Question
Solve the following equation and verify your answer: $\frac{2\text{x}+1}{3\text{x}-2}=\frac{5}{9}$
✓
Answer
$\frac{2\text{x}+1}{3\text{x}-2}=\frac{5}{9}$By cross multiplication:
$9(2\text{x+1})=5(3\text{x}-2)$
$\Rightarrow18\text{x}+9=15\text{x}-10$
$\Rightarrow18\text{x}-15\text{x}=-10$(By transposition)
$\Rightarrow3\text{x}=-19$
$\Rightarrow\text{x}=\frac{-19}{3}$
$\therefore\text{x}=\frac{-19}{3}$
Verification:
$\text{L.H.S.}=\frac{2\text{x}+1}{3\text{x}-2}=\frac{2\Big(\frac{-19}{3}\Big)+1}{3\Big(\frac{-19}{3}\Big)-2}$
$=\frac{\frac{-38}{3}+1}{\frac{-57}{3}-2}=\frac{\frac{-38+3}{3}}{\frac{-57-6}{3}}$
$=\frac{-35}{3}\times\frac{3}{-63}=\frac{5}{9}=\text{R.H.S.}$
$9(2\text{x+1})=5(3\text{x}-2)$
$\Rightarrow18\text{x}+9=15\text{x}-10$
$\Rightarrow18\text{x}-15\text{x}=-10$(By transposition)
$\Rightarrow3\text{x}=-19$
$\Rightarrow\text{x}=\frac{-19}{3}$
$\therefore\text{x}=\frac{-19}{3}$
Verification:
$\text{L.H.S.}=\frac{2\text{x}+1}{3\text{x}-2}=\frac{2\Big(\frac{-19}{3}\Big)+1}{3\Big(\frac{-19}{3}\Big)-2}$
$=\frac{\frac{-38}{3}+1}{\frac{-57}{3}-2}=\frac{\frac{-38+3}{3}}{\frac{-57-6}{3}}$
$=\frac{-35}{3}\times\frac{3}{-63}=\frac{5}{9}=\text{R.H.S.}$
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