Question

Please solve Question-4(ii)

Answer 1

Solution:

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Given & To find:

$\frac{4}{x} - 3 = \frac{5}{2x + 3}$

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$\frac{4}{x} -3 = \frac{5}{2x+3}$

=> $\frac{4 - 3x}{x} = \frac{5}{2x+3}$

=>$(4-3x)(2x+3) = (x)(5)$

=> $8x + 12 - 6x^2 - 9x = 5x$

=>$-6x^2 -9x + 8x +12 = 5x$

=> $6x^2 + 9x - 8x -12 = -5x$

=> $6x^2 +x - 12 +5x =0$

=> $6x^2 +6x -12 =0$

=> $6(x^2 + x -2) = 0$

=> $x^2 +x - 2 = 0$

=> $x^2 +2x - x -2 = 0$

=> $x(x+2) - (x+2) = 0$

=> $(x+1)(x+2) = 0$

we know that,

0 x anything is zero,.

Hence,.

either

x - 1 = 0
(or)
x+ 2 = 0

so,
x= 1 (or) x = -2

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                                 Hope it Helps !!

Answer 2

your solution has been provided to you