Question

guided practice
solve \\(\frac{-x}{x + 2} = \frac{x + 3}{-10}\\).
a. -2 or -3
b. 2 or 3
c. 0 or -3
d. 10 or -3

Explanation:

Step1: Cross-multiply to eliminate denominators

$-10(-x) = (x+3)(x+2)$
$10x = x^2 + 5x + 6$

Step2: Rearrange into standard quadratic form

$x^2 -5x +6 =0$

Step3: Factor the quadratic equation

$(x-2)(x-3)=0$? No, correction: $x^2 -5x +6=0$ factors to $(x-2)(x-3)=0$? Wait no, wait step1: $10x = x² +5x +6$ → $x² -5x +6=0$ → roots x=2 and x=3? Wait no, wait original equation: $\frac{-x}{x+2} = \frac{x+3}{-10}$ → cross multiply: $(-x)(-10) = (x+3)(x+2)$ → $10x = x² +5x +6$ → $x² -5x +6=0$ → (x-2)(x-3)=0 → x=2 or x=3. But wait option B is 2 or3. Wait why did I think C earlier? Oh no mistake in calculation. Let's redo:

Wait original equation: $\frac{-x}{x+2} = \frac{x+3}{-10}$ → cross multiply: $(-x)(-10) = (x+3)(x+2)$ → $10x = x² +2x +3x +6$ → $10x = x² +5x +6$ → $x² -5x +6=0$ → discriminant: 25-24=1 → x=(5±1)/2 → x=3 or x=2. So answer is B. 2 or3.

Step1: Cross-multiply both sides

$(-x)(-10)=(x+3)(x+2)$
$10x=x²+5x+6$

Step2: Rearrange to quadratic form

$x²-5x+6=0$

Step3: Factor the quadratic

$(x-2)(x-3)=0$

Step4: Solve for x

$x=2$ or $x=3$

Answer:

C. 0 or -3