Question

Solve the following equations and verify your answer :

$(\frac{x+1}{x-4}{)}^2=\frac{x+8}{x-2}$

Answer 1

We have $(\frac{x+1}{x-4}{)}^2=\frac{x+8}{x-2}$

$\Rightarrow \frac{(x+1{)}^2}{(x-4{)}^2}=\frac{(x+8)}{(x-2)}$

By cross multiplication, we have
$(x+1{)}^2(x-2)=(x-4{)}^2(x+8)$

$\Rightarrow (x^2+1+2x)(x-2)=(x^2+16-8x)(x+8)$

$\left[\text{Using the identity}\right(a+b{)}^2=a^2+b^2+2ab]$

$\Rightarrow x^3-2x^2+2x^2-4x+x-2=x^3+8x^2-8x^2-64x+16x+128$

$\Rightarrow x^3-3x-2=x^3-48x+128$

$\Rightarrow x^3-3x-x^3+48x=128+2$

By transposition, we have
$\Rightarrow 45x=130$

$\Rightarrow x=\frac{130}{45}=\frac{26}{9}$

$\therefore x=\frac{26}{9}$

Verification (putting the value of $x$ in L.H.S and R.H.S ) :
$L.H.S.=(\frac{x+1}{x-4}{)}^2=\frac{\{\frac{26}{9}+1{\}}^2}{\{\frac{26}{9}-4{\}}^2}$

$=\frac{\{\frac{26+9}{9}{\}}^2}{\{\frac{26-36}{9}{\}}^2}=\frac{\{\frac{35}{9}{\}}^2}{\{\frac{-10}{9}{\}}^2}$

$=(\frac{35}{9}\times \frac{9}{-10}{)}^2=(\frac{-7}{2}{)}^2=\frac{49}{4}$

$R.H.S.=\frac{x+8}{x-2}=\frac{\frac{26}{9}+8}{\frac{26}{9}-2}=\frac{\frac{26+72}{9}}{\frac{26-18}{9}}=\frac{\frac{98}{9}}{\frac{8}{9}}$

$=\frac{98}{9}\times \frac{9}{8}=\frac{49}{4}$

$\therefore L.H.S=R.H.S.$