Question

solve x:: $\frac{3}{x+1} +\frac{4}{x-1} = \frac{29}{4x -1} ; x\neq 1, -1, \frac{1}{4}$

Answer 1

Solution :

Now,

$\frac{3}{x+1} + \frac{4}{x-1} = \frac{29}{4x-1}$

$\to \frac{3(x-1)+4(x+1)}{(x+1)(x-1)} = \frac{29}{4x-1}$

$\to \frac{3x-3+4x+4}{(x+1)(x-1)} = \frac{29}{4x-1}$

$\to \frac{7x+1}{x^{2}-1} = \frac{29}{4x-1}$

→ (7x + 1) (4x - 1) = 29 (x² - 1)

→ 28x² - 3x - 1 = 29x² - 29

→ (29 - 28) x² + 3x - 29 + 1 = 0

→ x² + 3x - 28 = 0

→ x² + (7 - 4) x - 28 = 0

→ x² + 7x - 4x - 28 = 0

→ x (x + 7) - 4 (x + 7) = 0

→ (x + 7) (x - 4) = 0

∴ either, x + 7 = 0 or, x - 4 = 0

→ x = - 7, 4

which is the required solution.

Answer 2

3/x+1 + 4/x-1 = 29/4x-1

{3(x-1)+4(x+1)}/(x-1)(x+1)=29/4x-1

(3x-3+4x+4)/x^2-1 = 29/4x-1

7x+1/x^2-1=29/4x-1

(7x+1)(4x-1)=29(x^2-1)

28x^2-7x+4x-1=29x^2-29

-3x-1+29=29x^2-28x^2

-3x+28=x^2

x^2+3x-28=0

x^2+7x-4x-28=0

x(x+7)-4(x+7)=0

(x+7)(x-4)=0

x= -7 or x=4

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