Question

plz bro solve this question ....​

Answer 1

Answer

6 ± √10

Explanation

Let, x - 1 = a

x - 3 = b

Then, x - 1 - 1 = a - 1

→ x - 2 = a - 1

And, x - 3 - 1 = b - 1

→ x - 4 = b - 1

Now,

$\sf \frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = 2\frac{1}{3}$

Put value of (x - 1), (x - 2), (x - 3), (x - 4) in terms of a and b

$\sf\frac{a}{a - 1} + \frac{b}{b - 1} = \frac{7}{3}$

Take LCM

$\sf\frac{a(b-1) + b(a-1)}{(a-1)(b-1)} = \frac{7}{3}$

This gives

$\sf\frac{ab-a+ab-b}{ab-a-b+1} = \frac{7}{3}$

$\sf\frac{2ab-a-b}{ab-a-b+1} = \frac{7}{3}$

Now cross multiply

→ 6ab - 3a - 3b = 7ab - 7a - 7b + 7

→ 7a + 7b - 3a - 3b = 7ab - 6ab + 7

→ 4(a + b) = ab + 7

Now put a as (x - 1) and b as (x - 3)

→ 4(x - 1 + x - 3) = (x - 1)(x - 3) + 7

→ 4(2x - 4) = (x² - x - 3x + 3) + 7

→ 8x - 16 = x² - 4x + 3 + 7

→ x² - 4x - 8x + 10 + 16 = 0

→ x² - 12x + 26 = 0

Now solve it using quadratic equation

$\sf x = \frac{-(-12)\pm\sqrt{(-12)^2 - 4(26)(1)}}{2}$

$\sf x = \frac{12\pm\sqrt{144-104}}{2}$

$\sf x = \frac{12\pm\sqrt{40}}{2}$

$\sf x = \frac{12\pm2\sqrt{10}}{2}$

$\sf x = 6 \pm \sqrt{10}$