WORKSHEET ON SOLVING RATIONAL EQUATIONS

Problem 1 :

Solve for x :

⁴⁄₃ₓ + ⅓ = 1

Problem 2 :

Solve for a :

⁵ᵃ⁄₍ₐ ₊ ₄₎ = ⁵⁄₂

Problem 3 :

Solve for x :

⁽³ˣ ⁺ ¹²⁾⁄₍ₓ ₊ ₄₎ = ⁵⁄₃

Problem 4 :

Solve for m :

⅙ + ¹⁄₁₂ = ¹⁄m

Problem 5 :

Solve for x :

ˣ⁄₍ₓ ₊ ₃₎ = ⁸⁄₍ₓ ₊ ₆₎

Problem 6 : 

Solve for y :

⁽ʸ ⁺ ³⁾⁄₂y = ⅔

Problem 7 :

Solve for s :

⁹⁄₍₂s ₊ ₇₎ = ³⁄s

Problem 8 :

Solve for y :

⁽⁴ʸ ⁻ ¹⁾⁄₅y = ³⁄y

Problem 9 :

Solve for w :

⁴⁄₍₃w ₊ ₇₎ = ½

Problem 10 :

Solve for y :

¹²⁄y = ⁹⁄₍y ₋ ₃₎

Problem 11 :

Solve for x :

⁽³ˣ ⁻ ⁶⁾⁄₍₂ ₋ ₓ₎ = ³⁄₂

Problem 12 :

Solve for y :

⁽ʸ ⁺ ³⁾⁄₍y ₊ ₂₎ = ²⁄y + ¹⁄₍y ₊ ₂₎

Answers

1. Answer :

⁴⁄₃ₓ + ⅓ = 1

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

3x(⁴⁄₃ₓ + ⅓) = 1(3x)

3x(⁴⁄₃ₓ) + 3x(⅓) = 3x

4 + x = 3x

Subtract x from both sides.

4 = 2x

Divide both sides by 2.

2 = x

2. Answer :

⁵ᵃ⁄₍ₐ ₊ ₄₎ = ⁵⁄₂

Multiply both sides of the equation by (a + 4) to get rid of the denominator (a + 4) on the left side.

(a + 4)[⁵ᵃ⁄₍ₐ ₊ ₄₎] = (a + 4)(⁵⁄₂)

5a = ⁵⁽ᵃ ⁺ ⁴⁾⁄₂

Multiply both sides by 2.

2(5a) = 5(a + 4)

10a = 5(a + 4)

10a = 5a + 20

Subtract 5a from both sides.

5a = 20

Divide both sides by 5.

a = 4

3. Answer :

⁽³ˣ ⁺ ¹²⁾⁄₍ₓ ₊ ₄₎ = ⁵⁄₃

Multiply both sides of the equation by (x + 4) to get rid of the denominator (x + 4) on the left side.

(x + 4)[⁽³ˣ ⁺ ¹²⁾⁄₍ₓ ₊ ₄₎] = (x + 4)(⁵⁄₃)

3x + 12 = ⁵⁽ˣ ⁺ ⁴⁾⁄₃

Multiply both sides by 3.

3(3x + 12) = 5(x + 4)

9x + 36 = 5x + 20

Subtract 5x from both sides.

4x + 36 = 20

Subtract 36 from both sides.

4x = -16

Divide both sides by 4.

x = -4

4. Answer :

⅙ + ¹⁄₁₂ = ¹⁄m

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

Multiply both sides of the equation by 12 to get rid of the denominators 6 and 12 on the left side.

12(⅙ + ¹⁄₁₂) = 12(¹⁄m)

12(⅙) + 12(¹⁄₁₂) = ¹²⁄m

2 + 1 = ¹²⁄m

3 = ¹²⁄m

Multiply both sides by m.

3m = 12

Divide both sides of the equation by 3.

m = 4

5. Answer :

ˣ⁄₍ₓ ₊ ₃₎ = ⁸⁄₍ₓ ₊ ₆₎

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

(x + 3)[ˣ⁄₍ₓ ₊ ₃₎)] = (x + 3)[⁸⁄₍ₓ ₊ ₆₎]

x = ⁸⁽ˣ ⁺ ³⁾⁄₍ₓ ₊ ₆₎

x = ⁽⁸ˣ ⁺ ²⁴⁾⁄₍ₓ ₊ ₆₎

Multiply both sides by (x + 6).

(x + 6)(x) = 8x + 24

x² + 6x = 8x + 24

Subtract 8x from both sides.

x² - 2x = 24

Subtract 24 from both sides.

x² - 2x - 24 = 0

Solve by factoring.

x² - 6x + 4x - 24 = 0

x(x - 6) + 4(x - 6) = 0

(x - 6)(x + 4) = 0

x - 6 = 0  or  x + 4 = 0

x = 6  or  x = -4

6. Answer :

⁽ʸ ⁺ ³⁾⁄₂y = ⅔

Multiply both sides of the equation by 2y to get rid of the denominator 2y on the left side.

2y[⁽ʸ ⁺ ³⁾⁄₂y] = (2y)(⅔)

y + 3 = ⁴ʸ⁄₃

Multiply both sides by 3.

3(y + 3) = 4y

3y + 9 = 4y

Subtract 3y from both sides.

9 = y

7. Answer :

⁹⁄₍₂s ₊ ₇₎ = ³⁄s

Multiply both sides of the equation by (2s + 7) to get rid of the denominator (2s + 7) on the left side.

(2s + 7)[⁹⁄₍₂s ₊ ₇₎] = (2s + 7)[³⁄s]

9 = ⁽⁶ˢ ⁺ ²¹⁾⁄s

Multiply both sides by s

9s = 6s + 21

Subtract 6s from both sides.

3s = 21

Divide both sides by 3.

s = 7

8. Answer :

⁽⁴ʸ ⁻ ¹⁾⁄₅y = ³⁄y

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

5y[⁽⁴ʸ ⁻ ¹⁾⁄₅y] = 5y(³⁄y)

4y - 1 = 5(3)

4y - 1 = 15

Add 1 to both sides.

4y = 16

Divide both sides by 4.

y = 4

9. Answer :

⁴⁄₍₃w ₊ ₇₎ = ½

Multiply both sides of the equation by (3w + 7) to get rid of the denominator (3w + 7) on the left side.

(3w + 7)[⁴⁄₍₃w ₊ ₇₎] = (3w + 7)(½)

4 = ⁽³ʷ ⁺ ⁷⁾⁄₂

Multiply both sides by 2.

2(4) = 3w + 7

8 = 3w + 7

Subtract 7 from both sides.

1 = 3w

Divide both sides by 3.

⅓ = w

10. Answer :

¹²⁄y = ⁹⁄₍y ₋ ₃₎

Multiply both sides of the equation by y to get rid of the denominator y on the left side.

y(¹²⁄y) = y[⁹⁄₍y ₋ ₃₎]

12 = ⁹ʸ⁄₍y ₋ ₃₎

Multiply both sides by (y - 3).

(y - 3)(12) = -9y

12y - 36 = 9y

Subtract 9y from both sides.

3y - 36 = 0

Add 36 to both sides.

3y = 36

Divide both sides by 3.

y = 12

11. Answer :

⁽³ˣ ⁻ ⁶⁾⁄₍₂ ₋ ₓ₎ = ³⁄₂

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

(2 - x)[⁽³ˣ ⁻ ⁶⁾⁄₍₂ ₋ ₓ₎] = (2 - x)(³⁄₂)

3x - 6 = ⁽⁶ ⁻ ³ˣ⁾⁄₂

Multiply both sides by 2.

2(3x - 6) = 6 - 3x

6x - 12 = 6 - 3x

Add 3x to both sides.

9x - 12 = 6

Add 12 to both sides.

9x = 18

Divide both sides by 9.

x = 2

But, x = 2 makes one of the denominators zero in the given rational equation. So, x = 2 is an extraneous solution, hence it can not be considered as a solution to the given rational equation.

Therefore, there is NO SOLUTION for the given rational equation.

12. Answer :

⁽ʸ ⁺ ³⁾⁄₍y ₊ ₂₎ = ²⁄y + ¹⁄₍y ₊ ₂₎

Multiply both sides of the equation by (y + 2) to get rid of the denominators (y + 2) on both sides.

(y + 2)[⁽ʸ ⁺ ³⁾⁄₍y ₊ ₂₎] = (y + 2)[²⁄y + ¹⁄₍y ₊ ₂₎]

y + 3 = (y + 2)(²⁄y) + (y + 2)[¹⁄₍y ₊ ₂₎]

y + 3 = ⁽²ʸ ⁺ ⁴⁾⁄y + 1

Multiply both sides by y to get rid of the denominator y on the right side.

y(y + 3) = y[⁽²ʸ ⁺ ⁴⁾⁄y + 1]

y(y) + y(3) = y[⁽²ʸ ⁺ ⁴⁾⁄y] + y(1)

y² + 3y = 2y + 4 + y

y² + 3y = 3y + 4

Subtract 3y from both sides.

y² = 4

Take square root on both sides.

√y² = √4

y = ±2

y = -2  or  y = 2

We get two solutions y = 2 and x = -2 for the given rational equation.

But, y = -2 makes two of the denominators zero in the given rational equation. So, y = -2 is an extraneous solution, hence it is removed from the solutions.

Therefore, solution for the given rational equation is

y = 2

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