Question

Solve the given problem and find value of x :-

$\frac{1}{x}+\frac{1}{x-2}=3;x\ne 0,2$

Answer 1

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

$Multiplying$ $both$ $sides$ $by$ $x(x-2),$

$x-2$ $+$ $x$ $=$ $3x(x-2)$

$2x-2$ $=$ $3x^2$ $-$ $6x$

$3x^2-8x+2$ $=$ $0$

$Multiplying$ $the$ $equation$ $throughout$ $by$ $3,$

$9x^2-24x+6$ $=$ $0$

$(3x{)}^2-2\times (3x)\times 4+6$ $=$ $0$

$(3x{)}^2-2\times (3x)\times 4+4^2-4^2+6$ $=$ $0$

$(3x-4{)}^2-16+6$ $=$ $0$

$(3x-4{)}^2-10$ $=$ $0$

$(3x-4{)}^2$ $=$ $10$

$3x-4$ $=$ $\sqrt{10}$ $or$ $3x-4$ $=$ $-\sqrt{10}$

$Therefore,$

$x=\frac{\sqrt{10}+4}{3}$

$OR$

$x=\frac{-\sqrt{10}+4}{3}$