Question

solve the equation:-
$\frac{4}{x} - 3 = \frac{5}{2x + 3}$
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Answer 1

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Q:-solve the equation:-

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

$\huge\tt\underline\blue{ANSWER }$

------>>>>Here is your answer<<<<--------

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

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

$⟹ \frac{8x + 12 - 5x}{x(2x + 3)} = 3$

$⟹3x + 12 = 3x(2x + 3)$

$⟹3x + 12 = 6 {x}^{2} + 9x$

$⟹6 {x}^{2} + 6x - 12 = 0$

$⟹ {x}^{2} + x - 2 = 0$

$⟹ {x}^{2} + 2x - x - 2 = 0$

$⟹x(x + 2) - 1(x + 2) = 0$

$⟹(x + 2)(x - 1)$

$⟹x + 2 = 0 \: or \: x - 1 = 0$

$⟹x = - 2 \: or \: x = 1$

∴solutions of given equations are -2 and 1

HOPE IT HELPS YOU..

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Answer 2

⟹ \frac{4}{x} - 3 = \frac{5}{2x + 3}⟹x4−3=2x+35

⟹ \frac{4}{x} - \frac{5}{2x + 3} = 3⟹x4−2x+35=3

⟹ \frac{8x + 12 - 5x}{x(2x + 3)} = 3⟹x(2x+3)8x+12−5x=3

⟹3x + 12 = 3x(2x + 3)⟹3x+12=3x(2x+3)

⟹3x + 12 = 6 {x}^{2} + 9x⟹3x+12=6x2+9x

⟹6 {x}^{2} + 6x - 12 = 0⟹6x2+6x−12=0

⟹ {x}^{2} + x - 2 = 0⟹x2+x−2=0

⟹ {x}^{2} + 2x - x - 2 = 0⟹x2+2x−x−2=0

⟹x(x + 2) - 1(x + 2) = 0⟹x(x+2)−1(x+2)=0

⟹(x + 2)(x - 1)⟹(x+2)(x−1)

⟹x + 2 = 0 \: or \: x - 1 = 0⟹x+2=0orx−1=0

⟹x = - 2 \: or \: x = 1⟹x=−2orx=1

∴solutions of given equations are -2 and 1