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Linear Equations in One Variable — MATHS STD 8 — Question

Gujarat BoardEnglish MediumSTD 8MATHSLinear Equations in One Variable5 Marks
Question
Solve the following equation and verify your answer: $\frac{7\text{x}-2}{5\text{x}-1}=\frac{7\text{x}+3}{5\text{x}+4}$
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Answer

$\frac{7\text{x}-2}{5\text{x}-1}=\frac{7\text{x}+3}{5\text{x}+4}$
By cross multiplication:
$(7\text{x}-2)(5\text{x}+4)=(7\text{x}+3)(5\text{x}-1)$
$\Rightarrow35\text{x}^2+28\text{x}-10\text{x}-8=35\text{x}^2-7\text{x}+15\text{x}-$
$\Rightarrow35\text{x}^2+18\text{x}-8=35\text{x}^2+8\text{x}-3$
$\Rightarrow35\text{x}^2+18\text{x}-35\text{x}^2-8\text{x}=-3+8$ (By transpositior)
$\Rightarrow19\text{x}=-27$
$\Rightarrow\text{x}=\frac{-27}{19}$
$\therefore\text{x}=\frac{1}{2}$
Verification:
$\text{L.H.S.}=\frac{7\text{x}-2}{5\text{x}-1}=\frac{7\times\frac{1}{2}-2}{5\times\frac{1}{2}-1}=\frac{\frac{7}{2}-\frac{2}{1}}{\frac{5}{2}-1}$
$=\frac{\frac{7-4}{2}}{\frac{5-2}{2}}=\frac{\frac{3}{2}}{\frac{3}{2}}=\frac{3}{2}\times\frac{2}{3}=1$
$\text{R.H.S.}=\frac{7\text{x}+3}{5\text{x}+4}$
$=-\frac{7\times\frac{1}{2}+3}{5\times\frac{1}{2}+4}=\frac{\frac{7}{2}+3}{\frac{5}{2}+4}$
$=\frac{\frac{7+6}{2}}{\frac{5+8}{2}}=\frac{\frac{13}{2}}{\frac{13}{2}}=\frac{13}{2}\times\frac{2}{13}=1$
$\therefore\text{L.H.S.}=\text{R.H.S.}$

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