Envision Math Accelerated Grade 7 Volume 1 Student Edition Chapter 6 Solving Problems Using Equations And Inequalities Exercise 6.2

Envision Math Accelerated Grade 7 Volume 1 Student Edition Chapter 6 Solving Problems Using Equations And Inequalities

Question. Find out the relationship between the shapes.

Given: Three clues are given.

We have to find out the relationship between the shapes.

First, we will look for that shape which is common on both the left side and right side and then we will eliminate it and find out the relationship between the remaining shapes.

Let, C represents a circle, T represent a triangle and S represents a square.

Therefore, we get the following clues as

​4C  + 1T = 2S + 1T

3T + 2C = 2C + 1S

1S + 2C = 6T

​From the first clue we have

4C + 1T = 2S + 1T

4C + T−T = 2S + T − T (Subtracting T from both sides)

4C = 2S

\(\frac{4}{2}\)C=\(\frac{2}{2}\)S   (Dividing 2 from both sides)

S = 2C ———-(1)

From the second clue we have

3T+2C=2C+1S

3T + 2C − 2C = 2C + S − 2C     (Subtracting 2C from both sides)

S = 3T ———–(2)

From equation (1) and (2)

We have, S = 2C  and  S =   3T

Therefore, 2C = 3T———(3)

From equations (1), (2), and (3)

We get the relationship between the shapes

S=2C

S=3T

2C=3T

The relationship between the shapes are as follows: S = 2C, S = 3T, 2C = 3T

Envision Math Accelerated Grade 7 Volume 1 Student Edition Chapter 6 Solving Problems Using Equations And Inequalities Exercise 6.2

Envision Math Accelerated Grade 7 Chapter 6 Exercise 6.2 Answer Key

Question. Use properties of equality to solve the given equations

4C + 1T = 2S + 1T

3T + 2C = 2C + 1S

1S + 2C = 6T

We have to use properties of equality to reason about the equations.

Let, C represents a circle, T represent a triangle and S represents a square.

Therefore, we get the following clues as

​4C + 1T = 2S + 1T

3T + 2C = 2C + 1S

1S + 2C = 6T

From the first clue we have

4C + 1T = 2S + 1T

4C + T − T = 2S + T−T             (Subtracting T from both sides)

4C = 2S

\(\frac{4}{2}\)C =  \(\frac{2}{2}\) S                      (Dividing 2 from both sides)

S = 2C ———-(1)

From the second clue we have

3T + 2C = 2C + 1S

3T + 2C − 2C = 2C + S − 2C     (Subtracting 2C from both sides)

S = 3T ———–(2)

From equation (1) and (2)

We have, S = 2C and S = 3T

Therefore, 2C = 3T——–(3)

From equations (1), (2), and (3)

We get the relationship between the shapes:

S = 2C

S = 3T

2C = 3T

We used the subtraction and division properties of equality to solve the given equations.

Envision Math Grade 7 Chapter 6 Solving Equations And Inequalities Exercise 6.2 Solutions

Question. A shape equation C represents a circle, T represent a triangle and S represents a square.

Given: A shape equation is given.

Relationship between shapes:

S = 2C

S = 3T

2C = 3T

We have to complete the equation only using triangles.

Let, C represents a circle, T represent a triangle and S represents a square.

Therefore, we get

4C + S = ?

From solving and discussing exercise 1 we have the following relationship between the shapes.

S = 2C

S = 3T

2C = 3T

Therefore, ​ 4C = 6T

S = 3T

​Hence, we get

4C + S = 6T + 3T = 9T

Therefore, the complete equation is 4C + S = 9T

The complete equation is 4C + S = 9T

Question. Write a two-step equation is similar to solving a one-step equation.

We have to explain how solving a two-step equation is similar to solving a one-step equation.

Solving a two-step equation is similar to solving a one-step equation because they both use properties of equality.

The properties of equality can be applied the same way when solving two-step equations as when solving one-step equations.

Both methods use properties of equality to determine the value of the variable or to make the equation balance.

Solving a two-step equation is similar to solving a one-step equation because they both use properties of equality.

Question. Andrew rents bowling shoes for $4 and he bowls 2 games. Find out the cost of each game.

Given:

Andrew rents bowling shoes for $4 and he bowls 2 games.

The total money he spent is $22

Given bar diagrams are:

EnVisionmath 2.0 Accelerated Grade 7, Volume 1, Student Edition, Chapter 6.2 Solve Two- Step equations-Solve Page 338 Exercise 1 bar diagram 1

We have to complete the bar diagrams and find out the cost of each game.

The total money Andrew spent is $22

He rents the bowling shoes for $4 and played 2 games.

Therefore, we get the following bar diagrams:

EnVisionmath 2.0 Accelerated Grade 7, Volume 1, Student Edition, Chapter 6.2 Solve Two- Step equations-Solve Page 338 Exercise 1 bar diagram 2

The total cost is $22

Therefore, 4 + 2b = 22

4 + 2b − 4 = 22− 4     (Subtracting 4 from both sides)

2b = 18

\(\frac{2}{2}\) = \(\frac{18}{2}\)   (Dividing 2 from both sides)

b = 9

Therefore, the cost of each game is $9.

The complete bar diagrams are:

EnVisionmath 2.0 Accelerated Grade 7, Volume 1, Student Edition, Chapter 6.2 Solve Two- Step equations-Solve Page 338 Exercise 1 bar diagram 3

The cost of each game is $9.

Given:

Kristy ran 24 laps.

The total distance traveled by Kristy is 29.6 kilometers

She walked 0.2 km to the presentation table.

We have to determine the distance of each lap.

First, we will determine the total distance she traveled in 24 laps and then we will divide it by 24 to get the distance of each lap.

The total distance traveled by Kristy is 29.6 kilometers

She walked 0.2 km to the presentation table.

Therefore, total distance she covered in  24 laps = 29.6 − 0.2 = 29.4 km     (Subtraction)

Distance covered in 1 lap \(=\frac{29.4}{24}\)        (Division)

=1.225

Hence, the distance of each lap is 1.225 km

The two steps we used for solving the situation are subtraction and division.

The distance of each lap is 1.225 km. The two steps we used for solving the situation are subtraction and division.

Solving Problems Using Equations And Inequalities Grade 7 Exercise 6.2 Envision Math

Question. Solve the equation by using subtraction and division.

We have to explain what were the two steps we used to solve the equation.

The total cost is $22

Therefore, 4 + 2b = 22

4 + 2b−4 = 22 − 4       (Subtracting 4 from both sides)

2b = 18

\(\frac{2}{2}b\)=\(\frac{18}{2}\)              (Dividing 2 from both sides)

b = 9

Therefore, the cost of each game is $9.

We can see that the two steps we used to solve the equation are subtraction and division.

The two steps we used to solve the equation are subtraction and division.

Question. Write a two-step equation is similar to solving a one-step equation.

We have to explain how solving a two-step equation is similar to solving a one-step equation.

Solving a two-step equation is similar to solving a one-step equation because they both uses properties of equality.

The properties of equality can be applied the same way when solving two-step equations as when solving one-step equations.

Both methods use properties of equality to determine the value of the variable or to make the equation balance.

Two-step equations can be solved in two steps by using two different properties of equality.

Solving a two-step equation is similar to solving a one-step equation because they both use properties of equality.

Question. Use bar diagram for solve 4x – 3 = 13.

Given:

Bar diagram for 4x − 3 = 13 is given

EnVisionmath 2.0 Accelerated Grade 7, Volume 1, Student Edition, Chapter 6.2 Solve Two- Step equations-Solve Page 340 Exercise 2 bar diagram 1

we have to solve the bar diagram for x

4x−  3 = 13

4x − 3 + 3 = 13 + 3    (Adding 3 to both sides)

4x = 16

\(\frac{4x}{4}\)=\(\frac{16}{4}\) (Dividing 4 from both sides))

x = 4

The value of x is 4.

We used addition and division to determine the value of x from the bar diagram.

EnVisionmath 2.0 Accelerated Grade 7, Volume 1, Student Edition, Chapter 6.2 Solve Two- Step equations-Solve Page 340 Exercise 2 bar diagram 2
The value of x is 4.

Envision Math Grade 7 Exercise 6.2 Solution Guide

Question. Solve the equation 6p – 12 = 72 and said p = 14.

Given:

Clara solved a problem 6p − 12 = 72 and said p = 14.

We have to check whether Clara is correct or not.

First, we will substitute the value of p and calculate the left-hand side of the equation.

Given equation: 6p − 12 = 72 and p = 14

The left-hand side of the equation is 6p − 12

Substituting the value of p = 14 in 6p − 12 we get

6 (14) − 12      (Multiply)

=  84 − 12        (Subtract)

=  72

=   RHS

Since the left-hand side of the equation is equal to the right-hand side of the equation.

Therefore, the value of p is correct.

And hence, Clara is correct.

As the left-hand side of the equation is equal to the right-hand side of the equation. therefore, the value of p is correct. And hence, Clara is correct.

Question. Clyde is baking, and the recipe requires 1\(\frac{1}{3}\) cups of flour. Clyde has 2 cups of flour, but he is doubling the recipe to make twice as much. Find out how much more flour Clyde needs.

Given:  Clyde is baking, and the recipe requires  1\(\frac{1}{3}\)  cups of flour. Clyde has 2 cups of flour, but he is doubling the recipe to make twice as much.

We need to find out how much more flour Clyde needs.

Also, we need to write an equation to represent the problem.

Let c represent the amount of flour Clyde needs.

The number of cups required for the recipe is 1\(\frac{1}{3}\)

Clyde has two cups of flour.

He is doubling the recipe to make twice as much.

Here, c represents the amount of flour Clyde needs.

The equation that represents the problem will be

2 + c = 2 × 1\(\frac{1}{3}\)

The equation that represents the problem will be 2 + c = 2 × 1\(\frac{1}{3}\)

Given: Clyde is baking, and the recipe requires 1\(\frac{1}{3}\)cups of flour.

Clyde has 2 cups of flour, but he is doubling the recipe to make twice as much.

We need to find how much more flour does Clyde need. We need to solve the equation formed.

The equation formed will be 2 + c = 2 × 1\(\frac{1}{3}\)

Solving the given equation, we get

\(2+c=2 \times 1 \frac{1}{3}\) \(2+c=2 \times \frac{4}{3}\)

2 + c = \(\frac{8}{3}\)

c = \(\frac{8}{3}-2\)

c = \(\frac{8-6}{3}\)

c = \(\frac{2}{3}\)

\(\frac{2}{3}\) cups of flour Clyde needs to complete the recipe.

How To Solve Exercise 6.2 In Envision Math Grade 7

Question. Four times a number, n, added to 3 is 47. Write an equation that you can use to find the number.

Given: Four times a number, n, added to 3 is 47.

We need to write an equation that you can use to find the number.

Let the unknown number be n

The four times of the unknown number is added to three.

The result obtained is 47

The equation will be

4n + 3 = 47

The equation that we can use to find the number will be, 4n + 3 = 47

Given: Four times a number, n, added to 3 is 47.

We need to find the number represented by n

The equation formed from the given data is

4n + 3 = 47

Solving the equation, we get

​4n + 3 = 47

4n = 47 − 3

4n = 44

n  =  \(\frac{44}{4}\)

n = 11

The number represented by n is 11.

Envision Math 7th Grade Exercise 6.2 Step-By-Step Solutions

Question. Solve the equation 4x – 12 = 16.

We need to use the bar diagram to help you solve the equation  4x − 12 = 16.

The given equation is 4x − 12 = 16.

Solving the equation, we get

​4x − 12 = 16

4x = 16 + 12

4x = 28

x =  \(\frac{28}{4}\)

x = 7

Solving the given using the bar diagram, we get

​16 + 12 = 4x

28 = 4x

x = \(\frac{28}{4}\)

x = 7

The value of  x = 7

Question. Solve the given equation.

We need to complete the steps to solve the given equation.

We need to place the missing numbers in the given equation to solve it.

Thus, we get

\(\frac{1}{5}\)t+2−2 = 17−2

\(\frac{1}{5}\)t  = 15

5×\(\frac{1}{5}\)t = 5 × 15

t = 75

The value of  t = 75

Envision Math Accelerated Chapter 6 Exercise 6.2 Solving Equations And Inequalities

Question. Use the bar diagram to write an equation. Then solve for x.

We need to use the bar diagram to write an equation. Then solve for x.

The bar diagram given is

EnVisionmath 2.0 Accelerated Grade 7, Volume 1, Student Edition, Chapter 6.2 Solve Two- Step equations-Solve Page 341 Exercise 8

The equation formed from the bar diagram is

x + x + x = 7 + 5

Solving the given equation, we get

​x + x + x = 7 + 5

3x = 12

x = \(\frac{12}{3}\)

x = 4

The value of x = 4

Question. While shopping for clothes, Tracy spent $38 less than 3 times what Daniel spent. Find how much Daniel spent.

Given that, While shopping for clothes, Tracy spent $38 less than 3 times what Daniel spent.

We need to write and solve an equation to find how much Daniel spent.

Let x represent how much Daniel spent. Also, the amount Tracy spent is $10.

The equation represented by the given situation will be

3x − 38 = 10

Solving the given equation, we get

​3x − 38 = 10

3x = 10 + 38

3x = 48

x = \(\frac{48}{3}\)

x = 16

Daniel spent $16

Envision Math Accelerated Grade 7 Chapter 6 Exercise 6.2 Answers

Question. Solve the equation 0.5p – 3.45 = -1.2.

We need to solve the equation 0.5p − 3.45 = −1.2.

The given equation is 0.5p−3.45 = −1.2.

Solving the given we get

0.5p – 3.45 = -1.2.

0.5p = -1.2 + 3.45

0.5p = 2,25

p = \(\frac{2.25}{0.5}\)

p = \(\frac{22.5}{5}\)

p = 4.5

The value of p = 4.5

Question. Solve the equation \(\frac{n}{10}\) + 7 = 10

We need to solve the equation \(\frac{n}{10}\)+7 = 10

The given equation is n \(\frac{n}{10}\)+7 = 10

Solving the given we get

\(\frac{n}{10}\) + 7 = 10

\(\frac{n}{10}\) = 10− 7

\(\frac{n}{10}\) = 3

n = 3 × 10

n = 30

The value of n = 30

Question. A group of 4 friends went to the movies. In addition to their tickets, they bought a large bag of popcorn to share for $6.25.

Given:

A group of 4 friends went to the movies.

In addition to their tickets, they bought a large bag of popcorn to share for $6.25.

The total was $44.25

To find/solve

Write and solve an equation to find the cost of one movie ticket, m.

Use an equation to determine the cost of one ticket.

4m + 6.25 = 44.25

4m = 38

m = 9.50

The cost of one ticket is $9.50.

The equation is 4m + 6.25 = 44.25, The cost of one ticket is $9.50.

Given:

A group of 4 friends went to the movies.

In addition to their tickets, they bought a large bag of popcorn to share for $6.25.

The total was $44.25.

To find/solve

Draw a model to represent the equation.

4m + 6.25 = 44.25

4m = 38

m = 9.50

The bar diagram represents the equation

EnVisionmath 2.0 Accelerated Grade 7, Volume 1, Student Edition, Chapter 6.2 Solve Two- Step equations-Solve Page 342 Exercise 12 bar diagram 1

The bar diagram represents the equation

EnVisionmath 2.0 Accelerated Grade 7, Volume 1, Student Edition, Chapter 6.2 Solve Two- Step equations-Solve Page 342 Exercise 12 bar diagram 2

Given:

Oliver incorrectly solved the equation  2x + 4 = 10.

He says the solution is x = 7.

To find/solve

What is the correct solution?

The correct solution is

2x + 4 = 10

2x = 6

x = 3

The correct solution is x = 3.

Given:

Oliver incorrectly solved the equation 2x+4=10. He says the solution is x=7.

To find/solve

 What mistake might Oliver have made?

An expression is a finite combination of symbols that is well-formed according to rules that depend on the context.

Oliver did not use the inverse relationship of the equation.

He did not apply the properties of equality in order to balance the equation.

He added 4 on the right side of the equation instead of subtracting.

Oliver did not use the inverse relationship of the equation. He did not apply the properties of equality in order to balance the equation. He added 4 on the right side of the equation instead of subtracting.

Envision Math Grade 7 Chapter 6 Practice Problems Exercise 6.2

Question. What two properties of equality do you need to use to solve the equation?

An expression is a finite combination of symbols that is well-formed according to rules that depend on the context.

The two properties of equality that is needed to solve for the equation are the addition property of equality and the division property of equality.

Question. Find the solution of the equation 4.9x – 1.9 = 27.5

Given:

Use the equation 4.9x − 1.9 = 27.5

To find/solve

The solution is x =?

Determine the solution:

​4.9x −1.9 = 27.5

4.9x = 29.4

x = 29.4/4.9

x = 6

The solution is x = 6.

Question. At a party, the number of people who ate meatballs was 11 fewer than of the total number of people. Five people ate meatballs.

Given:

At a party, the number of people who ate meatballs was 11 fewer than of the total number of people.

Five people ate meatballs.

We form the equation as:

5=\(\frac{1}{3}\)x−11

Now we write a one-step equation that has the same solution:

\(\frac{1}{3}\)x = 16

Therefore, this is the required one-step equation that has the same solution.

Therefore, the required one-step equation with the same solution is \(\frac{1}{3}\)x=16

Envision Math Accelerated 7th Grade Exercise 6.2 Key

Question. In a week, Tracy earns $12.45 less than twice the amount Kalya earns. Tracy earns $102.45.

Given:

In a week, Tracy earns $12.45 less than twice the amount Kayla earns. Tracy earns $102.45.

We form the equation as:

102.45 = 2x − 12.45

Now we solve the equation

​2x = 102.45 + 12.45

2x = 114.9

x = \(\frac{114.9}{2}\)

x = 57.45

Therefore, Kayla earns $57.45.

Therefore, Kayla earns $57.45

Question. Solve the equation 2x + 4\(\frac{1}{5}\) = 9

Given: 2x+4\(\frac{1}{5}\) = 9

We solve the equation:

2x + 4\(\frac{1}{5}\) = 9

Now we use inverse operations and the subtraction property of equality to isolate the term with the variable.

​2x+ \(4 \frac{1}{5}-4\frac{1}{5}\)

= 9− 4\(\frac{1}{5}\)

2x = 4\(\frac{1}{5}\)

2x = \(\frac{41}{5}\)

Now we use inverse operations and the division property of equality to isolate the variable.

\(\frac{2x}{2}\) = \(\frac{41}{5}\).\(\frac{1}{2}\)

x = 4.1

Therefore, the required solution is x = 4.1

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