HOW TO SOLVE RATIONAL EQUATIONS

Example 1 :

⁵⁄ₓ - ⅓ = ¹⁄ₓ

Solution :

⁵⁄ₓ - ⅓ = ¹⁄ₓ

In this equation, there is one restriction :

x ≠ 0

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

x(⁵⁄ₓ - ⅓) = x(¹⁄ₓ)

x(⁵⁄ₓ) - x(⅓) = 1

5 - ˣ⁄₃ = 1

Subtract 5 from both sides.

-ˣ⁄₃ = -4

Multiply both sides by -3.

-3(-ˣ⁄₃) = -3(-4)

x = 12

Example 2 :

³⁄ₓ = ²⁄₍ₓ ₊ ₄₎

Solution :

³⁄ₓ = ²⁄₍ₓ ₊ ₄₎

In this equation, there are two restrictions :

x ≠ 0 and x ≠ -4

There is only one fraction on each side of the equation.

By cross multiplying,

3(x + 4) = 2x

3x + 12 = 2x

Subtract 2x from both sides.

x + 12 = 0

Subtract 12 from both sides.

x = -12

Example 3 :

³⁄₍ₓ ₊ ₂₎ + 5 = ⁴⁄₍ₓ ₊ ₂₎

Solution :

³⁄₍ₓ ₊ ₂₎ + 5 = ⁴⁄₍ₓ ₊ ₂₎

In this equation, there is one restriction :

x ≠ -2

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

(x + 2)[³⁄₍ₓ ₊ ₂₎ + 5] = (x + 2)[⁴⁄₍ₓ ₊ ₂₎]

(x + 2)[³⁄₍ₓ ₊ ₂₎] + 5(x + 2) = 4

3 + 5x + 10 = 4

5x + 13 = 4

Subtract 13 from both sides.

5x = -9

Divide both sides by 5.

x = ⁻⁹⁄₅

Example 4 :

²ˣ⁄₍ₓ ₊ ₄₎ - 3 = ⁻¹²⁄₍ₓ ₊ ₄₎

Solution :

²ˣ⁄₍ₓ ₊ ₄₎ - 3 = ⁻¹²⁄₍ₓ ₊ ₄₎

In this equation, there is one restriction :

x ≠ -4

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

(x + 4)[²ˣ⁄₍ₓ ₊ ₄₎ - 3] = (x + 4)[-12/(x + 4)]

(x + 4)[²ˣ⁄₍ₓ ₊ ₄₎] - 3(x + 4) = -12

2x - 3x - 12 = -12

-x - 12 = -12

Add 12 to both sides.

-x = 0

x = 0

Example 5 :

¹⁴⁄₍₂ ₋ ₓ₎ = ²⁄ₓ

Solution :

¹⁴⁄₍₂ ₋ ₓ₎ = ²⁄ₓ

In this equation, there are two restrictions :

x ≠ 2 and x ≠ 0

There is only one fraction on each side of the equation.

By cross multiplying,

14x = x(2 - x)

14x = 4 - 2x

Add 2x to both sides.

16x = 4

Divide both sides by 16.

x = 1/4

Example 6 :

²⁄₍ₓ ₋ ₄₎ + 2 = ⁶⁄₍ₓ ₋ ₄₎

Solution :

²⁄₍ₓ ₋ ₄₎ + 2 = ⁶⁄₍ₓ ₋ ₄₎

In this equation, there is one restriction :

x ≠ 4

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

(x - 4)[²⁄₍ₓ ₋ ₄₎ + 2] = (x - 4)[6/(x - 4)]

(x - 4)[²⁄₍ₓ ₋ ₄₎] + 2(x - 4) = 6

2 + 2x - 8 = 6

2x - 6 = 6

Add 6 to both sides.

2x = 12

Divide both sides by 2.

x = 6

Example 7 :

⁽ˣ ⁺ ²⁾⁄₍ₓ ₊ ₁₎ - x = ⁻⁶⁄₍ₓ ₊ ₁₎

Solution :

⁽ˣ ⁺ ²⁾⁄₍ₓ ₊ ₁₎ - x = ⁻⁶⁄₍ₓ ₊ ₁₎

In this equation, there is one restriction :

x ≠ -1

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

(x + 1)[⁽ˣ ⁺ ²⁾⁄₍ₓ ₊ ₁₎ - x] = (x + 1)[⁻⁶⁄₍ₓ ₊ ₁₎]

(x + 1)[⁽ˣ ⁺ ²⁾⁄₍ₓ ₊ ₁₎] - x(x + 1) = -6

x + 2 - x² - x = -6

2 - x² = -6

Subtract 2 from both sides.

-x² = -8

Multiply both sides by -1.

x² = 8

Take square root on both sides.

√x² = √8

x = ±2√2

x = -2√2  or  x = 2√2

Example 8 :

ˣ⁄₍ₓ ₊ ₂₎ + ²⁄₍ₓ ₊ ₂₎₍ₓ ₊ ₃₎ = ⁵⁄₍ₓ ₊ ₃₎

Solution :

ˣ⁄₍ₓ ₊ ₂₎ + ²⁄₍ₓ ₊ ₂₎₍ₓ ₊ ₃₎ = ⁵⁄₍ₓ ₊ ₃₎

In this equation, there are two restrictions :

x ≠ -2 and x ≠ -3

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

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

(x + 2)[ˣ⁄₍ₓ ₊ ₂₎] + (x + 2)[²⁄₍ₓ ₊ ₂₎₍ₓ ₊ ₃₎] = ⁽⁵ˣ ⁺ ¹⁰⁾⁄₍ₓ ₊ ₃₎

x + ²⁄₍ₓ ₊ ₃₎ = ⁽⁵ˣ ⁺ ¹⁰⁾⁄₍ₓ ₊ ₃₎

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

(x + 3)[x + ²⁄₍ₓ ₊ ₃₎] = (x + 3)[⁽⁵ˣ ⁺ ¹⁰⁾⁄₍ₓ ₊ ₃₎]

(x + 3)(x) + (x + 3)[²⁄₍ₓ ₊ ₃₎] = 5x + 10

x² + 3x + 2 = 5x + 10

Subtract 5x from both sides.

x² - 2x + 2 = 10

Subtract 10 from both sides.

x² - 2x - 8 = 0

Factor and solve.

x² - 4x + 2x - 8 = 0

x(x - 4) + 2(x - 4) = 0

(x - 4)(x + 2) = 0

x - 4 = 0  or  x + 2 = 0

x = 4  or  x = -2

We get two solutions x = 4 and x = -2 for the given rational equation.

Already, we know that there is a restriction x ≠ -2. 

So, x = -2 can not be considered as a solution.

Therefore, solution to the given rational equation is

x = 4

Example 9 :

⁽²ᵖ ⁻ ¹⁾⁄₍₂p ₊ ₅₎ = ⁽ᵖ ⁻ ¹⁾⁄₍p ₊ ₃₎

Solution :

⁽²ᵖ ⁻ ¹⁾⁄₍₂p ₊ ₅₎ = ⁽ᵖ ⁻ ¹⁾⁄₍p ₊ ₃₎

In this equation, there are two restrictions :

x ≠ ⁻⁵⁄₂ and x ≠ -3

There is only one fraction on each side of the equation.

By cross multiplying,

(2p - 1)(p + 3) = (p - 1)(2p + 5)

2p² + 6p - p - 3 = 2p² + 5p - 2p - 5

2p² + 5p - 3 = 2p² + 3p - 5

Subtract 2p² from both sides.

5p - 3 = 3p - 5

Subtract 3p from both sides.

2p - 3 = -5

Add 3 to both sides.

2p = -2

Divide both sides by 2.

p = -1

Example 10 :

⁽ʸ ⁺ ³⁾⁄₍y ₊ ₂₎ = ²⁄y + ¹⁄₍y ₊ ₂₎

Solution :

⁽ʸ ⁺ ³⁾⁄₍y ₊ ₂₎ = ²⁄y + ¹⁄₍y ₊ ₂₎

In this equation, there is one restriction :

y ≠ -2

Multiply both sides of the equation by (y + 2) to get rid of the denominator (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]

y² + 3y = 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 y = 2 for the given rational equation.

Already, we know that there is a restriction y ≠ -2. 

So, y = -2 can not be considered as a solution.

Therefore, solution to the given rational equation is

y = 2

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.