Question

what are the solutions to the equation \\(\frac{x}{x + 6} = \frac{2x - 3}{x + 12}\\)?
select all that apply.

a \\(x = -12\\)

b \\(x = -6\\)

c \\(x = -3\\)

d \\(x = 0\\)

e \\(x = \frac{3}{2}\\)

f \\(x = 3\\)

g \\(x = 6\\)

Explanation:

Step1: Cross-multiply to eliminate fractions

$x(x+12)=(2x-3)(x+6)$

Step2: Expand both sides

$x^2 + 12x = 2x^2 + 12x - 3x - 18$

Step3: Simplify right-hand side

$x^2 + 12x = 2x^2 + 9x - 18$

Step4: Rearrange to standard quadratic form

$0 = x^2 - 3x - 18$

Step5: Factor the quadratic

$0 = (x-6)(x+3)$

Step6: Solve for x

$x-6=0 \implies x=6$; $x+3=0 \implies x=-3$

Step7: Check for extraneous solutions

$x=-12$ and $x=-6$ make original denominators 0, so they are invalid. Verify $x=-3$: $\frac{-3}{-3+6}=\frac{-9}{-3+12} \implies \frac{-3}{3}=\frac{-9}{9} \implies -1=-1$ (valid). Verify $x=6$: $\frac{6}{6+6}=\frac{9}{6+12} \implies \frac{6}{12}=\frac{9}{18} \implies \frac{1}{2}=\frac{1}{2}$ (valid).

Answer:

C. $x = -3$
G. $x = 6$