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Quadratic Equations — Maths STD 10 — Question

Gujarat BoardEnglish MediumSTD 10MathsQuadratic Equations4 Marks
Question
Solve the following quadratic equations by factorization:
$\frac{3}{\text{x}+1}-\frac{1}{2}=\frac{2}{3\text{x}-1},$ $\text{x}\neq-1,\frac{1}{3}$
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Answer

$\frac{3}{\text{x}+1}-\frac{1}{2}=\frac{2}{3\text{x}-1}$
$\Rightarrow\frac{6-(\text{x}-1)}{2(\text{x}+1)}=\frac{2}{3\text{x}-1}$
$\Rightarrow\frac{6-\text{x}-1}{2\text{x}+2}=\frac{2}{3\text{x}-1}$
$ \Rightarrow(5-x)(3 x-1)=2(2 x+2)$
$ \Rightarrow 15 x-5-3 x^2+x=4 x+4$
$ \Rightarrow-3 x^2+16 x-5-4 x-4=0$
$ \Rightarrow-3 x^2+12 x+9=0$
$ \Rightarrow x^2-4 x+3=0$
$ \Rightarrow x^2-3 x-x+3=0$
$ \Rightarrow x(x-3)-1(x-3)=0$
$ \Rightarrow(x-1)(x-3)=0$
$ \Rightarrow x-1=0 \text { or } x-3=0$
$ \Rightarrow x=1 \text { or } x=3$
Hence, the factors are $3$ and $1$

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