Math, asked by naitik0059, 9 months ago

please solve this question ​

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Answered by renuagrawal393
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...

Answered by arunsomu13
0

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}

Hope this answer helps you, if so please mark Brainliest :)

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