algebraic expression for the given phrase is or \(\frac{n}{8}\) or n ÷ 8.
\(\frac{n}{8}\) or n ÷ 8

Question 15.
p multiplied by 4 _______
Answer:
Solution to this example is given below
The product of p and 4. The operation is multiplication (Final solution)
The algebraic expression is p × 4
p × 4

Question 16.
b plus 14 _______
Answer:
Solution to this example is given below
The sum of b and 14 The operation is addition. (Final solution)
The algebraic expression is b + 14
b + 14

Question 17.
90 times x ______
Answer:
Solution to this example is given below
The product of 90 and x. The operation is multiplication (Final solution)
The algebraic expression is 90 × x
90 × x

Expressions and Equations 6th Grade Answer Key Question 18.
a take away 16 ______
Answer:
Solution to this example is given below
The difference of a and 16 The operation is subtraction. (Final solution)
The algebraic expression is a – 16
a – 16

Question 19.
k less than 24 ______
Answer:
Solution to this example is given below
k lees than 24 The operation is subtraction. (Final solution)
The algebraic expression is 24 – k
24 – k

Question 20.
3 groups of w _____
Answer:
Solution to this example is given below
3 groups of w The operation is multiplication (Final solution)
The algebraic expression is 3 × w
3 × w

Question 21.
the sum of 1 and q ____
Answer:
The word sum indicates addition and so the algebraic expression for the given phrase is 1 + q.
1 + q

Question 22.
the quotient of 13 and z ____
Answer:
The word quotient indicates division and so the algebraic expression for the given phrase is \(\frac{13}{z}\) or 13 ÷ z.
\(\frac{13}{z}\) or 13 ÷ z.

Question 23.
c added to 45 ____
Answer:
Solution to this example is given below
c added to 45 The operation is addition. (Final solution)
The algebraic expression is c + 45
c + 45

Write a phrase in words for each algebraic expression.

Question 24.
m + 83 __________________
Answer:
m + 83 can be explained by the phrase, sum of m and 83.
Sum of m and 83.

Question 25.
42s _______
Answer:
42s can be explained by the phrase, Product of 42 and s.
Product of 42 and s.

Question 26.
\(\frac{9}{d}\) _______
Answer:
Solution to this example is given below
\(\frac{9}{d}\) The operation is Division. (Final solution)
The quotient of 9 and d

Grade 6 Go Math Answer Key Equivalent Expressions Question 27.
t – 29 _____
Answer:
t – 29 can be explained by the phrase, 29 less than t.
29 less than t.

Question 28.
2 + g ____
Answer:
Solution to this example is given below
2 + g The operation is Addition. (Final solution)
The sum of 2 and g

Question 29.
11x ______
Answer:
Solution to this example is given below
The operation is Multiplication. (Final solution)
The product of 11 and x

Question 30.
\(\frac{h}{12}\) _______
Answer:
Solution to this example is given below
\(\frac{h}{12}\) The operation is Division. (Final solution)
The quotient of h and 12

Question 31.
5 – k ____
Answer:
5 – k can be explained by the phrase, k less than 5.
k less than 5.

Sarah and Noah work at Read On Bookstore and get paid the same hourly wage. The table shows their work schedule for last week.

Answer:

Question 32.
Write an expression that represents Sarah’s total pay last week. Represent her hourly wage with w.
Answer:
She worked for 3 + 5 = 8 hours and her pay is $w per hour so lier total pay for 8 hours will be 8 × w = 8w.
Her total pay for 8 hours will be $8w.

Question 33.
Write an expression that represents Noah’s total pay last week. Represent his hourly wage with w.
Answer:
He worked for 8 hours and his pay is $w per hour so his total pay for 8 hours will be 8 × w = 8w.
His total pay for 8 hours will be $8w.

Question 34.
Are the expressions equivalent? Did Sarah and Noah earn the same amount last week? Use models to justify your answer.
Answer:
Yes, both the expressions are equivalent. This is because both of them worked for 8 hours each so earned a total of $8w each.

Question 35.
Critique Reasoning Lisa concluded that 3 • 2 and 3² are equivalent expressions. Is Lisa correct? Explain.
Answer:
Lisa is incorrect because 3 . 2 ≠ 3²
3 . 2 = 6 2 is a factor
3² = 3 × 3 2 is an exponent which multiplies the base by itself twice
= 9
Lisa is incorrect because the value of the expressions will not be equal

Lesson 11.1 Modeling with Expressions Answer Key Question 36.
Multiple Representations How could you represent the expressions x – 5 and x – 3 – 3 on a scale like the one you used in the Explore Activity? Would the scale balance?
Answer:
x – 5 and x – 3 – 3
By looking at the algebraic expressions and evaluating them, the scale is not balanced because the constant term on the right side of the expression is -6. while the constant term on the left side is -5.

In order to make the expressions balanced, the value of x for both sides will be different. The value of x on the left side must be 10 while the value of x on the right, side must be 11 to make the scale balance.

The value of x for both sides will be different to make the scale balanced.

Question 37.
Multistep Will, Hector, and Lydia volunteered at the animal shelter in March and April. The table shows the number of hours Will and Hector volunteered in March. Let x represent the number of hours Lydia volunteered in March.

a. Will’s volunteer hours in April were equal to his March volunteer hours plus Lydia’s March volunteer hours. Write an expression to represent Will’s volunteer hours in April.
Answer:
Will’s volunteer hours in April is equal to the 3 + x hours, as he volunteered for the sum of the number of hours
he did in March and the number of hours that Lydia volunteered for

b. Hector’s volunteer hours in April were equal to 2 hours less than his March volunteer hours plus Lydia’s March volunteer hours. Write an expression to represent Hector’s volunteer hours in April.
Answer:
Hector’s volunteer hours in April is equal to the (5 – 2) + x = 3 + x hours, as he volunteered for the sum of the number of hours he did in March minus 2 and the number of hours that Lydia volunteered for.

c. Did Will and Hector volunteer the same number of hours in April? Explain.
Answer:
Yes, in April both of them volunteered for the same number of hours as proved by their respective expressions.

Question 38.
The town of Rayburn received 6 more inches of snow than the town of Greenville. Let g represent the amount of snow in Greenville. Write an algebraic expression to represent the amount of snow in Rayburn.
Answer:
The town of Rayburn received 6 more inches of snow than the town of Greenville. Let g represent the amount of snow in Greenville, then the town of Rayburn received 6 + g inches of snow.
The town of Rayburn received 6 + g inches of snow

Question 39.
Abby baked 48 cookies and divided them evenly into bags. Let h represent the number of bags. Write an algebraic expression to represent the number of cookies in each bag.
Answer:
The number of cookies in each bag is represented by \(\frac{48}{b}\)
The algebraic expression is \(\frac{48}{b}\).

Question 40.
Eli is driving at a speed of 55 miles per hour. Let h represent the number of hours that Eli drives at this speed. Write an algebraic expression to represent the number of miles that Eli travels during this time.
Answer:
The algebraic expression that would represent the number of hours that Eli drives is 55h where h stands for the number of hours.
Expression is 55h.

Texas Go Math Grade 6 Lesson 11.1 H.O.T. Focus On Higher Order Thinking Answer Key

Question 41.
Represent Real-World Problems If the number of shoes in a closet is s, then how many pairs of shoes are in the closet? Explain.
Answer:
The algebraic expression that would represent the number of pairs of shoes in the closet is \(\frac{s}{2}\) where s stands for the number of shoes Since a pair of shoes consists of 2 shoes, that is why the total number of shoes in the closet will be divided by 2 to get the number of pairs of shoes.
Expression is \(\frac{s}{2}\)

Question 42.
Communicate Mathematical Ideas Is 12x an algebraic expression? Explain why or why not.
Answer:
The given expression is an algebraic expression because it consists of a variable and a coefficient. The coefficient is 12 and the variable is x. It represents the mathematical phrase, the product of 12 and x
Yes, it represents the product of 12 and x

Question 43.
Problem Solving Write an expression that has three terms, two different variables, and one constant.
Answer:
An expression that has three terms, two different variables, and one constant is 5x + 3y + 2 a real world problem to represent this expression is a total food bill of x sandwiches each costing $5 and y drinks each costing $3 and a tip of $2.

Question 44.
Represent Real-World Problems Describe a situation that can be modeled by the expression x – 8.
Answer:
A student scored x marks in his test but lost 8 due to silly mistakes so his total score resulted to be equal to x – 8.
x – 8

Question 45.
Critique Reasoning Ricardo says that the expression 1y + 4 is equivalent to the expression y + 4. Is he correct? Explain.
Answer:
Ricardo is correct and y + 4 = 1y + 4, because when there is no visible coefficient of the variable then this explicitly means that the coefficient of the variable is 1 and so it is not written.
Ricardo is correct.

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 11 Answer Key Generating Equivalent Algebraic Expressions.

Texas Go Math Grade 6 Module 11 Answer Key Generating Equivalent Algebraic Expressions

Essential Question
How can you generate equivalent algebraic expressions and use them to solve real-world problems?

Texas Go Math Grade 6 Module 11 Are You Ready? Answer Key
Complete these exercises to review the skills you will need for this chapter.

Evaluate.

Question 1.
11 + (20 – 13)
Answer:
Solution to this example is given below
11 + (20 – 13)
11 + (20 – 13) = 11 + 7 Perform operations inside parentheses. (Final solution)
= 18 Add.
= 18

Question 2.
(10 – 7) – (14 – 12)
Answer:
Solution to this example is given below
(10 – 7) – (14 – 12)
(10 – 7) – (14 – 12) = (10 – 7) – 2 Perform operations inside parentheses. (Final solution)
= 3 – 2 Perform operations inside parentheses.
= 1 Subtract.
= 1

Module 11 Go Math Grade 6 Answer Key Question 3.
(4 + 17) – (16 – 9)
Answer:
Solution to this example is given below
(4 + 17) – (16 – 9)
(4 + 17) – (16 – 9) = (4 + 17) – 7 Perform operations inside parentheses. (Final solution)
= 21 – 7 Perform operations inside parentheses.
= 14 Subtract.
= 14

Question 4.
(23 – 15) – (18 – 13)
Answer:
Given expression:
(23 – 15) – (18 – 13)
Simplify the expressions in the 2 parentheses of the given expression:
= (8) – (5)
Expand the parentheses:
= 8 – 5
Perform the indicated operation to evaluate the value of the expression:
= 3

Question 5.
8 × (4 + 5 + 7)
Answer:
Solution to this example is given below
8 × (4 + 5 + 7)
8 × (4 + 5 + 7) = 8 × 16 Perform operations inside parentheses. (Final solution)
= 128 Multiply.
= 128

Question 6.
(2 + 3) × (11 – 5)
Answer:
Solution to this example is given below
(2 + 3) × (11 – 5)
(2 + 3) × (11 – 5) = (2 – 3) × 6 Perform operations inside parentheses. (Final solution)
= 5 × 6 Perform operations inside parentheses.
= 30 Multiply.
= 30

Write a numerical expression for the word expression.

Question 7.
the difference between 42 and 19 ________
Answer:
Write a numerical expression for the difference between 42 and 19 Think: Difference means to subtract
42 – 19 Write from 42 taken 19 (Final solution)
42 – 19

Question 8.
the product of 7 and 12 ______
Answer:
Write a numerical expression for the product of 7 and 12 Think: Product means to multiply

7 × 12 Write 7 multiplied by 12 (Final solution)
7 × 12

Grade 6 Module 11 Answer Key Equivalent Expressions Question 9.
30 more than 20 _____________________
Answer:
The word more than indicates the addition of the 2 given numbers so the algebraic equivalent of the given statement is 30 + 20
30 + 20

Question 10.
100 decreased by 77 ________
Answer:
The word decreased indicates the subtraction of the smaller number from the larger number so the algebraic equivalent of the given statement is 100 – 77
100 – 77

Evaluate the expression.

Question 11.
3(8) – 15 ________
Answer:
Given expression:
3(8) – 15
Expand the parentheses of the given expression:
= 24 – 15
Perform the indicated operation to evaluate the value of the expression:
= 9
3(8) – 15 = 9

Question 12.
4(12) + 11 ________
Answer:
Given expression:
4(12) + 11
Expand the parentheses of the given expression:
= 48 + 11
Perform the indicated operation to evaluate the value of the expression.
= 59
4(12) + 11 = 59

Question 13.
3(7) – 4(2) ________
Answer:
Given expression:
3(7) – 4(2)
Expand both the parentheses of the given expression:
= 21 – 8
Perform the indicated operation to eva[uate the value of the expression:
= 13
3(7) – 4(2) = 13

Question 14.
4(2 + 3) – 12 ______
Answer:
Evaluate 4(2 + 3) – 12
4(2 + 3) – 12 = 4(5) – 12 Perform operations inside parentheses
= 20 – 12 Multiply (Final solution)
= 8 Subtract
= 8

Question 15.
9(14 – 5) – 42 ______
Answer:
Evaluate 9(14 – 5) – 42
9(14 – 5) – 42 = 9(9) – 42 Perform operations inside parentheses
= 81 – 42 Multiply (Final solution)
= 39 Subtract
= 39

Generating Equivalent Expressions Grade 6 Answer Key Question 16.
7(8) – 5(8) ________
Answer:
Evaluate 7(8) – 5(8)
7(8) – 5(8) = 56 – 40 Multiply (Final solution)
= 16 Subtract
= 16

Texas Go Math Grade 6 Module 11 Reading Start-Up Answer Key

Visualize Vocabulary

Use the review words to complete the graphic. You may put more numerical expressions than one word in each oval.

Understand Vocabulary

Complete the sentences using the preview words.

Question 1.
An expression that contains at least one variable is an ______
Answer:
An expression that contains at least one variable is an algebraic expression

Question 2.
A part of an expression that is added or subtracted is a _____.
Answer:
A part of an expression that is added or subtracted is a term

Question 3.
A ______________ is a specific number whose value does not change.
Answer:
A constant is a specific number whose value does not change.

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