SOLVING EQUATIONS WITH FRACTIONS WORKSHEET

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Problems 1-11 : Solve for x.

Problem 1 :

ˣ⁄₁₂ = β…”

Problem 2 :

²ˣ⁄₁₅ = βΆβ„β‚…

Problem 3 :

ˣ⁄₂ = Λ£β„₃ - 1

Problem 4 :

⁡ˣ⁄₄ - Λ£β„β‚‚ = β»Β³β„β‚„

Problem 5 :

ˣ⁄₂ - Β³Λ£β„β‚„ = ΒΎ - x

Problem 6 :

⁽ˣ ⁺ ¹⁾⁄₃ = β½Λ£ ⁻ ¹⁾⁄₅

Problem 7 :

ˣ⁄₂ - β…• = Λ£β„₃ + ΒΌ

Problem 8 :

ˣ⁄₂ - Β³Λ£β„β‚„ + β΅Λ£β„₆ = 21

Problem 9 :

x + 7 - ⁸ˣ⁄₃ = ΒΉβ·β„₆⁡ˣ⁄₂

Problem 10 :

⁽ˣ ⁺ ⁴⁾⁄₁₂ - β½Λ£ ⁻ β΅βΎβ„β‚β‚ˆ = 1

Problem 11 :

⁽³ˣ ⁻ ²⁾⁄₄ - β½Β²Λ£ ⁺ ³⁾⁄₃ = β…” - x

Problem 12 :

Subtracting two-third of a number from 5 results 3. Find the number.

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Answers

1. Answer :

ˣ⁄₁₂ = β…”

Least common multiple of the denominators 12 and 3 is 12.

Multiply both sides of the equation by 12 to get rid of the denominators 12 and 3.

12(ˣ⁄₁₂) = 12(β…”)

x = 4(2)

x = 8

2. Answer :

²ˣ⁄₁₅ = βΆβ„β‚…

Least common multiple of the denominators 15 and 5 is 15.

Multiply both sides of the equation by 15 to get rid of the denominators 15 and 5.

15(²ˣ⁄₁₅) = 15(⁢⁄₅)

2x = 3(6)

2x = 18

Divide both sides by 2.

x = 9

3. Answer :

ˣ⁄₂ = Λ£β„₃ - 1

Least common multiple of the denominators 2 and 3 is 6.

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

6(ˣ⁄₂) = 6(ˣ⁄₃ - 1)

3x = 6(ˣ⁄₃) - 6(1)

3x = 2x - 6

Subtract 2x from both sides.

x = -6

4. Answer :

⁡ˣ⁄₄ - Λ£β„β‚‚ = β»Β³β„β‚„

Least common multiple of the denominators 4 and 2 is 4.

Multiply both sides of the equation by 4 to get rid of the denominators 4 and 2.

4(⁡ˣ⁄₄ - Λ£β„β‚‚) = 4(⁻³⁄₄)

4(⁡ˣ⁄₄) - 4(ˣ⁄₂) = -3

5x - 2x = -3

3x = -3

Divide both sides by 3.

x = -1

5. Answer :

ˣ⁄₂ - Β³Λ£β„β‚„ = ΒΎ - x

Least common multiple of the denominators 2 and 4 is 4.

Multiply both sides of the equation by 4 to get rid of the denominators.

4(ˣ⁄₂ - Β³Λ£β„β‚„) = 4(ΒΎ - x)

4(ˣ⁄₂) - 4(³ˣ⁄₄) = 4(ΒΎ) - 4(x)

2x - 3x = 3 - 4x

-x = 3 - 4x

Add 4x to both sides.

3x = 3

Divide both sides by 3.

x = 1

6. Answer :

⁽ˣ ⁺ ¹⁾⁄₃ = β½Λ£ ⁻ ¹⁾⁄₅

Least common multiple of the denominators 3 and 5 is 15.

Multiply both sides of the equation by 15 to get rid of the denominators 3 and 5.

15[⁽ˣ ⁺ ¹⁾⁄₃] = 15[⁽ˣ ⁻ ¹⁾⁄₅]

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

5x + 5 = 3x - 3

Subtract 3x from both sides.

2x + 5 = -3

Subtract 5 from both sides.

2x = -8

Divide both sides by 4.

x = -4

7. Answer :

ˣ⁄₂ - β…• = Λ£β„₃ + ΒΌ

Least common multiple of the denominators 2, 5, 3 and 4 is 60.

Multiply each side of the above equation by 60 to get rid of the denominators. 

60(ˣ⁄₂ - β…•) = 60(ˣ⁄₃ + ΒΌ)

Using Distributive Property,

60(ˣ⁄₂) - 60(β…•) = 60(ˣ⁄₃) + 60(ΒΌ)

30x - 12 = 20x + 15

Subtract 20x from both sides. 

10x - 12 = 15

Add 12 to each side. 

10x = 27

Divide each side by 10.

x = ²⁷⁄₁₀

8. Answer :

ˣ⁄₂ - Β³Λ£β„β‚„ + β΅Λ£β„₆ = 21

Least common multiple of the denominators 2, 4 and 6 is 12.

Multiply each side of the above equation by 12 to get rid of the denominators.

12(ˣ⁄₂ - Β³Λ£β„β‚„ + β΅Λ£β„₆) = 12(21)

Using Distributive Property,

12(ˣ⁄₂) - 12(³ˣ⁄₄) + 12(⁡ˣ⁄₆) = 252

6x - 9x + 10x = 252

7x = 252

x = 36

9. Answer :

x + 7 - ⁸ˣ⁄₃ = ΒΉβ·β„₆⁡ˣ⁄₂

Least common multiple of the denominators 3, 6 and 2 is 6.

Multiply each side of the above equation by 6 to get rid of the denominators. 

6(x + 7 - ⁸ˣ⁄₃) = 6(¹⁷⁄₆⁡ˣ⁄₂)

Using Distributive Property,

6(x) + 6(7) - 6(⁸ˣ⁄₃) = 6(¹⁷⁄₆) - 6(⁡ˣ⁄₂)

6x + 42 - 16x = 17 - 15x

-10x + 42 = 17 - 15x

Add 15x to each side. 

5x + 42 = 17

Subtract 42 from each side. 

5x = -25

Divide each side by 5.

x = -5

10. Answer :

⁽ˣ ⁺ ⁴⁾⁄₁₂ - β½Λ£ ⁻ β΅βΎβ„β‚β‚ˆ = 1

Least common multiple of the denominators 12 and 18 is 36.

Multiply each side of the above equation by 36 to get rid of the denominators. 

36[⁽ˣ ⁺ ⁴⁾⁄₁₂ - β½Λ£ ⁻ β΅βΎβ„β‚β‚ˆ] = 36(1)

Using Distributive Property,

36[⁽ˣ ⁺ ⁴⁾⁄₁₂] - 36[⁽ˣ ⁻ β΅βΎβ„β‚β‚ˆ] = 36

3(x + 4) - 2(x - 5) = 36

3x + 12 - 2x + 10 = 36

x + 22 = 36

Subtract 22 from each side. 

x = 14

11. Answer :

⁽³ˣ ⁻ ²⁾⁄₄ - β½Β²Λ£ ⁺ ³⁾⁄₃ = β…” - x

Least common multiple of the denominators 4 and 3 is 12. 

Multiply each side of the above equation by 12 to get rid of the denominators. 

12[⁽³ˣ ⁻ ²⁾⁄₄ - β½Β²Λ£ ⁺ ³⁾⁄₃] = 12(β…” - x)

Distribute. 

12[⁽³ˣ ⁻ ²⁾⁄₄] - 12[⁽²ˣ ⁺ ³⁾⁄₃] = 12(β…”) - 12(x)

3(3x - 2) - 4(2x + 3) = 4(2) - 12x

9x - 6 - 8x - 12 = 8 - 12x

x - 18 = 8 - 12x

Add 12x to each side. 

13x - 18 = 8

Add 18 to each side. 

13x = 26

Divide each side by 13.

x = 2

12. Answer :

Let x be the number.

It is given that subtracting two-third of a number from 5 results 3.

5 - (β…”)x = 3

5 - ²ˣ⁄₃ = 3

Multiply both sides by 3 to get rid of the denominator.

3(5 - ²ˣ⁄₃) = 3(3)

3(5) - 3(²ˣ⁄₃) = 9

15 - 2x = 9

Subtract 15 from both sides.

-2x = -6

Divide both sides by -2.

x = 3

The number is 3.

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