Question

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

Answer 1

$\frac{x-4}{x-5}+\frac{x-6}{x-7}=\frac{10}{3}\frac{(x-4)(x-7)+(x-6)(x-5)}{(x-5)(x-7)}=\frac{10}{3}3[x^2-4x-7x+28+x^2-6x-5x+30]=10[x^2-12x+35]$
$3[2x^2-22x+58]=10[x^2-12x+35]6x^2-66x+174=10x^2-120x+3504x^2-54x+176=02x^2-27x+88=0$
$x=\frac{27\pm \sqrt{{27}^2-4\left(2\right)\left(88\right)}}{2\left(2\right)}x=\frac{27\pm 5}{4}x=\frac{32}{4},\frac{22}{4}x=8,5.5$