Question

$\sf \dfrac {2x+5}{3x-3}=\dfrac {5x+8}{4x-5}$ $\ltexts {Solve\:it}$
answer only if u..give it correct please..​

Answer 1

⠀⠀${\rm{\pink{\underline{\underline{★Answer★}}}}}$

${\rm{\red{\underline{\underline{Given\: Equation : }}}}}$

$\rm\dfrac {2x+5}{3x-3}=\dfrac {5x+8}{4x-5}$

${\rm{\purple{\underline{\underline{Solution}}}}}$

$\rm\dfrac {2x+5}{3x-3}=\dfrac {5x+8}{4x-5}$

Cross multiplying the terms, we get :-

$\longrightarrow\rm4x - 5(2x + 5) = 3x - 3(5x + 8)$

$\longrightarrow\rm 8x {}^{2} + 20x - 10 {x}^{} - 25 = 15x {}^{2} + 24x - 15x - 24$

$\longrightarrow\rm 8x {}^{2} + 10x - 25 = 15x {}^{2} + 9x - 24$

Taking all the terms to the L.H.S, we get :-

$\longrightarrow\rm 8x {}^{2} - 15x {}^{2} + 10x - 9x - 25 + 24 = 0$

$\longrightarrow\rm - 7x {}^{2} + x - 1 = 0$

Multiplying all the terms by ( - ), we get :-

$\longrightarrow\rm - ( - 7x {}^{2} + x - 1) = 0$

$\longrightarrow\rm 7x {}^{2} - x + 1 = 0$

By using the quadratic equation, we get :-

$\longrightarrow\rm x = \frac{- b± \sqrt{} (b {}^{2} - 4ac)}{2a}$

Here,

a = 7

b = 1

c = 1

Now, substituting the values,

$\longrightarrow\rm x = \frac{ - 1± \sqrt{} [( 1) {}^{2} - 4 \times 7 \times 1]}{2(7)}7$

$\longrightarrow\rm x = \frac{ - 1 ±\sqrt{1 - 28} }{14}$

$\longrightarrow\rm x = \frac{ - 1 \: ±\sqrt{} - 27}{14}$

No real solutions because the discriminant is negative.

RESULT:- NO SOLUTION

Answer 2

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