Solve the following quadratic equation by factorization:

$\frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3}, x ≠3, 5$

Given:

Given quadratic equation is $\frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3}, x ≠3, 5$.

To do:

We have to solve the given quadratic equation by factorization. 

Solution:

$\frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3}$

$\frac{(x-2)(x-5)+(x-4)(x-3)}{(x-3)(x-5)}=\frac{10}{3}$

$\frac{x^2-5x-2x+10+x^2-3x-4x+12}{x^2-5x-3x+15}=\frac{10}{3}$

$\frac{2x^2-14x+22}{x^2-8x+15}=\frac{10}{3}$

$3(2x^2-14x+22)=10(x^2-8x+15)$   (on cross multiplication)

$3\times2(x^2-7x+11)=10(x^2-8x+15)$

$3x^2-21x+33=5x^2-40x+75$

$(5-3)x^2+(-40+21)x+75-33=0$

$2x^2-19x+42=0$

$2x^2-12x-7x+42=0$

$2x(x-6)-7(x-6)=0$

$(2x-7)(x-6)=0$

$2x-7=0$ or $x-6=0$

$2x-7=0$ or $x-6=0$

$2x=7$ or $x=6$

$x=\frac{7}{2}$ or $x=6$

The values of $x$ are $\frac{7}{2}$ and $6$.  

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Updated on: 10-Oct-2022

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