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Quadratic Equations — Maths STD 10 — Question

Gujarat BoardEnglish MediumSTD 10MathsQuadratic Equations3 Marks
Question
Solve the following quadratic equations by factorization:
$\frac{\text{x}-1}{\text{x}-2}+\frac{\text{x}-3}{\text{x}-4}=3\frac{1}{3},$ $\text{x}\neq2, 4$
✓

Answer

We have been given
$\frac{\text{x}-1}{\text{x}-2}+\frac{\text{x}-3}{\text{x}-4}=3\frac{1}{3}$
$3\left(x^2-5 x+4+x^2-5 x+6\right)=10\left(x^2-6 x+8\right) $
$4 x^2-30 x+50=0 $
$2 x^2-15 x+25=0 $
$2 x^2-10 x-5 x+25=0 $
$2 x(x-5)-5(x-5)=0 $
$(2 x-5)(x-5)=0$
Therefore,
$2x - 5 = 0$
$2x = 5$
$\text{x}=\frac{5}{2}$
or, $x - 5 = 0$
$x = 5$
Hence, $\text{x}=\frac{5}{2}$ or $x = 5$

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