Question

Solve and Check ur ans.

$\tt\dfrac{x^2-(x+2)(x+3)}{7x+1}=\dfrac{2}{3}$​

Answer 1

SOLUTION :

$\tt\leadsto\dfrac{x^2-(x+2)(x+3)}{7x+1}=\dfrac{2}{3}$

$\tt\leadsto\dfrac{x^2-(x^2+5x+6)}{7x+1}=\dfrac{2}{3}$

$\tt\leadsto\dfrac{ x^2- x^2-5x-6)}{7x+1}=\dfrac{2}{3}$

(Here x² and -x² are cut)

$\tt\leadsto{3(-5x-6)=2(7x+1)}$

[By Cross Multiplication]

$\tt\leadsto{-15x-18=14x+2}$

$\tt\leadsto{-15x-14x=2+18}$

$\tt\leadsto{-29x=20}$

$\tt\leadsto{x=}\dfrac{-20}{29}$

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★Verification:

$\tt\mapsto\dfrac{x^2-(x^2+5x+6)}{7x+1}=\dfrac{2}{3}$

$\sf\underline\red{Putting~the~value~of~x :}$

$\sf\mapsto{\dfrac{\bigg(\dfrac{-20}{29}\bigg)^{2}- \bigg[\dfrac{-20}{29}^{2}+5 \times \dfrac{-20}{29}+ 6 \bigg ]}{7 \times\dfrac{-20}{9}+ 1 } = \dfrac{2}{3}}$

$\sf\mapsto{\dfrac{\cancel{{\bigg(\dfrac{-20}{29}\bigg )}^2} - \cancel{{\bigg(\dfrac{ -20}{29}\bigg )}^2} +\dfrac{100}{29} - 6 }{\dfrac{-140}{29}+1}= \dfrac{2}{3}}$

$\sf\mapsto{\dfrac{\dfrac{100}{29} -6}{\dfrac{- 140}{29}+1}= \dfrac{2}{3}}$

$\sf\mapsto{\dfrac{\dfrac{100-174}{29} }{\dfrac{- 140+29}{29}}= \dfrac{2}{3}}$

$\sf\mapsto{\dfrac{\dfrac{-74}{29} }{\dfrac{- 111}{29}}= \dfrac{2}{3}}$

$\mapsto\dfrac{-74}{29}\times\dfrac{-29}{111}=\dfrac{2}{3}$

$\mapsto\dfrac{74}{111}=\dfrac{2}{3}$

$\mapsto\dfrac{37 \times2}{37 \times3}=\dfrac{2}{3}$

$\mapsto\dfrac{2}{3}=\dfrac{2}{3}$

$\sf\underline\green{Hence, Verified.}$

L.H.S. = R.H.S.