## Everyday Mathematics 6th Grade Answer Key Unit 6 Equivalent Expressions and Solving Equations

### Everyday Mathematics Grade 6 Home Link 6.1 Answers

**Approximating Solutions**

For each equation, try to get as close as possible to the exact solution. Use the suggested test numbers to get started. Round numbers to the nearest thousandth.

Question 1.

Equation: r^{2} + r = 15

My closest number: ___________

Answer:

The given equation is:

rÂ² + r = 15

So,

The completed table of the values of r, rÂ², and rÂ² + r is:

Hence, from the above table,

We can conclude that the closest exact number for the given equation is: 3.41

Question 2.

Equation: x^{2} – 2x = 23

My closest number: __________

Answer:

The given equation is:

xÂ² – 2x = 23

So,

The completed table of the values of x, xÂ², 2x, and xÂ² – 2x is:

Hence, from the above,

We can conclude that the closest exact number for the given equation is: 5.9

**Practice**

Rewrite each expression in exponential notation.

Question 3.

9 Ã— 9 Ã— 9 __________

Answer:

The given expression is:

9 Ã— 9 Ã— 9

Hence,

The representation of the given expression in the exponential form is: 9^{3}

Question 4.

7 Ã— 7 Ã— 7 Ã— 7 Ã— 7 _________

Answer:

The given expression is:

7 Ã— 7 Ã— 7 Ã— 7 Ã— 7

Hence,

The representation of the given expression in the exponential form is: 7^{5}

Question 5.

6.2 Ã— 6.2 ________

Answer:

The given expression is:

6.2 Ã— 6.2

Hence,

The representation of the given expression in the exponential form is: 6.2^{2}

### Everyday Math Grade 6 Home Link 6.2 Answer Key

**Solution Sets**

Question 1.

The solution set is {all numbers less than 7}. Circle inequalities with this solution set.

j > 4

7 < j

7 > j

j < 7

Answer:

The given inequalities are:

j > 4

7 < j

7 > j

j < 7

Hence, from the above,

We can conclude that the solution sets for all the numbers less than 7 are:

j < 7, and 7 > j

Question 2.

a. The solution set is {all numbers greater than 10}. Circle inequalities with this solution set.

m + 10 < 11

11 < m + 1

6 > 5 + m

6 > 5m

Answer:

The given inequalities are:

a) m + 10 < 11

b) 11 < m + 1

c) 6 > 5 + m

d) 6 > 5m

So,

a)

Subtract with 10 on both sides

So,

m + 10 – 10 < 11 – 10

m < 1

b)

Subtract with 1 on both sides

So,

11 – 1 < m + 1 – 1

10 < m

m > 10

c)

Subtract with 5 on both sides

So,

6 – 5 > 5 + m – 5

1 > m

m < 1

d)

6 > 5m

So,

5m < 6

m < \(\frac{6}{5}\)

Hence, from the above,

We can conclude that the solution set for all numbers greater than 10 is: m > 10

b. Explain how you found your answer for Problem 2a.

Answer:

We have to solve the given inequalities so that only variable will be present either it is given on the left side or the right side

So,

We will get the inequality in the form of

Variable < constant or Variable > constant or Variable = constant

Hence, in the above way,

We can conclude that we have found the answer to problem 2a.

Question 3.

Record the solution sets for the equations below.

a. 3x = 45

Solution set: __________

Answer:

The given expression is:

3x = 45

Divide with 3 on both sides

So,

\(\frac{3}{3}\)x = \(\frac{45}{3}\)

x = 15

Hence, from the above,

We can conclude that the solution set for the given expression is: {15}

b. x + 138 = 204

Solution set: ____________

Answer:

The given expression is:

x + 138 = 204

Subtract with 138 on both sides

So,

x + 138 – 138 = 204 – 138

x = 66

Hence, from the above,

We can conclude that the solution set for the given expression is: {66}

Question 4.

Write the letter of the solution set that matches each number sentence.

x Ã· 4 = 8Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ________Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â A. {All numbers}

\(\frac{4}{x}\) = 8Â Â Â Â Â _________Â Â Â Â Â Â Â Â Â Â Â Â Â Â B. {0}

10 – x = 7Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â __________Â Â Â Â Â Â Â Â Â Â Â Â Â Â C. { }

3x + x = 16Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â _________Â Â Â Â Â Â Â Â Â Â Â Â Â Â D. {\(\frac{1}{2}\)}

5x = 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â __________Â Â Â Â Â Â Â Â Â Â Â Â Â E. {- \(\frac{1}{2}\), \(\frac{1}{2}\)}

12 âˆ— x = x âˆ— 12Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ________Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â F. {3}

0.5 = |x|Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â _________Â Â Â Â Â Â Â Â Â Â Â Â Â Â G. {32}

x – 5 = xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â __________Â Â Â Â Â Â Â Â Â Â Â Â Â H. {4}

Answer:

The given expressions are:

a)

x Ã· 4 = 8

So,

x = 8 Ã— 4

x = 32

So,

The solution set is: {32}

b)

\(\frac{4}{x}\) = 8

So,

x = \(\frac{4}{8}\)

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

So,

The solution set is: \(\frac{1}{2}\)

c)

10 – x = 7

So,

x = 10 – 7

x = 3

So,

The solution set is: {3}

d)

3x + x = 16

4x = 16

x = \(\frac{16}{4}\)

x = 4

So,

The solution set is: {4}

e)

5x = 0

x = \(\frac{0}{5}\)

x = 0

So,

The solution set is: {0}

f)

12 * x = x * 12

12x = 12x

So,

The solution set is: {All numbers}

g)

0.5 = |x|

We know that,

|x| = x for x > 0

|x| = -x for x < 0

\(\frac{1}{2}\) = |x|

x = \(\frac{1}{2}\) or x = –\(\frac{1}{2}\)

So,

The solution set is: {\(\frac{1}{2}\), –\(\frac{1}{2}\)}

h)

x – 5 = x

x – x = 5

0 = 5

So,

The solution set is: No solution or {}

Hence, from the above,

The letter of the solution set that matches each number sentence is:

**Practice**

Divide.

Question 5.

8.8 Ã· 2 = _______

Answer:

The given expression is:

8.8 Ã· 2

So,

8.8 Ã· 2 = (8 + 0.8) Ã· 2

= (8 Ã· 2) + (0.8 Ã· 2)

= 4 + 0.4

= 4.4

Hence, from the above,

We can conclude that the value of the given expression is: 4.4

Question 6.

0.95 Ã· 5 = _______

Answer:

The given expression is:

0.95 Ã· 5

So,

0.95 Ã· 5 = (0.90 + 0.5) Ã· 5

= (0.90 Ã· 5) + (0.5 Ã· 5)

= 0.18 + 0.1

= 0.19

Hence, from the above,

We can conclude that the value of the given expression is: 0.19

Question 7.

98 Ã· 0.2 = _____

Answer:

The given expression is:

98 Ã· 0.2

So,

98 Ã· 0.2

= 980 Ã· 2

= (1,000 – 20) Ã· 2

= (1,000 Ã· 2) – (20 Ã· 2)

= 500 – 10

= 490

Hence, from the above,

We can conclude that the value of the given expression is: 490

Question 8.

198 Ã· 0.2 = ________

Answer:

The given expression is:

198 Ã· 0.2

So,

198 Ã· 0.2

= 1980 Ã· 2

= (2,000 – 20) Ã· 2

= (2,000 Ã· 2) – (20 Ã· 2)

= 1,000 – 10

= 990

Hence, from the above,

We can conclude that the value of the given expression is: 990

### Everyday Mathematics Grade 6 Home Link 6.3 Answers

**Modeling and Solving Number Stories**

Use bar models to solve these equations. Check your answers.

Question 1.

4a + 12 = 96

Solution: _______

Check: _________

Answer:

The given expression is:

4a + 12 = 96

We know that,

A bar model is a pictorial representation of a problem or concept where bars or boxes are used to represent the known and unknown quantities.

So,

Subtract with 12 on both sides

4a + 12 – 12 = 96 – 12

4a = 84

a = \(\frac{84}{4}\)

a = 21

Now,

The representation of the given expression as a bar model is:

Hence, from the above,

We can conclude that the value of a is: 21

Question 2.

6d + 7 = d + 22

Solution: _______

Check: _________

Answer:

The given expression is:

6d + 7 = d + 22

We know that,

A bar model is a pictorial representation of a problem or concept where bars or boxes are used to represent the known and unknown quantities.

So,

Subtract with 7 on both sides

6d + 7 – 7 = d + 22 – 7

6d = d + 15

6d – d = 15

5d = 15

d = \(\frac{15}{5}\)

d = 3

Now,

The representation of the given expression as a bar model is:

Hence, from the above,

We can conclude that the value of d is: 3

Use bar models to solve the problems.

Question 3.

Jane is 6 years older than twice Martinâ€™s age.

Let s be Martinâ€™s age.

The sum of Janeâ€™s and Martinâ€™s ages is 39.

a)

Write an expression to represent Janeâ€™s age ________________

b)

Write an equation to represent the situation. _____________

c)

How old are Jane and Martin?

Jane: __________

Martin: __________

Answer:

a)

It is given that

Jane is 6 years older than twice Martin’s age

It is also given that s is Martin’s age

So,

The expression to represent Jane’s age is: 2s + 6

b)

It is also given that

The sum of Jane’s and Martin’s age is 39

So,

Jane’s age + Martin’s age = 39

From part (a),

Jane’s age = 2s + 6

So,

2s + 6 +s = 39

3s + 6 = 39

Subtract with 6 on both sides

So,

3s + 6 – 6 = 39 – 6

3s = 33

s = \(\frac{33}{3}\)

s = 11 years

So,

Jane’s age = 2s + 6

= 2 (11) + 6

= 22 + 6

= 28 years

Hence, from the above,

We can conclude that

The age of Jane is: 28 years

The age of Martin is: 11 years

Try This

Question 4.

Dan is thinking of a number. He doubles his number and adds 15. He multiplies his number by 5 and gets the same answer. Let n be Danâ€™s number.

a)

Write an equation to represent the situation. ____________

b)

Danâ€™s number: __________

Answer:

a)

It is given that

Dan doubles his number and adds 15.

It is also given that he multiplies his number by 5 and gets the same answer

Now,

Let n be Dan’s number

So,

So,

The representation of “Dan doubles his number and adds 15” is: 2n + 15

The representation of “He multiplies his number byÂ and gets the same answer” is: 5n = 2n + 15

Hence,

The equation to represent the above situation is: 5n = 2n + 15

b)

From part (a),

The equation that represents the given siyuation is:

5n = 2n + 15

Subtract with 2n on both sides

So,

5n – 2n = 2n + 15 – 2n

3n = 15

n = \(\frac{15}{3}\)

n = 5

Hence, from the above,

We can conclude that Dan’s number is: 5

**Practice**

Solve.

Question 5.

5 Ã— (6.8 – 2) = ________

Answer:

The given expression is:

5 Ã— (6.8 – 2)

= 5 Ã— 4.8

= 24.0

= 24

Hence, from the above,

We can conclude that the value of the given expression is: 24

Question 6.

8 Ã· 2 Ã— 3.5 = ________

Answer:

The given expression is:

8 Ã· 2 Ã— 3.5

According to the BODMAS rule,

We have to first make the division and then multiplication (D – Division, M – Multiplication in the BODMAS)

So,

\(\frac{8}{2}\) Ã— 3.5

= 4 Ã— 3.5

= 14.0

= 14

Hence, from the above,

We can conclude that the value of the given expression is: 14

Question 7.

9.43 – 4.5 + 1.7 = _________

Answer:

The given expression is:

9.43 – 4.5 + 1.7

According to the BODMAS rule,

We have to first make the addition and then subtraction (A – Addition, S – Subtraction in the BODMAS)

So,

(9.43 + 1.7) – 4.5

= 11.13 – 4.5

= 6.63

Hence, from the above,

We can conclude that the value of the given expression is: 6.63

### Everyday Math Grade 6 Home Link 6.4 Answer Key

**Pan-Balance Problems**

In each figure, the two pans are balanced. Solve these pan-balance problems.

Question 1.

One triangle weighs as much as __________ squares.

Answer:

In the given figure,

Let each triangle be x

Let each square be y

It is given that the two pans are balanced

So,

LHS = RHS

Now,

2x + 2y = 8y

Subtract with 2y on both sides

2x + 2y – 2y = 8y – 2y

2x = 6y

\(\frac{x}{y}\) = ]latex]\frac{6}{2}[/latex]

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

x = 3y

1 Triangle = 3 squares

So,

From the above expression,

We can say that one triangle weighs as much as 3 squares

Hence, from the above,

We can conclude that 1 Triangle weighs as much as 3 Squares

Question 2.

One ball weighs as much as _______ coin(s).

Note: Remember that 2 means

Answer:

In the given figure,

Let each circle be x

Let each coin be y

It is given that the two pans are balanced

So,

LHS = RHS

Now,

2x + 7y = 4y + 5x

Rearrange the like terms at one side

So,

5x – 2x = 7y – 4y

3x = 3y

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

\(\frac{x}{y}\) = \(\frac{1}{1}\)

x = y

So,

1 Circle = 1 Coin

So,

From the above expression,

We can say that one circle weighs as much as 1 coin

Hence, from the above,

We can conclude that 1 Circle weighs as much as 1 Coin

Question 3.

Two cantaloupes weigh as much as _________ apples.

Answer:

In the given figure,

Let each cantaloupe be x

Let each apple be y

It is given that the two pans are balanced

So,

LHS = RHS

Now,

\(\frac{1}{2}\)x = 9y

Multiply with 2 on both sides

So,

\(\frac{2}{2}\)x = 9 (2)y

x = 18y

Multiply with 2 again on both sides

So,

2x = 18 (2)y

2x = 36y

So,

2 Cantaloupes = 36 Apples

So,

From the above expression,

We can say that two cantaloupes weigh as much as 36 apples

Hence, from the above,

We can conclude that 2 Cantaloupes weigh as much as 36 Apples

Question 4.

One cube weighs as much as _________ coin(s).

Answer:

In the given figure,

Let each cube be x

Let each coin be y

It is given that the two pans are balanced

So,

LHS = RHS

Now,

4x + 7y = 13y + x

Rearrange the like terms at one side

So,

4x – x = 13y – 7y

3x = 6y

\(\frac{x}{y}\) = \(\frac{6}{3}\)

\(\frac{x}{y}\) = \(\frac{2}{1}\)

x = 2y

So,

1 Cube = 2 Coins

So,

From the above expression,

We can say that one cube weighs as much as 2 coins

Hence, from the above,

We can conclude that 1 Cube weighs as much as 2 Coins

Question 5.

One cube weighs as much as _______ marbles.

Answer:

In the given figure,

Let each cube be x

Let each marble be y

It is given that the two pans are balanced

So,

LHS = RHS

Now,

3x + 10y = 13y + x

Rearrange the like terms at one side

So,

3x – x = 13y – 10y

2x = 3y

\(\frac{x}{y}\) = \(\frac{3}{2}\)

x = \(\frac{3}{2}\)y

So,

1 Cube = \(\frac{3}{2}\) Marbles

So,

From the above expression,

We can say that one cube weighs as much as \(\frac{3}{2}\) marbles

Hence, from the above,

We can conclude that 1 Cube weighs as much as \(\frac{3}{2}\) Marbles

**Practice**

Solve.

Question 6.

1 = \(\frac{3}{5}\) Ã— ________

Answer:

Let the missing term be x in the given expression

So,

1 = \(\frac{3}{5}\) Ã— x

Multiply by \(\frac{5}{3}\) on both sides

So,

x = \(\frac{5}{3}\)

Hence, from the above,

We can conclude that the missing term in the given expression is: \(\frac{5}{3}\)

Question 7.

\(\frac{1}{6}\) Ã— ______ = 1

Answer:

Let the missing term in the given expreesion be x

So,

\(\frac{1}{6}\) Ã— x = 1

Multiply with 6 on both sides

\(\frac{6}{6}\)x = 1 (6)

So,

x = 6

Hence, from the above,

We can conclude that the missing term in the given expression is: 6

Question 8.

________ Ã— 1 \(\frac{1}{2}\) = 1

Answer:

Let the missing term in the given expression be x

So,

x Ã— 1\(\frac{1}{2}\) = 1

We know that,

1\(\frac{1}{2}\) = \(\frac{3}{2}\)

So,

x Ã— \(\frac{3}{2}\) = 1

Multiply with \(\frac{2}{3}\) on both sides

So,

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

Hence, from the above,

We can conclude that the missing term in the given expression is: \(\frac{2}{3}\)

Question 9.

1 = 1 \(\frac{5}{6}\) Ã— _________

Answer:

Let the missing term in the given expression be x

So,

1 = 1\(\frac{5}{6}\) Ã— x

We know that,

1\(\frac{5}{6}\) = \(\frac{11}{6}\)

So,

1 = \(\frac{11}{6}\) Ã— x

Multiply with \(\frac{6}{11}\) on both sides

So,

x = \(\frac{6}{11}\)

Hence, from the above,

We can conclude that the missing term in the given expression is: \(\frac{6}{11}\)

### Everyday Mathematics Grade 6 Home Link 6.5 Answers

**Solving Pan-Balance Problems**

Question 1.

These two pan balances are in perfect balance.

a. Use the relationships in the pan balances shown above to determine which of the pan balances below are balanced. Circle the ones that are in balance.

Answer:

b. For any pan balance above that you did not a circle, add, or cross out objects to balance the pans.

Answer:

Let each hexagon be x

Let each semicircle be y

Let each triangle be z

Now,

It is given that these 2 pan balances are in perfect balance

So,

x = 2y —–> From pan 1 ———> (1)

3z = y ——> From Pan 2 ———> (2)

a)

The given pan balances are:

Now,

From Pan 1,

2x = 2y + x

Subtract with x on both sides

2x – x = 2y + x – x

x = 2y

From Pan 2,

2x + y = 2y

Subtract with y on both sides

2x + y – y = 2y – y

2x = y

From Pan 3,

3z + x = 2z + y

Rearrange the terms at both sides

So,

3z – 2z = y – x

z = y – x

From Pan 4,

3z + 2y = x + y

Rearrange the terms at both sides

So,

x – 3z = 2y – y

x – 3z = y

2y – 3z = y (From eq (2))

3z = 2y – y

y = 3z

Now,

When we compare eq (1) and eq (2) with the equations of 4 pans,

We can observe that Pan 1 and Pan 4 are the balanced

Hence, from the above,

We can conclude that the pans that are in balance are:

b)

From part (a),

We can observe that Pan 2 and Pan 4 are not balanced as of eq (1) and eq (2)

Now,

The pan equations of Pan2 and Pan 3 are:

2x = y —–> Pan 2’s equation

z = y – x —–> Pan 3’s equation

Now,

To balance the Pan 2’s equation,

We have to add x and y at the right side of the Pan 2’s equation to compare with eq (1)

So,

2x = y + x + y

Hence,

For Pan 2 to be balanced,

We have to add a hexagon and a semicircle at the right side of the pan 2

Now,

To balance the Pan 3’s equation,

We have to add 2z at the left side and x at the right side of the Pan 3’s equation to compare with eq (2)

So,

2z + z = y – x + x

Hence,

For Pan 3 to be balanced,

We have to add 2 triangles at the left side and 1 hexagon at the right side of the pan 3

Question 2.

Find the value of the missing number that will balance each set of pans below. The same number is missing from both sides of a pan balance.

Answer:

The given pan balances are:

a)

Let the missing numer be x

Since the pans are balanced,

LHS = RHS

So,

15 Ã— x = 5 Ã— x + 30

15x = 5x + 30

15x – 5x = 30

10x = 30

x = \(\frac{30}{10}\)

x = 3

Hence, from the above,

We can conclude that the missing number is: 3

b)

Let the missing number be x

Since the pans are balanced,

LHS = RHS

So,

x Ã· 6 = x – 20

Multiply with 6 on both sides

x =Â 6(x – 20)

x = 6x – 120

6x – x = 120

5x = 120

x = \(\frac{120}{5}\)

x = 24

Hence, from the above,

We can conclude that the missing number is: 24

Question 3.

Makeup two of your own missing-number pan balances.

Fill in the missing numbers for the pan-balance problems you made.

Answer:

a)

Let the missing number be y

Since the pan is balanced,

LHS = RHS

So,

The equation is:

5y = 20

y = \(\frac{20}{5}\)

y = 4

Hence, from the above,

We can conclude that the missing number is: 4

b)

Let the missing number be x

Since the pan is balanced,

LHS = RHS

So,

The equation is:

x + 3 = 5

Subtract with 3 on both sides

x + 3 – 3 = 5 – 3

x = 2

Hence, from the above,

We can conclude that the missing number is: 2

**Practice**

Solve.

Question 4.

4.3 Ã— 7 = _____

Answer:

The given expression is:

4.3 Ã— 7

= (4 + 0.3) Ã— 7

= (4 Ã— 7 ) + (0.3 Ã— 7)

= 28 + 2.1

= 30.1

Hence, from the above,

We can conclude that the value of the given expression is: 30.1

Question 5.

0.2 Ã— 1.5 = ______

Answer:

The given expression is:

0.2 Ã— 1.5

= 0.2 Ã— (1 + 0.5)

= (0.2 Ã— 1) + (0.2 Ã— 0.5)

= 0.2 + 0.1

= 0.3

Hence, from the above,

We can conclude that the value of the given expression is: 0.3

Question 6.

1.9 Ã— 2.3 = ________

Answer:

The given expression is:

1.9 Ã— 2.3

= (2 – 0.1) Ã— (2 + 0.3)

= (2 Ã— 2) + (2 Ã— 0.3) – (0.1 Ã— 2) – (0.1 Ã— 0.3)

= 4 + 0.6 – 0.2 – 0.03

= 4 + 0.37

= 4.37

Hence, from the above,

We can conclude that the value of the given expression is: 4.37

### Everyday Math Grade 6 Home Link 6.6 Answer Key

**Simplifying Expressions**

Question 1.

Simplify each expression by combining the like terms.

a. 42x + 12xÂ Â Â Â __________

Answer:

The given expression is:

42x + 12x

= x (42 + 12)

= x(54)

= 54x

Hence,

42x + 12x = 54x

b. 17w – 8wÂ Â Â Â Â __________

Answer:

The given expression is:

17w – 8w

= w (17 – 8)

= w (9)

= 9w

Hence,

17w – 8w = 9w

c. 25.42e – 23.3eÂ Â __________

Answer:

The given expression is:

25.42e – 23.3e

= e (25.42 – 23.3)

= e (2.12)

= 2.12e

Hence,

25.42e – 23.3e = 2.12e

d. 88h + 30.5hÂ Â Â Â __________

Answer:

The given expression is:

88h + 30.5h

= h (88 + 30.5)

= h (118.5)

= 118.5h

Hence,

88h + 30.5h = 118.5h

Question 2.

Simplify. Check that your expressions are equivalent.

a. 12m + 24m

Answer:

The given expression is:

12m + 24m

= m (12 + 24)

= m (36)

= 36m

Hence,

12m + 24m = 36m

b. 90a – 30a

Answer:

The given expression is:

90a – 30a

= a (90 – 30)

= a (60)

= 60a

Hence,

90a – 30a = 60a

c. 14b + 15b + 8

Answer:

The given expression is:

14b + 15b + 8

= b (14 + 15) + 8

= b (29) + 8

= 29b + 8

Hence,

14b + 15b + 8 = 29b + 8

d. 58d + 25 – 22d

Answer:

The given expression is:

58d + 25 – 22d

= d ( 58 – 22) + 25

= d (36) + 25

= 36d + 25

Hence,

58d + 25 – 22d = 36d + 25

e. 3(14 + 15f) + 79

Answer:

The given expression is:

3 (14 + 15f) + 79

= 3 (14) + 3 (15f) + 79

= 45f + 42 + 79

= 45f + 121

Hence,

3 (14 + 15f) + 79 = 45f + 121

f. 20(18 + 5t) – 47 + 28t

Answer:

The given expression is:

20 (18 + 5t) – 47 + 28t

= 20 (18) + 20 (5t) – 47 + 28t

= 360 + 100t – 47 + 28t

= t (100 + 28) + 360 – 47

= t (128) + 313

= 128t + 313

Hence,

20 (18 + 5t) – 47 + 28t = 128t + 313

**Practice**

Insert =, <, or >.

Question 3.

-5 _____ -10

Answer:

The given expression is:

-5 ______ -10

We know that,

5 is less than 10

Hence, from the above,

We can conclude that

-5 > -10

Question 4.

0.23 _____ 0.009

Answer:

The given expression is:

0.23 _____ 0.009

We know that,

0.23 is greater than 0.009

Hence, from the above,

We can conclude that

0.23 > 0.009

Question 5.

-11 ______ 1

Answer:

The given expression is:

-11 _____ 1

We know that,

-11 is less than 1

Hence, from the above,

We can conclude that

-11 < 1

Question 6.

0.092 ______ 0.0920

Answer:

The given expression is:

0.092 _____ 0.0920

We know that,

0.92 is equal to 0.0920 (Since if the last digit is 0, then it does not have any value after the decimal point)

Hence, from the above,

We can conclude that

0.092 = 0.0920

### Everyday Mathematics Grade 6 Home Link 6.7 Answers

**Exploring Equivalent Equations**

Question 1.

Use the Commutative Property (turn-around rule) to create an equivalent expression in which like terms are next to each other.

a. 12k + 2 Ã— 3 + 3k + 1 __________

Answer:

The given expression is:

12k + 2 Ã— 3 + 3k + 1

Using the Commutative property (Turn -around rule),

The representation of the given expression is:

12k + 3k + 3 Ã— 2 + 1

b. Combine like terms from Problem 1a and write a simplified equivalent expression.

Answer:

From paroblem 1a,

The given expression is:

12k + 2 Ã— 3 + 3k + 1

= 12k + 6 + 3k + 1

= k (12 + 3) + 6 + 1

= k (15) + 7

= 15k + 7

Hence, from the above,

We can conclude that the simplified expression for the expression mentioned in problem 1a is: 15k + 7

Write the expressions in the simplest form.

Question 2.

4a + 5 – a + 10Â Â Â __________

Answer:

The given expression is:

4a + 5 – a + 10

= a (4 – 1) + 5 + 10

= a (3) + 15

= 3a + 15

Hence, from the above,

We can conclude that the simplified expression for the given expression is: 3a + 15

Question 3.

10(b + 5) + 15 + wÂ Â _________

Answer:

The given expression is:

10 (b + 5) + 15 + w

= 10 (b) + 10 (5) + 15 + w

= 10b + 50 + 15 + w

= 10b + w + 65

Hence, from the above,

We can conclude that the simplified expression for the given expression is: 10b + w + 65

For Problems 4â€“5, identify the equations that are equivalent to the given equation. Circle ALL that apply.

Question 4.

6x + (7 – 2) Ã— x = 8 + 3x – 4

A. 6x + 5x = 8 + 3x – 4

B. 6x + (7 – 2) Ã— x = 3x + 12

C. 11x = 3x + 4

Answer:

The given expression is:

6x + (7 – 2) Ã— x = 8 + 3x – 4

6x + 5 Ã— x = 8 + 3x – 4

6x + 5x = 8 + 3x – 4

11x = 3x + 4

Hence, from the above,

We can conclude that the equations that are equivalent to the given equation are: A and C

Question 5.

b + 2b – 10 = 10(b + 5) + 15 + b

A. 3b – 10 = 10(b + 5) + 15 + b

B. b + 2b – 10 = 10b + 50 + 15b

C. b(1 + 2) – 10 = 10b + 50 + 15 + b

Answer:

The given expression is:

b + 2b – 10 = 10 (b + 5) + 15 + b

3b – 10 = 10 (b + 5) + 15 + b

3b – 10 = 10 (b) + 10 (5) + 15 + b

3b – 10 = 10b + 50 + 15 + b

b (1 + 2) – 10 = 10b + 50 + 15 + b

Hence, from the above,

We can conclude that the equivalent equations for the given equation are: A and C

Question 6.

Use a bar model or pan-balance model to solve one of the equations you circled in Problem 4.

x = _________

Answer:

One of the quations that is circled in problem 4 is:

11x = 3x + 4

Now,

Subtract with 3x on both sides

11x – 3x = 3x + 4 – 3x

8x = 4

x = \(\frac{4}{8}\)

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

Hence,

The representation of bar model for the equation we have taken is:

**Practice
**Multiply

Question 7.

2\(\frac{1}{3}\) Ã— \(\frac{3}{7}\) = _________

Answer:

The given expression is:

2\(\frac{1}{3}\) Ã— \(\frac{3}{7}\)

= \(\frac{7}{3}\) Ã— \(\frac{3}{7}\)

= \(\frac{7 Ã— 3}{3 Ã— 7}\)

= \(\frac{1}{1}\)

= 1

Hence, from the above,

We can conclude that the value of the given expression is: 1

Question 8:

1\(\frac{2}{3}\) Ã— 2\(\frac{1}{2}\) = __________

Answer:

The given expression is:

1\(\frac{2}{3}\) Ã— 2\(\frac{1}{2}\)

= \(\frac{5}{3}\) Ã— \(\frac{5}{2}\)

= \(\frac{5 Ã— 5}{3 Ã— 2}\)

= \(\frac{25}{6}\)

Hence, from the above,

We acn conclude that the value of the given expression is: \(\frac{25}{6}\)

Question 9.

_______ = 3\(\frac{7}{10}\) Ã— 2\(\frac{1}{4}\)

Answer:

The given expression is:

3\(\frac{7}{10}\) Ã— 2\(\frac{1}{4}\)

= \(\frac{37}{10}\) Ã— \(\frac{9}{4}\)

= \(\frac{37 Ã— 9}{10 Ã— 4}\)

= \(\frac{333}{40}\)

Hence, from the above,

We can conclude that the value of the given expression is: \(\frac{333}{40}\)

Question 10:

5\(\frac{3}{4}\) Ã— 4\(\frac{2}{5}\) = ______

Answer:

The given expression is:

5\(\frac{3}{4}\) Ã— 4\(\frac{2}{5}\)

= \(\frac{23}{4}\) Ã— \(\frac{22}{5}\)

= \(\frac{23 Ã— 22}{4 Ã— 5}\)

= \(\frac{506}{20}\)

Hence, from the above,

We can conclude that the value of the given expression is: \(\frac{506}{20}\)

### Everyday Math Grade 6 Home Link 6.8 Answer Key

**Comparing Racing Times**

Katya runs at a rate of 6.25 meters per second. Her younger cousin, Liova, runs 2.5 meters per second. Because Katya runs faster than Liova, she gives Liova a 100-meter head start in a 200-meter race.

Question 1.

Using the variable t to represent the number of seconds, write two expressionsâ€”one for Katya and one for Liovaâ€”that model how far from the start line they will be after t seconds.

Expression for Katya: _______________

Expression for Liova: _________________

Answer:

a)

It is given that Katya runs at a rate of 6.25 meters per second and her younger cousin, Liova, runs 2.5 meters per second.

We know that,

Speed = \(\frac{Distance}{Time}\)

It is given that use the variable t for time

So,

The expression that represents how far Katya is from the start line is:

Distance (d) = Speed (Time)

d = 6.25t

Hence,

The expression of how far Katya traveled from the start line is:

d = 6.25t

b)

It is also given that Katya runs faster than Liova, and she gives Liova a 100-meter head start in a 200-meter race.

So,

The expression that represents how far Liova is from the start line is:

Distance (d) = Speed (Time)

d = 100 + 2.5t

Hence,

The expression of how far Liova is from the start line is:

d = 100 + 2.5t

Question 2.

Use your expressions from Problem 1 to figure out who will win the race.

Show your work and explain your answer.

Winner: _____________

Answer:

From problem 1,

We know that,

Katya runs 200 meters and she gives Liova a 100 meters headstart

Now,

The expression of how far Katya is from the start line is:

d = 6.25t

Now,

t = \(\frac{d}{6.25}\)

t = \(\frac{200}{6.25}\)

t = 32 seconds

The expression of how far Liona is from the start line is:

d = 100 + 2.5t

Now,

200 = 100 + 2.5t

2.5t = 200 – 100

2.5t = 100

t = \(\frac{100}{2.5}\)

t = 40 seconds

When we compare the time taken by Katya and Liova to complete the race,

We can observe that

The time took by Katya < The time took by Liova

Hence, from the above,

We can conclude that Katya wins the race

**Practice**

Solve.

Question 3.

5% of 66 is _______.

Answer:

The given expression is:

5% of 66

So,

5% of 66

= \(\frac{5}{100}\) Ã— 66

= 0.05 Ã— 66

= 3.30

Hence, from the above,

We can conclude that the value of the given expression is: 3.30

Question 4.

18% of 50 is ___________ .

Answer:

The given expression is:

18% of 50

So,

18% of 50

= \(\frac{18}{100}\) Ã— 50

= 0.18 Ã— 50

= 9

Hence, from the above,

We can conclude that the value of the given expression is: 9

Question 5.

45% of 120 is _______.

Answer:

The given expression is:

45% of 120

So,

45% of 120

= \(\frac{45}{100}\) Ã— 120

= \(\frac{45}{10}\) Ã— 12

= \(\frac{45 Ã— 12}{10}\)

= 54

Hence, from the above,

We can conclude that the value of the given expression is: 54

Question 6.

90% of 8,000 is _________.

Answer:

The given expression is:

90% of 8,000

So,

90% of 8,000

= \(\frac{90}{100}\) Ã— 8,000

= 90 Ã— 80

= 7,200

Hence, from the above,

We can conclude that the value of the given expression is: 7,200

### Everyday Mathematics Grade 6 Home Link 6.9 Answers

Using Inverse Operations

Question 1.

Linda has a secret number. She doubles the number, adds 5, and then subtracts 7. Her result is 8. What was her original secret number? Explain what you did to find her secret number.

Answer:

It is given that Linda has a secret number.

It is also given that she doubles the number, adds 5, and then subtracts 7. Her result is 8.

Now,

Let the secret number of Linda be x

So,

Double of Linda’s secret number is: 2x

So,

2x + 5 – 7 = 8

2x – 2 = 8

Add with 2 on both sides

2x – 2 + 2 = 8 + 2

2x = 10

x = \(\frac{10}{2}\)

x = 5

Hence, from the above,

We can conclude that the secret number of Linda is: 5

The process of how I find Linda’s secret number:

I subtracted 5 from -7 to get -2 and added 2 on both sides to make 0 on the left side and make 10 on the right side and I make half of the value of 10 to get the result i.e., 5

For Problems 2â€“5, solve the equations using the inverse-operations strategy. Show all of your steps and check your work.

Question 2.

257 = a – 105

Check:

Answer:

The given expression is:

257 = a – 105

Add with 105 on both sides

So,

257 + 105 = a – 105 + 105

362 = a

Hence, from the above,

We can conclude that the value of ‘a’ from the given expression is: 362

Question 3.

12 = \(\frac{r}{4}\)

Check:

Answer:

The given expression is:

12 = \(\frac{r}{4}\)

Multiply with 4 on both sides

So,

12 (4) = \(\frac{4r}{4}\)

r = 48

Hence, from the above,

We can conclude that the value of r from the given expression is: 48

Question 4.

j + 3 \(\frac{3}{4}\) = 8

Check:

Answer:

The given expression is:

j + 3\(\frac{3}{4}\) = 8

We know that,

3\(\frac{3}{4}\) = \(\frac{15}{4}\)

So,

j + \(\frac{15}{4}\) = 8

Subtract with \(\frac{15}{4}\) on both sides

So,

j + \(\frac{15}{4}\) – \(\frac{15}{4}\) = 8 – \(\frac{15}{4}\)

j = \(\frac{17}{4}\)

Hence, from the above,

We can conclude that the value of j from the given expression is: \(\frac{17}{4}\)

Question 5.

6.72 = 4u

Check:

Answer:

The given expression is:

6.72 = 4u

Divide by 4 on both sides

So,

\(\frac{6.72}{4}\) = \(\frac{4u}{4}\)

1.68 = u

u = 1.68

Hence,f rom the above,

We can conclude that the value of u from the given expression is: 1.68

**Practice**

Write a unit rate for each rate below.

Question 6.

55 pages in 10 minutes _________

Answer:

The given rate is:

55 pages in 10 minutes

So,

The number of pages in 1 minute (Unit rate) = \(\frac{55}{10}\)

= 5.5

Hence, from the above,

We can conclude that the unit rate is: 5.5 pages per minute

Question 7.

$46.50 for 6 hours _______

Answer:

The given rate is:

$46.50 for 6 hours

So,

The number of dollars per hour (Unit rate) = \(\frac{46.50}{6}\)

= 7.75

Hence, from the above,

We can conclude that the unit rate is: $7.75 per hour

### Everyday Math Grade 6 Home Link 6.10 Answer Key

**Solving Pan-Balance Equations**

Question 1.

Build an equation with two operations that are equivalent to the equation k = 19. Record the operations that you use to create each equation below.

Original equation:

Operation (in words)

Answer:

Question 2.

Check that 19 is a solution to your equations.

Answer:

From Question 1,

The equations are:

k + 1 = 20 —— (1)

5k = 95 ——— (2)

Now,

Put k = 19

So,

In eq (1),

19 + 1 = 20

20 = 20

In eq (2),

5 (19) = 95

95 = 95

Hence, from the baove,

We can conclude that 19 is a solution to your equations

Question 3.

Find the mistake in the work below.

Original pan-balance equation:

Operation (in words)

Subtract 10.

Divide by 2.

Describe the mistake and how to correct it.

Answer:

The mistake is:

In the original Pan-Balance equation,

We have to subtract with 10 on both sides

But, it is added on the right side of the Pan-Balance equation

Nw,

The given original Pan-Balance equation is:

2x + 10 = 28

Subtract with 10 on both sides

So,

2x + 10 – 10 = 28 – 10

2x = 18

x = \(\frac{18}{2}[/altex]

x = 9

Hence, from the above,

We can conclude that after correcting the mistake in the original Pan-Balance equation,

The value of x is: 9

Question 4.

Record the operations you use to create equivalent equations and solve the equation.

Original equation:

Operation (in words)

Answer:

### Everyday Mathematics Grade 6 Home Link 6.11 Answers

**Solving Multistep Home Equations**

Solve the equations below. Use each of these strategies once:

A. Trial and Error

B. Bar Model

C. Pan-Balance Model

D. Inverse-Operations Strategy

Plan ahead to make sure you use the strategy or model that you think works better for each equation.

Question 1.

54 = 4m + 2

Strategy: _________

Solution: ________

Answer:

The given equation is:

54 = 4m + 2

Now,

The strategy that is used to solve the given equation is: Bar Model

So,

The Bar Model representation for the given equation is:

Hence, from the above,

We can conclude that the value of m for the given equation is: 13

Question 2.

6n + 8 = 10n + 4

Strategy: _________

Solution: ________

Answer:

The given equation is:

6n + 8 = 10n + 4

Now,

The strategy that is used to solve the given equation is: Inverse Operations Strategy

So,

Now,

Subtract with 8 on both sides

6n + 8 – 8 = 10n + 4 – 8

6n = 10n – 4

Rearrange the like terms on both sides

So,

10n – 6n = 4

4n = 4

n = [latex]\frac{4}{4}\)

n = 1

Hence, from the above,

We can conclude that the value of n for the given equation is: 1

Question 3.

3p + 3 = 2p + 4.5

Strategy: _________

Solution: ________

Answer:

The given equation is:

3p + 4 = 2p + 4.5

Now,

The strategy that is used to solvethe given equation is Pan-Balance Model

So,

Now,

The representation of the Pan-Balance Model for the given equation is;

Hence, from the above,

We can conclude that the value of p is: 1.5

Question 4.

\(\frac{2}{3}\)q + 8 = q – 10

Strategy: _________

Solution: ________

Answer:

The given equation is:

\(\frac{2}{3}\)q + 8 = q – 10

Now,

The strategy that is used to solve the given equation is Trial -error method

So,

Now,

By using trial and error method,

Let the value of q be: 54

So,

\(\frac{2}{3}\) (54) + 8 = 54 – 10

\(\frac{2 Ã— 54}{3}\) + 8 = 44

36 + 8 = 44

44 = 44

Hence, from the above,

We can conclude that the value of q from the given equation is: 54

**Practice**

List the numbers in order from least to greatest.

Question 5.

\(\frac{9}{4}\), 2.35, \(\frac{1}{8}\), 1.5, \(\frac{3}{8}\)

Answer:

The given numbers are:

\(\frac{9}{4}\), 2.35, \(\frac{1}{8}\), 1.5, \(\frac{3}{8}\)

Now,

The representation of all the numbers in the decimal form is:

2.25, 2.35, 0.125, 1.5, 0.375

Hence, from the above,

The list of numbers in order from the least to the greatest is:

0.125, 0.375, 1.5, 2.25, 2.35