Question
Solve for x.
$\frac{1}{\text{x}-3}-\frac{1}{\text{x}+5}=\frac{1}{6},$ $\text{x}\neq3,-5$
$\frac{1}{\text{x}-3}-\frac{1}{\text{x}+5}=\frac{1}{6},$ $\text{x}\neq3,-5$
✓
Answer
$\frac{1}{\text{x}-3}-\frac{1}{\text{x}+5}=\frac{1}{6},$ $\text{x}\neq3,-5$
$\frac{1}{\text{x}-3}-\frac{1}{\text{x}+5}=\frac{1}{6}$
$\Rightarrow\frac{\text{x}+5-\text{x}+3}{(\text{x}-3)(\text{x}+5)}=\frac{1}{6}$
$\Rightarrow\frac{8}{(\text{x}-3)(\text{x}+5)}=\frac{1}{6}$
$\Rightarrow 48 = x^2 + 2x - 15$
$\Rightarrow x^2 + 2x - 15 - 48 = 0$
$\Rightarrow x^2 + 2x - 63 = 0$
$\Rightarrow x^2 + 9x - 7x - 63 = 0$
$\Rightarrow x(x + 9) - 7(x + 9) = 0$
$\Rightarrow (x - 7)(x + 9) = 0$
$\Rightarrow x = 7, -9$
$\frac{1}{\text{x}-3}-\frac{1}{\text{x}+5}=\frac{1}{6}$
$\Rightarrow\frac{\text{x}+5-\text{x}+3}{(\text{x}-3)(\text{x}+5)}=\frac{1}{6}$
$\Rightarrow\frac{8}{(\text{x}-3)(\text{x}+5)}=\frac{1}{6}$
$\Rightarrow 48 = x^2 + 2x - 15$
$\Rightarrow x^2 + 2x - 15 - 48 = 0$
$\Rightarrow x^2 + 2x - 63 = 0$
$\Rightarrow x^2 + 9x - 7x - 63 = 0$
$\Rightarrow x(x + 9) - 7(x + 9) = 0$
$\Rightarrow (x - 7)(x + 9) = 0$
$\Rightarrow x = 7, -9$
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