Question
Solve the following quadratic equations by factorization:
$\frac{1}{\text{x}+4}-\frac{1}{\text{x}-7}=\frac{11}{30},$ $\text{x}\neq4,7$
$\frac{1}{\text{x}+4}-\frac{1}{\text{x}-7}=\frac{11}{30},$ $\text{x}\neq4,7$
✓
Answer
$\frac{1}{\text{x}+4}-\frac{1}{\text{x}-7}=\frac{11}{30},$ $\text{x}\neq4,7$
$\frac{\text{x}-7-\text{x}-4}{(\text{x}+4)(\text{x}-7)}=\frac{11}{30}$
$\frac{-11}{(\text{x}+4)(\text{x}-7)}=\frac{11}{30}$
$\frac{-1}{(\text{x}+4)(\text{x}-7)}=\frac{1}{30}$ (dividing both side by 11)
$(x + 4)(x - 7) = -30$
$x^2 + 4x - 7x - 28 + 30 = 0$
$\Rightarrow x^2 - 3x + 2 = 0$
$\begin{Bmatrix}\because2=(-2)\times(-1)\\-3=-2-1\end{Bmatrix}$
$\Rightarrow x^2- x - 2x + 2 =0$
$\Rightarrow x(x - 1) - 2(x - 1) = 0$
$\Rightarrow (x - 1)(x - 2) = 0$
Either x - 1 = 0, then x = 1
Or x - 2 = 0, then x = 2
$\therefore$ x = 1, 2
$\frac{\text{x}-7-\text{x}-4}{(\text{x}+4)(\text{x}-7)}=\frac{11}{30}$
$\frac{-11}{(\text{x}+4)(\text{x}-7)}=\frac{11}{30}$
$\frac{-1}{(\text{x}+4)(\text{x}-7)}=\frac{1}{30}$ (dividing both side by 11)
$(x + 4)(x - 7) = -30$
$x^2 + 4x - 7x - 28 + 30 = 0$
$\Rightarrow x^2 - 3x + 2 = 0$
$\begin{Bmatrix}\because2=(-2)\times(-1)\\-3=-2-1\end{Bmatrix}$
$\Rightarrow x^2- x - 2x + 2 =0$
$\Rightarrow x(x - 1) - 2(x - 1) = 0$
$\Rightarrow (x - 1)(x - 2) = 0$
Either x - 1 = 0, then x = 1
Or x - 2 = 0, then x = 2
$\therefore$ x = 1, 2
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