Question

1/x+3 + 1/x+2 = 2/x+4
SOLVE☝️​

Answer 1

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

$\sf \dfrac{1}{x + 3} + \dfrac{1}{x + 2} = \dfrac{2}{x + 4}$

Taking LCM left side

$\sf \large= > \dfrac{(x + 2) + (x + 3)}{(x + 3)(x + 2)} = \dfrac{2}{x + 4}$

$\sf \large= > \dfrac{x + 2 + x + 3}{x(x + 2) + 3(x + 2)} = \dfrac{2}{x + 4}$

$\sf \large= > \dfrac{2x + 5}{ {x}^{2} + 2x + 3x + 6 } = \dfrac{2}{x + 4}$

$\sf \large= > \dfrac{2x + 5}{ {x}^{2} + 5x + 6} = \dfrac{2}{x + 4}$

Now,cross multiply to both sides:

$\sf \large= > (2x + 5)(x + 4) = 2 {x}^{2} + 10x + 12$

$\sf \large= > x(2x + 5) + 4(2x + 5) = 2 {x}^{2} + 10x + 12$

$\sf \large= > 2 {x}^{2} + 5x + 8x + 20 = 2 {x}^{2} + 10x + 12$

We see that 2x² lies on both side so,it gets automatically cancelled:

$\sf \large= >5x + 8x + 20 = 10x + 12$

$\sf \large= > 13x + 20 = 10x + 12$

$\sf \large= > 13x - 10x = 12 - 20$

$\sf \large= > 3x = - 8$

$\sf \large= > x = - \bold{\dfrac{8}{3} }$

Check:-

$\sf \large= > \dfrac{1}{x +3 } + \dfrac{1}{x + 2} = \dfrac{2}{x + 4}$

taking LHS

$\sf \large= > \dfrac{1}{ - \dfrac{8}{3} + 3 } + \dfrac{1}{ \dfrac{ - 8}{3} + 2}$

$\sf \large= > \dfrac{1}{ \dfrac{ - 8 + 9}{3} } + \dfrac{1}{ \dfrac{ - 8 + 6}{3} }$

$\sf \large= > \dfrac{1}{ \dfrac{1}{3} } + \dfrac{1}{ \dfrac{ - 2}{3} }$

$\sf \large= > 3 - \dfrac{3}{2}$

$\sf \large= > \dfrac{6 - 3}{2} = \dfrac{3}{2}$

Now,taking RHS

$\sf \large= > \dfrac{2}{x + 4}$

$\sf \large= > \dfrac{2}{ \dfrac{ - 8}{3} + 4}$

$\sf \large= > \dfrac{2}{ \dfrac{ - 8 + 12}{3} } = \dfrac{2}{ \dfrac{4}{3} }$

$\sf \large= > \dfrac{6}{4} = \dfrac{3}{2}$

Hence ,LHS =RHS(verified)