Question
Solve the following equation and verify your answer: $\frac{\text{x}+3}{\text{x}-3}+\frac{\text{x}+2}{\text{x}-2}=2$
✓
Answer
$\frac{\text{x}+3}{\text{x}-3}+\frac{\text{x}+2}{\text{x}-2}=2$
$=\frac{(\text{x}+3)(\text{x}-2)+(\text{x}+2)(\text{x}-3)}{(\text{x}-3)(\text{x}-2)}=2$
$=\frac{\text{x}^2-2\text{x}+3\text{x}-6+\text{x}^2-3\text{x}+2\text{x}-6}{\text{x}^2-2\text{x}-3\text{x}+6}=2$
$\Rightarrow\frac{2\text{x}^2-12}{\text{x}^2-5\text{x}+6}=\frac{2}{1}$
By cross multiplication: $2\text{x}^2-12=2\text{x}^2-10\text{x}+12$
$\Rightarrow2\text{x}^2-2\text{x}^2+10\text{x}=12+12$ (By transposition)
$\Rightarrow10\text{x}=24$
$\Rightarrow\text{x}=\frac{24}{10}=\frac{12}{5}$
$\therefore\text{x}=\frac{12}{5}$
Verification: $\text{L.H.S.}=\frac{\text{x}+3}{\text{x}-3}+\frac{\text{x}+2}{\text{x}-2}=\frac{\frac{12}{5}+3}{\frac{12}{5}-3}+\frac{\frac{12}{5}+2}{\frac{12}{5}-2}$
$=\frac{\frac{12+12}{5}}{\frac{12-15}{5}}+\frac{\frac{12+10}{5}}{\frac{12-10}{5}}=\frac{\frac{27}{5}}{\frac{-3}{5}}+\frac{\frac{22}{5}}{\frac{2}{5}}$
$=\frac{27}{5}\times\frac{5}{-3}+\frac{22}{5}\times\frac{5}{2}$
$=-9+11=2=\text{R.H.S.}$
$=\frac{(\text{x}+3)(\text{x}-2)+(\text{x}+2)(\text{x}-3)}{(\text{x}-3)(\text{x}-2)}=2$
$=\frac{\text{x}^2-2\text{x}+3\text{x}-6+\text{x}^2-3\text{x}+2\text{x}-6}{\text{x}^2-2\text{x}-3\text{x}+6}=2$
$\Rightarrow\frac{2\text{x}^2-12}{\text{x}^2-5\text{x}+6}=\frac{2}{1}$
By cross multiplication: $2\text{x}^2-12=2\text{x}^2-10\text{x}+12$
$\Rightarrow2\text{x}^2-2\text{x}^2+10\text{x}=12+12$ (By transposition)
$\Rightarrow10\text{x}=24$
$\Rightarrow\text{x}=\frac{24}{10}=\frac{12}{5}$
$\therefore\text{x}=\frac{12}{5}$
Verification: $\text{L.H.S.}=\frac{\text{x}+3}{\text{x}-3}+\frac{\text{x}+2}{\text{x}-2}=\frac{\frac{12}{5}+3}{\frac{12}{5}-3}+\frac{\frac{12}{5}+2}{\frac{12}{5}-2}$
$=\frac{\frac{12+12}{5}}{\frac{12-15}{5}}+\frac{\frac{12+10}{5}}{\frac{12-10}{5}}=\frac{\frac{27}{5}}{\frac{-3}{5}}+\frac{\frac{22}{5}}{\frac{2}{5}}$
$=\frac{27}{5}\times\frac{5}{-3}+\frac{22}{5}\times\frac{5}{2}$
$=-9+11=2=\text{R.H.S.}$
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