Question

please solve this question ​

Answer 1

Answer:

$\frac{ {(2x - 3)}^{2} - 4 {(x - 1)}^{2} }{(x - 1)(2x - 3)} = 3 \\ \frac{4 {x}^{2} - 12x + 9 - 4( {x}^{2} - 2x + 1)}{2 {x}^{2} - 2x + 3 - 3x } = 3 \\ \frac{4 {x}^{2} - 12x + 9 - 4 {x}^{2} + 8x - 4}{2 {x}^{2} - 5x + 3} = 3 \\ \frac{ - 4x + 5}{2 {x}^{2} - 5x + 3} = 3 \\ - 4x + 5 = 6 {x}^{2} - 15x + 9 \\ \bold{ 6 {x}^{2} - 11x + 4 = 0} \\ 6 {x}^{2} - 8x - 3x + 4 = 0 \\ 2x(3x - 4) - 1(3x - 4) = 0 \\ (2x - 1)(3x - 4) = 0 \\ \\ 2x - 1 = 0 \\ = > 2x = 1 = > \bold{x = \frac{1}{2} } \\ 3x - 4 = 0 \\ = > 3x = 4 = > \bold{x = \frac{4}{3} } \\ hope \: it \: helps \: u...$

Answer 2

Answer:

$x=\frac{4}{3}$ OR $x=\frac{1}{2}$

Step-by-step explanation:

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

$\frac{(2x-3)^{2}-4(x-1)^{2} }{(x-1)(2x-3)}$

$\frac{(4x^{2}+9-12x) -4x^{2}-4+8x }{(2x^{2}+3-5x )}$

$\frac{5-4x}{(2x^{2}+3-5x )}=3\\5-4x=3(2x^{2}+3-5x )\\5-4x=6x^{2}+9-15x \\6x^{2}+4-11x=0$

On solving the quadratic, you get:-

$x=\frac{4}{3}$ OR $x=\frac{1}{2}$

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