SOLVE RATIONAL EQUATIONS

Solve the following rational equations, where the given values are excluded.

Problem 1 :

3/(x + 4) = 9/(x - 2)

where x ≠ -4 and x ≠ 2

Solution :

In this equation, there are two restrictions:

x ≠ -4 and x ≠ 2

3/(x + 4) = 9/(x - 2)

By cross multiplication,

3(x - 2) = 9(x + 4)

3x - 6 = 9x + 36

3x - 9x = 36 + 6

-6x = 42

x = -42/6

x = -7

Problem 2 :

4x/(x - 1) = x/(x² - 1)

Where x ≠ -1 and x ≠ 1

Solution :

In this equation, there are two restrictions:

x ≠ -1 and x ≠ 1

By cross multiplication,

4x(x² - 1) = x(x - 1)

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

4x(x + 1) = x

4x² + 4x - x = 0

4x²+ 3x = 0

x(4x + 3) = 0

x = 0 and x = -3/4

Problem 3 :

3/(x² - 4) = 2/(x + 2) + x/(x - 2)

Solution :

3/(x² - 4) = 2(x - 2) + x(x + 2) / (x - 2) (x + 2)

3/(x² - 4) = (2x - 4 + x² + 2x) / (x - 2) (x + 2)

3/(x² - 4) = (4x - 4 + x²) / (x² - 4)

3 = x² + 4x - 4

x² + 4x - 4 - 3 = 0

x² + 4x - 7 = 0

Use quadratic formula

x = -b ± √b² - 4ac/2a

a = 1, b = 4, c = -7

x = -4 ± √4² - 4(-7)(1)/2

x = -4 ± 2√11/2

x = -2 ± √11

Problem 4 :

(3x - 2)/(x - 2) = [6/(x² - 4)] + 1

Where x ≠ -2 and x ≠ 2

Solution :

In this equation, there are two restrictions:

x ≠ -2 and x ≠ 2

(3x - 2) / (x - 2) = [6 + (x² - 4)] / (x + 2)(x - 2)

By cross multiplication,

(3x - 2) (x + 2) = 6 + x² - 4

3x² - 2x + 6x - 4 = 6 + x² - 4

3x² + 4x - 4 = x² + 2

2x² + 4x - 6 = 0

2x² - 2x + 6x - 6 = 0

2x² + 4x - 6 = 0

x² + 2x - 3 = 0

(x - 1) (x + 3) = 0

x - 1 = 0 or x + 3 = 0

x = 1 or x = -3

Problem 5 :

x/(x + 2) = (3x + 1)/(x - 1) + 4/(x² + x - 2)

where x ≠ -2 and x ≠ 1

Solution :

In this equation, there are two restrictions:

x ≠ -2 and x ≠ 1

x/(x + 2) = (3x + 1)/(x - 1) + [4/(x² + x - 2)]

x/(x + 2) = [(3x + 1)/(x - 1) + 4]/(x - 1)(x + 2)

x(x - 1) / (x - 1) (x + 2) = [(3x + 1)(x + 2) + 4] / (x - 1)(x + 2)

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

x² - x = 3x² + x + 6x + 2 + 4

x² - x - 3x² - 7x + 6 = 0

-2x² - 8x - 6 = 0

2x² + 8x + 6 = 0

x² + 4x + 3 = 0

(x + 1)(x + 3) = 0

x + 1 = 0 or x + 3 = 0

x = -1 or x = -3