Question

solve and check:
\\(\frac{1}{x + 3} = \frac{x + 10}{x - 2}\\)
from least to greatest, the solutions are \\(x = \square\\) and \\(x = \square\\).

Explanation:

Step1: Cross-multiply to eliminate fractions

$$1 \cdot (x-2) = (x+10) \cdot (x+3)$$

Step2: Expand both sides

$$x - 2 = x^2 + 13x + 30$$

Step3: Rearrange to quadratic form

$$x^2 + 12x + 32 = 0$$

Step4: Factor the quadratic

$$(x+4)(x+8) = 0$$

Step5: Solve for x

Set each factor to 0:
$x+4=0 \implies x=-4$
$x+8=0 \implies x=-8$

Step6: Check for extraneous solutions

For $x=-4$:
Left side: $\frac{1}{-4+3}=-1$
Right side: $\frac{-4+10}{-4-2}=\frac{6}{-6}=-1$, valid.
For $x=-8$:
Left side: $\frac{1}{-8+3}=-\frac{1}{5}$
Right side: $\frac{-8+10}{-8-2}=\frac{2}{-10}=-\frac{1}{5}$, valid.

Answer:

From least to greatest, the solutions are $x = -8$ and $x = -4$.