Question

SOLVE
3/x+2 - 1/x+1 = 2/x+3​

Answer 1

hlo buddy

here is your answer....

Answer 2

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

Now, taking LCM to hand side

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

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

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

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

Now,cross multiply to both sides:

$\sf \large= > (2x + 1)(x + 3) = 2( {x}^{2} + 3x + 2)$

$\sf \large= > 2x(x + 3) + 1(x + 3) = 2 {x}^{2} + 6x + 4$

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

Now,2x²+6 lies on both sides so,it automatically gets cancelled

$\sf \large= > x + 3 = 4$

$\sf \large= > x = 4 - 3$

$\sf \large= > \bold{\red{x = 1}}$

Check:

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

Taking LHS

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

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

$\sf \large= > \dfrac{6 - 3}{6} = \dfrac{3}{6} = \dfrac{1}{2}$

Taking RHS

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

Now,LHS =RHS (verified)