Question

please answer this question.
Answer is x=1. But how I don't know.​

Answer 1

★Given:-

• $\displaystyle \sf \frac{x^2 + 5x+6}{x^2 + 4x +4} = \frac{3x+1}{5x-2}$

★To find:-

• Value of x.

★Solution:-

$\displaystyle :\implies \sf \frac{x^2 + 5x+6}{x^2 + 4x +4} = \frac{3x+1}{5x-2}$

$\displaystyle :\implies \sf \frac{x^2+3x+2x+6}{x^2+4x+4} = \frac{3x+1}{5x-2}$

As (x+2)² = x²+4x+4

$\displaystyle :\implies \sf \frac{x^2+3x+2x+6}{(x+2)^2} = \frac{3x+1}{5x-2}$

$\displaystyle :\implies \sf \frac{x(x+3)+2(x+3)}{(x+2)^2} = \frac{3x+1}{5x-2}$

$\displaystyle :\implies \sf \frac{(x+2)(x+3)}{(x+2)^2} =\frac{3x+1}{5x -2}$

$\displaystyle :\implies \sf \frac{x+3}{x+2} = \frac{3x+1}{5x-2}$

Where [x≠-2]

$\\ \displaystyle :\implies \sf (x+3)(5x-2) = (x+2)(3x+1)$

$\displaystyle :\implies \sf 5x^2 - 2x+15x -6=3x^2 + x+6x+2$

$\displaystyle :\implies \sf 5x^2 - 3x^2 -2x+15x - x -6x - 6 -2 = 0$

$\displaystyle :\implies \sf 2x^2+6x-8 = 0$

$\displaystyle :\implies \sf 2(x^2 + 3x-4) = 0$

$\displaystyle :\implies \sf x^2+3x-4 = 0$

$\displaystyle :\implies \sf x^2 + 4x - x - 4 =0$

$\displaystyle :\implies \sf x(x+4)-1(x+4) = 0 \\\\:\implies \sf (x-1)(x+4) = 0$

$\displaystyle :\implies \underline{\boxed{\sf x = 1,-4}}$

Hence,

The value of x is 1,-4.

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