Question

x-2/x-3+x-4/x-5=10/3

Answer 1

Answer:

$x = 6 \: Or \: x = \frac{7}{2}$

Step-by-step explanation:

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

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

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

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

$\implies \frac{2(x^{2}-7x+11}{x^{2}-8x+15}=\frac{10}{3}$

$\implies \frac{(x^{2}-7x+11}{x^{2}-8x+15}=\frac{10}{2\times 3}$

$\implies \frac{(x^{2}-7x+11}{x^{2}-8x+15}=\frac{5}{ 3}$

$\implies 3(x^{2}-7x+11)=5(x^{2}-8x+15)$

$\implies 3x^{2}-21x+33=5x^{2}-40x+75$

$\implies 0 = 5x^{2}-40x+75-3x^{2}+21x-33$

$\implies 2x^{2}-19x+42=0$

$\implies 2x^{2}-12x-7x+42=0$

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

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

$\implies x-6 = 0 \: Or \: 2x-7 =0$

$\implies x = 6 \: Or \: x = \frac{7}{2}$

Therefore,

$x = 6 \: Or \: x = \frac{7}{2}$

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