Question

Find all the possible of x for which of the below is true:

Answer 1

Answer:

x = $\frac{1}{2}$ , x = 1

Step-by-step explanation:

Given

$\frac{2x}{x+3}$ + $\frac{3}{x-3}$ = $\frac{8}{x^2-9}$ ← denominator is a difference of squares

$\frac{2x}{x+3}$ + $\frac{3}{x-3}$ = $\frac{8}{(x+3)(x-3)}$

Multiply through by (x + 3)(x - 3) to clear the fractions

2x(x - 3) + 3(x + 3) = 8 ← distribute left side and simplify

2x² - 6x + 3x + 9 = 8

2x² - 3x + 9 = 8 ( subtract 8 from both sides )

2x² - 3x + 1 = 0 ← in standard form

(2x - 1)(x - 1) = 0 ← in factored form

Equate each factor to zero and solve for x

2x - 1 = 0 ⇒ 2x = 1 ⇒ x = $\frac{1}{2}$

x - 1 = 0 ⇒ x = 1