Question

solve the following:-

5x/3x-3+6/x+2=5/3​

Answer 1

Step-by-step explanation:

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Answer 2

$\tt\frac{5x}{3x - 3} + \frac{6}{x + 2} = \frac{5}{3}$

$\star\tt{\large{\underline{\underline{\blue{According\:to\:question}}}}}$

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

Multiplying both sides,

$3(x - 1) \: (x + 2) \: ( \tt\frac{5x}{3(x - 1)} + \frac{6}{x + 2} ) \\ \\ = ( \frac{5}{3} ) \: \: \: [3( x- 1) \: \: x + 2]$

$\implies$$5x (x+2) +3(x-1) (6) = 5(x-1) (x+2)$

$\implies$$5x^2 + 10x + 18(x-1) = 5(x^2+x-2)$

$\implies$$5x^2 + 10x + 18x -18 = 5x^2 + 5x -10$

$\implies$$28x -18 = 5x -10$

$\implies$$23x = 8$

$\implies$x = $\tt\frac{8}{23}$

$\huge\sf{\bold{\blue{\underline{Verification:}}}}$

$\implies$$\dfrac{3\bigg(\dfrac{8}{23}\bigg)}{3\bigg(\dfrac{8}{23}\bigg)\ -\ 3}\ +\ \dfrac{6}{\bigg(\dfrac{8}{23}\bigg)}\ +\ 2\ =\ \dfrac{5}{3}\\\\\\\dfrac{\dfrac{40}{23}}{\dfrac{-45}{23}}\ +\ \dfrac{6}{\dfrac{54}{23}}\ +\ \dfrac{5}{3}\\\\\\\dfrac{-8}{9}\ +\ \dfrac{23}{9}\ +\ \dfrac{5}{3}\\\\\\\dfrac{5}{3}\ =\ \dfrac{5}{3}$

Hence, verified (R.H.S = L.H.S)