Question
Solve the following quadratic equations by factorization:
$\frac{3}{\text{x}+1}-\frac{1}{2}=\frac{2}{3\text{x}-1},$ $\text{x}\neq-1,\frac{1}{3}$
$\frac{3}{\text{x}+1}-\frac{1}{2}=\frac{2}{3\text{x}-1},$ $\text{x}\neq-1,\frac{1}{3}$
✓
Answer
$\frac{3}{\text{x}+1}-\frac{1}{2}=\frac{2}{3\text{x}-1}$
$\Rightarrow\frac{6-(\text{x}+1)}{2(\text{x}+1)}=\frac{2}{3\text{x}-1}$
$\Rightarrow\frac{6-\text{x}-1}{2\text{x}+2}=\frac{2}{3\text{x}-1}$
$\Rightarrow (5 - x)(3x - 1) = 2(2x + 2)$
$\Rightarrow 15x - 5 - 3x^2 + x = 4x + 4$
$\Rightarrow -3x^2 + 16x - 5 - 4x - 4 = 0$
$\Rightarrow -3x^2 + 12x - 9 = 0$
$\Rightarrow 3x^2 - 12x + 9 = 0$
$\Rightarrow x^2 - 4x + 3 = 0$
$\Rightarrow x^2 - 3x - x + 3 = 0$
$\Rightarrow x(x - 3) - 1(x - 3) = 0$
$\Rightarrow (x - 1)(x - 3) = 0$
$\Rightarrow x - 1 = 0 or x - 3 = 0$
$\Rightarrow x = 1 or x = 3$
Hence, the factors are 3 and 1
$\Rightarrow\frac{6-(\text{x}+1)}{2(\text{x}+1)}=\frac{2}{3\text{x}-1}$
$\Rightarrow\frac{6-\text{x}-1}{2\text{x}+2}=\frac{2}{3\text{x}-1}$
$\Rightarrow (5 - x)(3x - 1) = 2(2x + 2)$
$\Rightarrow 15x - 5 - 3x^2 + x = 4x + 4$
$\Rightarrow -3x^2 + 16x - 5 - 4x - 4 = 0$
$\Rightarrow -3x^2 + 12x - 9 = 0$
$\Rightarrow 3x^2 - 12x + 9 = 0$
$\Rightarrow x^2 - 4x + 3 = 0$
$\Rightarrow x^2 - 3x - x + 3 = 0$
$\Rightarrow x(x - 3) - 1(x - 3) = 0$
$\Rightarrow (x - 1)(x - 3) = 0$
$\Rightarrow x - 1 = 0 or x - 3 = 0$
$\Rightarrow x = 1 or x = 3$
Hence, the factors are 3 and 1
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