Question

please solve anyone very hard question.​

Answer 1

Question :- Solve for x :-

$\rm \: \dfrac{3}{x + 1} + \dfrac{5}{x + 3} = \dfrac{8}{x + 2}$

$\large\underline{\sf{Solution-}}$

Given equation is

$\rm \: \dfrac{3}{x + 1} + \dfrac{5}{x + 3} = \dfrac{8}{x + 2}$

On taking LCM on left side of the equation,

$\rm \: \dfrac{3(x + 3) + 5(x + 1)}{(x + 1)(x + 3)} = \dfrac{8}{x + 2}$

$\rm \: \dfrac{3x + 9 + 5x + 5}{ {x}^{2} + 3x + x + 3} = \dfrac{8}{x + 2}$

$\rm \: \dfrac{8x + 14}{ {x}^{2} + 4x + 3} = \dfrac{8}{x + 2}$

On cross multiplication, we get

$\rm \: (8x + 14)(x + 2) = 8( {x}^{2} + 4x + 3)$

$\rm \: 8x(x + 2) + 14(x + 2) = 8 {x}^{2} + 32x + 24$

$\rm \: 8 {x}^{2} + 16x + 14x + 28 = 8 {x}^{2} + 32x + 24$

$\rm \: \cancel{8 {x}^{2}} + 30x + 28 = \cancel{ 8 {x}^{2}} + 32x + 24$

On transposition, we get

$\rm \: 30x - 32x = 24 - 28$

$\rm \: - 2x = - 4$

$\rm \: x = \dfrac{ - 4}{ - 2}$

$\rm\implies \:x = 2$

Verification :-

Consider the equation

$\rm \: \dfrac{3}{x + 1} + \dfrac{5}{x + 3} = \dfrac{8}{x + 2}$

On substituting x = 2, we get

$\rm \: \dfrac{3}{2 + 1} + \dfrac{5}{2 + 3} = \dfrac{8}{2 + 2}$

$\rm \: \dfrac{3}{3} + \dfrac{5}{5} = \dfrac{8}{4}$

$\rm \: 1 + 1 = 2$

$\rm \: 2 = 2$

Hence, Verified

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

What to do:

We have to solve the given equation

Solution:

$\mathsf{\frac{3}{x+1} +\frac{5}{x+3} =\frac{8}{x+2} }\\\\\mathsf{Or,\frac{3(x+3)+5(x+1)}{(x+1)(x+3)}=\frac{8}{x+2} }\\\\\mathsf{Or,\frac{3x+9+5x+5}{(x+1)(x+3)} =\frac{8}{x+2} }\\\\\mathsf{Or,\frac{8x+14}{x^2+4x+3} =8(x^2+4x+3)}$

$\mathsf{Or,(x+2)(8x+14)=8(x^2+4x+3)}\\\\\mathsf{Or,8x(x+2)+14(x+2)=8x^2+32x+24}\\\\\mathsf{Or,8^2+16x+14x+28=8^2+32x+24}\\\\\mathsf{Or,8x^2+30x+28=8^2+32x+24}\\\\\mathsf{Or,30x-32x=24-28}\\\\\mathsf{Or,-2x=-4}\\\\\mathsf{Or,x=\frac{4}{2} }\\\\\mathsf{Or,x=2}$