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Quadratic Equations — Maths STD 10 — Question

CBSE 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
$\frac{\text{x}-1}{\text{x}-2}+\frac{\text{x}-3}{\text{x}-4}=3\frac{1}{3}$
$3(x^2 - 5x + 4 + x^2 - 5x + 6) = 10(x^2 - 6x + 8)$
$4x^2 - 30x + 50 = 0$
$2x^2- 15x + 25 = 0$
$2x^2- 10x - 5x + 25 = 0$
$2x(x - 5) - 5(x - 5) = 0$
$(2x - 5)(x - 5) = 0$
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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