Question

Solve for x:
$\sf \frac{x + 1}{x - 1} + \frac{x - 2}{x + 2} = 3\; where \; x \neq 1,-2$

Answer 1

$\frac{(x + 1)(x + 2) + (x - 2)(x - 1)}{(x - 1)(x + 2)} =3 \\ \frac{ {x}^{2} + x + 2x + 2 - 2x + {x}^{2} - x + 2 }{ {x}^{2} - x + 2x - 2} = 3 \\ \frac{2 {x}^{2} }{ {x}^{2} - x + 2x - 2} = 3 \\ 2 {x}^{2} = 3( {x}^{2} - x + 2x - 2) \\ 2 {x}^{2} = 3 {x}^{2} - 3x + 6x - 6 \\ 0 = {x}^{2} + 3x - 6$

you can solve the equation and you get the answer !! sorry I couldn't do till the end , I have my exam tomorrow...

(mark as brainliest if you are satisfied !)

Answer 2

Given Equation

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

Step by step explanation

$\dashrightarrow \: \: \sf \dfrac{ \bigg(\dfrac{x + 1}{x - 1} (x - 1)(x + 2) \bigg)+ \normalsize \bigg( \dfrac{x - 2}{x + 2} (x - 1)(x + 2) \bigg)} {(x - 1)(x + 2) } = 3$

$\dashrightarrow \: \: \sf \dfrac{ \bigg(\dfrac{x + 1}{\cancel{x - 1}} \: \: \cancel{(x - 1)} \: (x + 2) \bigg)+ \normalsize \bigg( \dfrac{x - 2}{\cancel{x + 2}} \: (x - 1) \: \cancel{(x + 2) } \bigg)} {(x - 1)(x + 2) } = 3$

$\dashrightarrow \: \: \sf \dfrac{ \bigg(\dfrac{x + 1}{{1}} \: \: (x + 2) \bigg)+ \normalsize \bigg( \dfrac{x - 2}{1} \: (x - 1) \: \bigg)} {(x - 1)(x + 2) } = 3$

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

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

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

$\dashrightarrow \: \: \sf \dfrac{{x}^{2} \cancel{+ 3x }+ 2x+{x}^{2} \: \: \cancel{ - 3x} + 3}{ {x}^{2} + 2x - x - 2} = 3$

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

$\dashrightarrow \: \: \sf {2x}^{2} + 2x+ 3= 3 ({x}^{2} + x- 2)$

$\dashrightarrow \: \: \sf {2x}^{2} + 2x+ 3= 3 {x}^{2} +3 x- 6$

$\dashrightarrow \: \: \sf0= 3 {x}^{2} - {2x}^{2} +3 x - 2x- 6 + 3$

$\dashrightarrow \: \: \sf0= 2{x}^{2} +x- 3$

$\dashrightarrow \: \: \sf2{x}^{2} +x- 3 = 0$

$\dashrightarrow \: \: \sf2{x}^{2} + 3x - 2x- 3 = 0$

$\dashrightarrow \: \: \sf x(2x+ 3 )- (2x + 3) = 0$

$\dashrightarrow \: \: \sf (x - 1)(2x+ 3 ) = 0$

Value of x :-

• x - 1 = 0
• x = 1

or

• 2x + 3 = 0
• 2x = -3
• x = -3/2
• x = 1.5

as it is told value of x is not equals to 1

$\therefore \: \: \underline{ \sf{value \: of \: x = \bf1.5}}$