Solve the following quadratic equation by factorization:

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

Given:

Given quadratic equation is $\frac{x-4}{x-5}+\frac{x-6}{x-7}=\frac{10}{3}, x ≠5,7$.

To do:

We have to solve the given quadratic equation by factorization. 

Solution:

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

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

$\frac{x^2-4x-7x+28+x^2-6x-5x+30}{x^2-5x-7x+35}=\frac{10}{3}$

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

$3(2x^2-22x+58)=10(x^2-12x+35)$   (on cross multiplication)

$3\times2(x^2-11x+29)=10(x^2-12x+35)$

$3x^2-33x+87=5x^2-60x+175$

$(5-3)x^2+(-60+33)x+175-87=0$

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

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

$2x(x-8)-11(x-8)=0$

$(2x-11)(x-8)=0$

$2x-11=0$ or $x-8=0$

$2x-11=0$ or $x-8=0$

$2x=11$ or $x=8$

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

The values of $x$ are $\frac{11}{2}$ and $8$. 

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Updated on: 10-Oct-2022

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