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Linear Equations in One Variable — Maths STD 8 — Question

Maharashtra BoardEnglish MediumSTD 8MathsLinear Equations in One Variable5 Marks
Question
Solve the following equation and verify your answer:
$\Big(\frac{\text{x}+1}{\text{x}+2}\Big)^2=\frac{\text{x}+2}{\text{x}+4}$
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Answer

$\Big(\frac{\text{x}+1}{\text{x}+2}\Big)^2=\frac{\text{x}+2}{\text{x}+4}$
$\Rightarrow\frac{(\text{x}+1)^2}{(\text{x}+2)^2}$
$=\frac{(\text{x}+1)}{(\text{x}+4)}$
By cross multiplication,
$=(\text{x}+1)^2\times(\text{x}+4)=(\text{x}+2)^2(\text{x}+2)$
$=(\text{x}^2+2\text{x}+1)(\text{x}+4)=(\text{x}^2+4\text{x}+4)(\text{x}+2)$
$\Rightarrow\text{x}^3+4\text{x}^2+2\text{x}^2+8\text{x}+\text{x}+ 4\\=\text{x}^3+2\text{x}^2+4\text{x}^2+8\text{x}+4\text{x}+8$
$\Rightarrow\text{x}^3+6\text{x}^2+9\text{x}+4=\text{x}^3+6\text{x}^2+12\text{x}+8$
$\Rightarrow\text{x}^3+6\text{x}^2+9\text{x}-\text{x}^3-6\text{x}^2-12\text{x}=8-4$
(By transposition)
$\Rightarrow-3\text{x}$
$=4$
$\Rightarrow\text{x}=\frac{4}{-3}$
$\therefore\text{x}=\frac{-4}{3}$
Verification:
$\text{L.H.S}=\bigg(\frac{\text{x}+1}{\text{x}+2}\bigg)^2=\Bigg(\frac{\frac{-4}{3}+1}{\frac{-4}{3}+2}\Bigg)$
$=\Bigg(\frac{\frac{-4+3}{3}}{\frac{-4+6}{3}}\Bigg)^2=\Bigg(\frac{\frac{-1}{3}}{\frac{2}{3}}\Bigg)^2$
$=\bigg(\frac{\frac{-1}{3}}{3}\times\frac{3}{2}\bigg)^2=\bigg(\frac{-1}{2}\bigg)^2=\frac{1}{4}$
$\text{R.H.S}=\frac{\text{x}+2}{\text{x}+4}=\frac{\frac{-4}{3}+2}{\frac{-4}{3}+2}=\frac{\frac{-4+6}{3}}{\frac{-4+12}{3}}$
$=\frac{\frac{2}{3}}{\frac{8}{3}}=\frac{2}{3}\times\frac{3}{8}=\frac{1}{4}$
$\therefore\text{L.H.S}=\text{R.H.S}$

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