# Solve the Following Quadratic Equations by Factorization: - Mathematics

Solve the following quadratic equations by factorization: $\frac{x - 4}{x - 5} + \frac{x - 6}{x - 7} = \frac{10}{3}; x \neq 5, 7$

#### Solution

$\frac{x - 4}{x - 5} + \frac{x - 6}{x - 7} = \frac{10}{3}$

$\Rightarrow \frac{\left( x - 4 \right)\left( x - 7 \right) + \left( x - 6 \right)\left( x - 5 \right)}{\left( x - 5 \right)\left( x - 7 \right)} = \frac{10}{3}$

$\Rightarrow \frac{x^2 - 11x + 28 + x^2 - 11x + 30}{x^2 - 12x + 35} = \frac{10}{3}$

$\Rightarrow \frac{2 x^2 - 22x + 58}{x^2 - 12x + 35} = \frac{10}{3}$

$\Rightarrow 3\left( 2 x^2 - 22x + 58 \right) = 10\left( x^2 - 12x + 35 \right)$

$\Rightarrow 6 x^2 - 66x + 174 = 10 x^2 - 120x + 350$

$\Rightarrow 4 x^2 - 54x + 176 = 0$

$\Rightarrow 2 x^2 - 27x + 88 = 0$

$\Rightarrow 2 x^2 - 11x - 16x + 88 = 0$

$\Rightarrow x\left( 2x - 11 \right) - 8\left( 2x - 11 \right) = 0$

$\Rightarrow \left( x - 8 \right)\left( 2x - 11 \right) = 0$

$\Rightarrow x - 8 = 0 \text { or } 2x - 11 = 0$

$\Rightarrow x = 8 or x = \frac{11}{2}$

Hence, the factors are 8 and $\frac{11}{2}$.

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 10 Maths