Question

Find the value of x in
$3 \div x - 1 + 1 \div x + 1 = 1 \div 2$
Where x Not equal to 0,1,-1​

Answer 1

$\frac{3}{x-1}+\frac{1}{x+1}=\frac{1}{2}$

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

$\frac{3x+3+x-1}{x^{2}-1}=\frac{1}{2}$

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

$\frac{4x+2}{x^{2}-1}-\frac{1}{2}$

$\frac{8x+4}{2(x^{2}-1)}-\frac{x^{2}-1}{2(x^{2}-1)}=0$

$\frac{-x^{2}+8x+5}{2(x^{2}-1)}=0$

we get

$-x^{2}+8x+5=0$

$x^{2}-8x-5=0$

$x^{2}-8x=5$

$x^{2}-8x+16=5+16$

$(x-4)^{2}=21$

$(x-4)=±\sqrt{21}$

$x_{1}=4+\sqrt{21}$

$x_{2}=4-\sqrt{21}$