Question
Solve the following quadratic equations by factorization:
$\frac{1}{\text{x}-1}-\frac{1}{\text{x}+5}=\frac{6}{7},$ $\text{x}\neq1,-5$
$\frac{1}{\text{x}-1}-\frac{1}{\text{x}+5}=\frac{6}{7},$ $\text{x}\neq1,-5$
✓
Answer
$\frac{1}{\text{x}-1}-\frac{1}{\text{x}+5}=\frac{6}{7}$ $(\text{x}\neq1,-5)$
$\frac{\text{x}+5-\text{x}+1}{(\text{x}-1)(\text{x}+5)}=\frac{6}{7}$
$\Rightarrow\frac{6}{(\text{x}-1)(\text{x}+5)}=\frac{6}{7}$
$\Rightarrow\frac{1}{(\text{x}-1)(\text{x}+5)}=\frac{1}{7}$ (Dividing by 6)
$(x - 1)(x + 5) = 7 \Rightarrow x^2 + 5x - x - 5 = 7$
$\begin{Bmatrix}\because-12=6\times(-2)\\+4=6-2\end{Bmatrix}$
$\Rightarrow x^2 + 4x - 5 - 7 = 0$
$\Rightarrow x^2 + 4x - 12 = 0$
$\Rightarrow x^2 + 6x - 2x - 12 = 0$
$\Rightarrow x(x + 6) - 2(x + 6) = 0$
$\Rightarrow (x + 6)(x - 2) = 0$
Either $x + 6 = 0$, then $x = -6$
or $x - 2 = 0$, then $x = 2$
$\therefore$ $x = 2, -6$
$\frac{\text{x}+5-\text{x}+1}{(\text{x}-1)(\text{x}+5)}=\frac{6}{7}$
$\Rightarrow\frac{6}{(\text{x}-1)(\text{x}+5)}=\frac{6}{7}$
$\Rightarrow\frac{1}{(\text{x}-1)(\text{x}+5)}=\frac{1}{7}$ (Dividing by 6)
$(x - 1)(x + 5) = 7 \Rightarrow x^2 + 5x - x - 5 = 7$
$\begin{Bmatrix}\because-12=6\times(-2)\\+4=6-2\end{Bmatrix}$
$\Rightarrow x^2 + 4x - 5 - 7 = 0$
$\Rightarrow x^2 + 4x - 12 = 0$
$\Rightarrow x^2 + 6x - 2x - 12 = 0$
$\Rightarrow x(x + 6) - 2(x + 6) = 0$
$\Rightarrow (x + 6)(x - 2) = 0$
Either $x + 6 = 0$, then $x = -6$
or $x - 2 = 0$, then $x = 2$
$\therefore$ $x = 2, -6$
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