Question

Solve for x: (4x-3/2x+1)-10(2x+1/4x-3)=3

Answer 1

: here is your answer hope that help you

Answer 2

x = $\frac{1}{8}$ or  $\frac{-4}{3}$

Step-by-step explanation:

Given,

$\frac{4x-3}{2x+1} -10( \frac{2x+1}{4x-3}) =3$

Multiplying by 10

$\frac{4x-3}{2x+1} - \frac{20x+10}{4x-3} =3$

L.C.M. will be (2x + 1) (4x - 3)

$\frac{(4x-3)(4x-3) - (20x+10) (2x+1)}{(2x+1)(4x-3)} = 3$

$\frac{(16x^{2} + 9- 24x)- (40x^{2}+20x+20x +10) }{(8x^{2}+4x-6x-3) } = 3$

$\frac{(16x^{2} + 9- 24x)- (40x^{2}+40x +10) }{(8x^{2}-2x-3) } = 3$

$\frac{16x^{2} + 9- 24x- 40x^{2}-40x -10 }{8x^{2}-2x-3 } = 3$

$\frac{-24x^{2}-64x-1 }{8x^{2}-2x-3 } = 3$

-24x² - 64x -1 = 3 (8x² - 2x - 3)

-24x² - 64x -1 = 24x² - 6x - 9

-24x² - 64x -1 -24x² + 6x + 9 = 0

-48x² - 58x + 8 = 0

48x² + 58x - 8 = 0

Taking out 2 as common:

2 (24x² + 29x - 4) = 0

Splitting the middle term:

24x² + 32x - 3x - 4 = 0

8x (3x + 4) -1 (3x + 4) = 0

(8x - 1) (3x + 4) = 0

8x - 1 = 0

∴ x = $\frac{1}{8}$

3x + 4 = 0

∴ x = $\frac{-4}{3}$

x = $\frac{1}{8}$ or  $\frac{-4}{3}$

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