# ISOLATING THE VARIABLE WORKSHEET

Problems 1-5 : In each case, solve the equation by isolating the variable.

Problem 1 :

5x - 3 = 22

Problem 2 :

²ˣ⁄₃ + 5 = 13

Problem 3 :

ˣ⁄₃ + ⁷⁄₂ = 0

Problem 4 :

2(p + 3) - 3(1 - 5p) = 3(3p + 7) - (9p - 16)

Problem 5 :

8(y - 1) + y(y - 1) = y2 + 2(3y + 5)

Problem 6 :

If k = -5 and ⁽²ˣ ⁻ ⁷⁾⁄₃ = k, find the value of x.

Problem 7 :

If the circumference of a circle is 24π cm, find the radius.

Problem 8 :

If the perimeter of a rectangle is 24 in., find the length of the rectangle in terms of its width.

Problem 9 :

If 6x - 7 = -5. find the value of 12x + 1 without isolating the variable x.

Problem 10 :

If 14x + 3 = 25. find the value of 21x - 7 without isolating the variable x.

Problem 11 :

If 5 is decreased from 7 times of a number, the result would be 37. Find the number.

Problem 12 :

13 is added to 5 times of a number results 3. Find the number results when 5 is added to 7 times of the same number.

Problem 13 :

It is given that 3(a + 5) = 2(b + 9) - 3. Find the value of ᵃ⁄b.

Problem 14 :

Adam is thrice as old as his son now. 5 years ago, Adam's age was 1 more than four times of his son's age. Find the present age of Adam and son.

5x - 3 = 22

5x = 25

Divide both sides by 5.

x = 5

²ˣ⁄₃ + 5 = 13

Subtract 5 from both sides.

²ˣ⁄₃ = 8

Multiply both sides by 3.

2x = 24

Divide both sides by 2.

x = 12

ˣ⁄₃ + ⁷⁄₂ = 0

Least common multiple of the decominators (3, 2) is 6.

Multiply both sides by the equation by 6 to get rid of the denominators 3 and 2.

6(ˣ⁄₃ + ⁷⁄₂) = 6(0)

6(ˣ⁄₃) + 6(⁷⁄₂) = 0

2x + 3(7) = 0

2x + 14 = 0

Subtract 14 from both sides.

2x = -14

Divide both sides by 2.

x = -7

2(p + 3) - 3(1 - 5p) = 3(3p + 7) - (9p - 16)

Using Distributive Property,

2p + 6 - 3 + 15p = 9p + 21 - 9p + 16

Combine the like terms.

17p + 3 = 37

Subtract 3 from both sides.

17p = 34

Divide both sides by 17.

p = 2

8(y - 1) + y(y - 1) = y2 + 2(3y + 5)

Using Distributive Property,

8y - 8 + y2 - y = y2 + 6y + 10

Combine the like terms.

y2 + 7y - 8 = y2 + 6y + 10

Subtract y2 from both sides.

7y - 8 = 6y + 10

Subtract 6y from both sides.

y - 8 = 10

y = 18

⁽²ˣ ⁻ ⁷⁾⁄₃ = k

Substitute k = -5.

⁽²ˣ ⁻ ⁷⁾⁄₃ = -5

Multiply both sides by 3.

2x - 7 = -15

2x = -8

Divide both sides by 2.

x = -4

Circumference of the circle = 24π cm

2πr = 24π

r = 12

The radius of the circle is 12 cm.

Perimeter of the rectangle = 24 in.

2(l + w) = 24

Divide both sides by 2.

l + w = 12

Subtract w from both sides.

l = 12 - w

length = 12 - width

6x - 7 = -5

6x = 2

Multiply both sides by 2.

12x = 4

12x + 1 = 5

14x + 3 = 25

Subtract 3 from both sides.

14x = 22

Divide both sides by 2.

7x = 11

Multiply both sides by 3.

21x = 33

Subtract 7 from both sides.

21x - 7 = 26

Let x be the required number.

Given : Decreasing 5 from 7 times of a number results 37.

7x - 5 = 37

7x = 42

Divide both sides by 7.

x = 6

The required number is 2.

Let y be the number.

Given : 13 is added to 5 times of a number results 12.

5y + 13 = 3

Subtract 13 from both sides.

5y = -10

Divide both sides by 5.

y = -2

Multiply both sides by 7.

7y = -14

7y + 5 = -9

3(a + 5) = 2(b + 9) - 3

Using Distributive Property,

3a + 15 = 2b + 18 - 3

3a + 15 = 2b + 15

Subtract 15 from both sides.

3a = 2b

Divide both sides by b.

³ᵃ⁄b = 2

Multiply both sides by .

ᵃ⁄b =

Let x be the present of the son.

Then the present age of Adam is 3x.

Age of the son 5 years ago = x - 5

Age of Adam 5 years ago = 3x - 5

Given : Adam's age was 1 more than four times of his son's age.

3x - 5 = 4(x - 5) + 1

Using Distributive Property,

3x - 5 = 4x - 20 + 1

3x - 5 = 4x - 19

Subtract 4x from both sides.

-x - 5 = -19

-x = -14

Multiply both sides by -1.

x = 14

3x =  3(14) = 42

The present age of Adam is 42 years and that of his son is 14 years.

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