# SOLVING EQUATIONS WITH VARIABLE IN DENOMINATOR WORKSHEET

Problems 1-9 : Solve for x.

Problem 1 :

³⁄₍ₓ ₊ ₂₎ + ¹⁄ₓ = 0

Problem 2 :

¹⁄₆ₓ + ¾ = ¹⁄₃ₓ

Problem 3 :

¹⁄ₓ = ⁶⁄₅ₓ + 1

Problem 4 :

⁽ˣ ⁻ ⁴⁾⁄₅ₓ = ¹⁄₅ₓ + 1

Problem 5 :

¹⁄₍ₓ ₋ ₈₎ - 1 = ⁷⁄₍ₓ ₋ ₈₎

Problem 6 :

1 = ⁽ˣ ⁺ ²⁾⁄₍ₓ ₋ ₄₎⁽⁷ˣ ⁻ ⁴²⁾⁄₍ₓ ₋ ₄₎

Problem 7 :

ˣ⁄₍ₓ ₊ ₈₎ = 1 + ⁶⁄₍ₓ ₊ ₁₎

Problem 8 :

⁽ˣ ⁻ ²⁾⁄₍ₓ ₊ ₃₎ - 1 = ³⁄₍ₓ ₊ ₂₎

Problem 9 :

⁵⁄₍ₓ ₋ ₅₎ = ³⁄₍ₓ ₊ ₅₎

Problem 10 :

Difference between 5 and reciprocal of a positive integer is ¹⁴⁄₃. Find the positive integer.

³⁄₍ₓ ₊ ₂₎ + ¹⁄ₓ = 0

The equation above has two denominators.

(x + 2) and x

To get rid of the first denominator (x + 2), multiply both sides of the equation by (x + 2).

(x + 2)[³⁄₍ₓ ₊ ₂₎ + ¹⁄ₓ] = (x + 2)(0)

(x + 2)[³⁄₍ₓ ₊ ₂₎] + (x + 2)(¹⁄ₓ) = 0

3 + ⁽ˣ ⁺ ²⁾⁄ₓ = 0

Multiply both sides of the equation by x to get rid of the denominator x.

x[3 + ⁽ˣ ⁺ ²⁾⁄ₓ] = x(0)

(x + 2)(3) + x[⁽ˣ ⁺ ²⁾⁄ₓ] = 0

3x + 6 + (x + 2) = 0

3x + 6 + x + 2 = 0

4x + 8 = 0

Subtract 8 from both sides.

4x = -8

Divide both sides by 4.

x = -2

¹⁄₆ₓ + ¾ = ¹⁄₃ₓ

In the equation above, there are three different denominators 6x, 4 and 3x.

Least common multiple of (6, 4, 3) = 12.

Since x is multiplied in two of the denominators, multiply both sides of the equation by 12x to get rid of all the denominators.

12x(¹⁄₆ₓ + ¾) = 12x(¹⁄₃ₓ)

12x(¹⁄₆ₓ) + 12x(¾) = 12x(¹⁄₃ₓ)

2(1) + 3x(3) = 4(1)

2 + 9x = 4

Subtract 2 from both sides.

9x = 2

Divide both sides by 9.

x = ²⁄₉

¹⁄ₓ = ⁶⁄₅ₓ + 1

Multiply both sides of the equation by 5x to get rid of the denominators x and 5x.

5x(¹⁄ₓ) = 5x(⁶⁄₅ₓ + 1)

5 = 5x(⁶⁄₅ₓ) + 5x(1)

5 = 6 + 5x

Subtract 6 from both sides.

-1 = 5x

Divide both sides by 5.

⁻¹⁄₅ = x

⁽ˣ ⁻ ⁴⁾⁄₅ₓ = ¹⁄₅ₓ + 1

Multiply both sides of the equation by 5x to get rid of the denominator 5x.

5x[⁽ˣ ⁻ ⁴⁾⁄₅ₓ] = 5x[¹⁄₅ₓ + 1]

x - 4 = 5x(¹⁄₅ₓ) + 5x(1)

x - 4 = 1 + 5x

Subtract x from both sides.

-4 = 1 + 4x

Subtract 1 from both sides.

-5 = 4x

Divide both sides by 4.

⁻⁵⁄₄ = x

¹⁄₍ₓ ₋ ₈₎ - 1 = ⁷⁄₍ₓ ₋ ₈₎

Multiply both sides of the equation by (x - 8) to get rid of the denominator (x - 8).

(x - 8)[¹⁄₍ₓ ₋ ₈₎ - 1] = (x - 8)[⁷⁄₍ₓ ₋ ₈₎]

(x - 8)[¹⁄₍ₓ ₋ ₈₎] - (x - 8) = 7

1 - (x - 8) = 7

1 - x + 8 = 7

-x + 9 = 7

Subtract 9 from both sides.

-x = -2

Multiply both sides by -1.

x = 2

1 = ⁽ˣ ⁺ ²⁾⁄₍ₓ ₋ ₄₎⁽⁷ˣ ⁻ ⁴²⁾⁄₍ₓ ₋ ₄₎

Multiply both sides by (x - 4) to get rid of the denominator (x - 4).

(x - 4)(1) = (x - 4)[⁽ˣ ⁺ ²⁾⁄₍ₓ ₋ ₄₎⁽⁷ˣ ⁻ ⁴²⁾⁄₍ₓ ₋ ₄₎]

Using Distributive Property,

x - 4 = (x - 4)[⁽ˣ ⁺ ²⁾⁄₍ₓ ₋ ₄₎] + (x - 4)[⁽⁷ˣ ⁻ ⁴²⁾⁄₍ₓ ₋ ₄₎]

x - 4 = (x + 2) + (7x - 42)

x - 4 = x + 2 + 7x - 42

x - 4 = 8x - 40

Subtract x from both sides.

-4 = 7x - 40

36 = 7x

Divide both sides by 7.

³⁶⁄₇ = x

ˣ⁄₍ₓ ₊ ₈₎ = 1 + ⁶⁄₍ₓ ₊ ₁₎

Multiply both sides of the equation by (x + 8) to get rid of the denominator (x + 8) on the left side.

(x + 8)[ˣ⁄₍ₓ ₊ ₈₎] = (x + 8)[1 + ⁶⁄₍ₓ ₊ ₁₎]

x = (x + 8)(1) + (x + 8)[⁶⁄₍ₓ ₊ ₁₎]

x = x + 8 + ⁶⁽ˣ ⁺ ⁸⁾⁄₍ₓ ₊ ₁₎

Multiply both sides by (x + 1) to get rid of the denominator (x + 1) on the right side.

(x + 1)(x) = (x + 1)[x + 8 + ⁶⁽ˣ ⁺ ⁸⁾⁄₍ₓ ₊ ₁₎]

(x + 1)(x) = (x + 1)(x) + (x + 1)(8) + (x + 1)[⁶⁽ˣ ⁺ ⁸⁾⁄₍ₓ ₊ ₁₎]

Subtract (x + 1)(x) from both sides.

0 = (x + 1)(8) + (x + 1)[⁶⁽ˣ ⁺ ⁸⁾⁄₍ₓ ₊ ₁₎]

0 = 8x + 8 + 6(x + 8)

0 = 8x + 8 + 6x + 48

0 = 14x + 56

Subtract 56 from both sides.

-56 = 14x

Divide both sides by 14.

-56/14 = x

-4 = x

⁽ˣ ⁻ ²⁾⁄₍ₓ ₊ ₃₎ - 1 = ³⁄₍ₓ ₊ ₂₎

Multiply both sides of the equation by (x + 3) to get rid of the denominator (x + 3) on the left side.

(x + 3)[⁽ˣ ⁻ ²⁾⁄₍ₓ ₊ ₃₎ - 1] = (x + 3)[³⁄₍ₓ ₊ ₂₎]

(x + 3)[⁽ˣ ⁻ ²⁾⁄₍ₓ ₊ ₃₎] - 1(x + 3) = (x + 3)[³⁄₍ₓ ₊ ₂₎]

(x - 2) - x - 3 = ³⁽ˣ ⁺ ³⁾⁄₍ₓ ₊ ₂₎

x - 2 - x - 3 = ³⁽ˣ ⁺ ³⁾⁄₍ₓ ₊ ₂₎

-5 = ³⁽ˣ ⁺ ³⁾⁄₍ₓ ₊ ₂₎

Multiply both sides of the equation by (x + 2) to get rid of the denominator (x + 2) on the right side.

(x + 2)(-5) = (x + 2)[³⁽ˣ ⁺ ³⁾⁄₍ₓ ₊ ₂₎]

-5x - 10 = 3(x + 3)

-5x - 10 = 3x + 9

-10 = 8x + 9

Subtract 9 from both sides.

-19 = 8x

Divide both sides by 8.

⁻¹⁹⁄₈ = x

⁵⁄₍ₓ ₋ ₅₎ = ³⁄₍ₓ ₊ ₅₎

In the above equation, only one fraction is found on each side of the fraction. So, the equation can be solved by cross multiplication.

That is, numerator on the left side to be multiplied by denominator on the right side and numerator on the right side has to be multiplied by denominator on the left side.

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

Using Distributive Property,

5x + 25 = 3x - 15

Subtract 3x from both sides.

2x + 25 = -15

Subtract 25 from both sides.

2x = -40

Divide both sides by 2.

x = -20

Let x be the positive integer.

Then, its reciprocal is ¹⁄ₓ.

Given : Difference between 5 and the positive integer is 14/3.

5 - ¹⁄ₓ = ¹⁴⁄₃

Multiply both sides by 3x to get rid of the denomimators x and 3.

3x(5 - ¹⁄ₓ) = 3x(¹⁴⁄₃)

3x(5) - 3x(¹⁄ₓ) = 14x

15x - 3 = 14x

Subtract 14x from both sides.

x - 3 = 0

x = 3

The positive integer is 3.

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