Question
Solve the following equation and verify your answer: $\frac{\text{y}-(7-8\text{y})}{9\text{y}-(3+4\text{y})}=\frac{2}{3}$
✓
Answer
$\frac{\text{y}-(7-8\text{y})}{9\text{y}-(3+4\text{y})}=\frac{2}{3}$
$\Rightarrow\frac{\text{y}-7+8\text{y}}{9\text{y}-3-4\text{y}}=\frac{2}{3}$
$\Rightarrow\frac{9\text{y}-7}{5\text{y}-3}=\frac{2}{3}$
By cross multiplication:
$\Rightarrow27\text{y}-21=10\text{y}-6$
$\Rightarrow27\text{y}-21=10\text{y}-6+21$ (By transposition)
$\Rightarrow17\text{y}=15$
$\Rightarrow\text{y}=\frac{15}{17}$
$\therefore\text{y}=\frac{15}{17}$Verification:
$\text{L.H.S.}=\frac{\text{y}-(7-8\text{y})}{9\text{y}-(3+4\text{y})}=\frac{\text{y-7+8}\text{y}}{9\text{y}-3-4\text{y}}=\frac{9\text{y}-7}{5\text{y}-3}$
$=\frac{9\Big(\frac{15}{17}\Big)-7}{5\Big(\frac{15}{17}\Big)-3}=\frac{\frac{135}{17}-7}{\frac{75}{17}-3}$
$=\frac{\frac{135-199}{17}}{\frac{75-51}{17}}=\frac{\frac{16}{17}}{\frac{24}{17}}=\frac{16}{17}\times\frac{17}{24}$
$=\frac{2}{3}=\text{R.H.S.}$
$\Rightarrow\frac{\text{y}-7+8\text{y}}{9\text{y}-3-4\text{y}}=\frac{2}{3}$
$\Rightarrow\frac{9\text{y}-7}{5\text{y}-3}=\frac{2}{3}$
By cross multiplication:
$\Rightarrow27\text{y}-21=10\text{y}-6$
$\Rightarrow27\text{y}-21=10\text{y}-6+21$ (By transposition)
$\Rightarrow17\text{y}=15$
$\Rightarrow\text{y}=\frac{15}{17}$
$\therefore\text{y}=\frac{15}{17}$Verification:
$\text{L.H.S.}=\frac{\text{y}-(7-8\text{y})}{9\text{y}-(3+4\text{y})}=\frac{\text{y-7+8}\text{y}}{9\text{y}-3-4\text{y}}=\frac{9\text{y}-7}{5\text{y}-3}$
$=\frac{9\Big(\frac{15}{17}\Big)-7}{5\Big(\frac{15}{17}\Big)-3}=\frac{\frac{135}{17}-7}{\frac{75}{17}-3}$
$=\frac{\frac{135-199}{17}}{\frac{75-51}{17}}=\frac{\frac{16}{17}}{\frac{24}{17}}=\frac{16}{17}\times\frac{17}{24}$
$=\frac{2}{3}=\text{R.H.S.}$
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