Question
Solve the following equation and verify your answer:
$\frac{(\text{x}+2)(2\text{x}-3)-2\text{x}^2+6}{\text{x}-5}=2$
$\frac{(\text{x}+2)(2\text{x}-3)-2\text{x}^2+6}{\text{x}-5}=2$
✓
Answer
$\frac{(\text{x}+2)(2\text{x}-3)-2\text{x}^2+6}{\text{x}-5}=\frac{2}{1}$
By cross multiplication:
$(\text{x}+2)(2\text{x}-3)-2\text{x}^2+6=2(\text{x}-5)$
$2\text{x}^2-3\text{x}+4\text{x}-6-2\text{x}^2+6=2\text{x}-10$
$\Rightarrow-\text{x}=-10$
$\Rightarrow\text{x}=10$
$\therefore\text{x}=10$
Verification:
$\text{L.H.S.}=\frac{(\text{x}+2)(2\text{x}-3)-2\text{x}^2+6}{\text{x}-5}$
$=\frac{(10+2)(2\times10-3)-2(10)^2+6}{10-5}$
$=\frac{12(20-3)-2\times(100)+6}{5}$
$=\frac{12\times17-200+6}{5}$
$=\frac{204-200+6}{5}$
$=\frac{10}{5}=2=\text{R.H.S.}$
By cross multiplication:
$(\text{x}+2)(2\text{x}-3)-2\text{x}^2+6=2(\text{x}-5)$
$2\text{x}^2-3\text{x}+4\text{x}-6-2\text{x}^2+6=2\text{x}-10$
$\Rightarrow-\text{x}=-10$
$\Rightarrow\text{x}=10$
$\therefore\text{x}=10$
Verification:
$\text{L.H.S.}=\frac{(\text{x}+2)(2\text{x}-3)-2\text{x}^2+6}{\text{x}-5}$
$=\frac{(10+2)(2\times10-3)-2(10)^2+6}{10-5}$
$=\frac{12(20-3)-2\times(100)+6}{5}$
$=\frac{12\times17-200+6}{5}$
$=\frac{204-200+6}{5}$
$=\frac{10}{5}=2=\text{R.H.S.}$
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