Question

SOLVE
3/x+2 - 1/x+1 = 2/x+3​​

Answer 1

GIVEN :-

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

TO FIND :-

• Value of x.

SOLUTION :-

$\\ \implies\sf \: \dfrac{3}{x + 2} - \dfrac{1}{x + 1} = \dfrac{2}{x + 3} \\ \\ \\ \implies\sf \: \dfrac{3(x + 1) - 1(x + 2)}{(x + 2)(x + 1)} = \dfrac{2}{x + 3} \\ \\ \\ \implies \sf \: \dfrac{3x + 3 - x - 2}{x(x + 1) + 2(x + 1)} = \dfrac{2}{x + 3} \\ \\ \\ \implies \sf \: \dfrac{2x + 1}{ {x}^{2} + x + 2x + 2 } = \dfrac{2}{x + 3} \\ \\ \\ \implies \sf \: \dfrac{2x + 1}{ {x}^{2} + 3x + 2} = \dfrac{2}{x + 3}$

$\implies \sf \: (2x + 1)(x + 3) = 2( {x}^{2} + 3x + 2) \\ \\ \\ \implies \sf \: 2x(x + 3) + 1(x + 3) = 2 {x}^{2} + 6x + 4 \\ \\ \\ \implies \sf \: \cancel{2 {x}^{2} } + 6x + x + 3 = \cancel{2 {x}^{2}} + 6x + 4 \\ \\ \\ \implies \sf \: 7x + 3 = 6x + 4 \\ \\ \\ \implies \sf \: 7x - 6x = 4 - 3 \\ \\ \\ \implies \boxed{ \mathfrak{x = 1}}$

VERIFICATION :-

$\\ \sf \: \dfrac{3}{x + 2} - \dfrac{1}{x + 1} = \dfrac{2}{x + 3} \\ \\ \\ \sf \: \dfrac{3}{1 + 2} - \dfrac{1}{1 + 1} = \dfrac{2}{1 + 3} \\ \\ \\ \sf \: \cancel \dfrac{3}{3} - \dfrac{1}{2} = \cancel \dfrac{2}{4} \\ \\ \\ \sf \: 1 - \dfrac{1}{2} = \dfrac{1}{2} \\ \\ \\ \sf \: \dfrac{1}{2} = \dfrac{1}{2} \: \: \: \: \: (verified )$