Question

Please answer this linear equation ​

Answer 1

$\bf{\dfrac{1}{x+3}+\dfrac{1}{x+2} = \dfrac{2}{x+4}}\\\\\\\\ \rm{Taking\:LCM \: on \: Left\: Hand \: Side,}$

$\sf{\dfrac{(x+2)+(x+3)}{(x+3)(x+2)} = \dfrac{2}{x+4}}\\\\\\\\\longrightarrow \: \: \sf{\dfrac{x+2+x+3}{x(x+2)+3(x+2)} = \dfrac{2}{x+4}}\\\\\\\\\longrightarrow \: \: \sf{\dfrac{2x+5}{x^{2} + 2x+3x + 6} = \dfrac{2}{x+4}}$

$\longrightarrow \: \: \sf{\dfrac{2x+5}{x^{2} +5x+6}= \dfrac{2}{x+4}}$

On Cross Multiplication,

$\\\sf{(2x+5)(x+4)=2(x^{2} +5x+6)}\\\\\\\longrightarrow \: \: \sf{2x(x+4)+5(x+4) = 2(x^{2} + 5x + 6)}\\\\\\\\\longrightarrow \: \: \sf{2x^{2} +8x + 5x + 20= 2x^{2} + 10x + 12}\\\\\\\\\longrightarrow \: \: \sf{2x^{2} + 13x + 20 = 2x^{2} +10x+12 }$

Cancelling 2x² on both sides,

$\\\\\sf{13x+20=10x+12}\\\\\\\implies \sf{13x-10x = 12-20}\\\\\\\implies \sf{3x = -8}\\\\\\\implies \boxed{\bf{x = \dfrac{-8 }{3}}}$

Answer 2

Answer:

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