Question

Evaluate:
(i) $\frac{4}{x-1}+\frac{5}{y-1}=2,\frac{8}{x-1}+\frac{15}{y-1}=3,x\ne 1,y\ne 1$
(ii)$\frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}{4},\frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}=\frac{-1}{8},3x+y\ne 0,3x-y\ne 0$

Answer 1

$\left(i\right)\frac{4}{x-1}+\frac{5}{y-1}=2$ .........$\left(1\right)$

$\frac{8}{x-1}+\frac{15}{y-1}=3$ ........$\left(2\right)$

Let $u=\frac{1}{x-1},v=\frac{1}{y-1}$

Equations $\left(1\right)$ and $\left(2\right)$ becomes

$\Rightarrow 4u+5v=2$ ........$\left(3\right)$

$8u+15v=3$ .........$\left(4\right)$

$\Rightarrow 8\times \left(3\right)-4\times \left(4\right)$ we get

$\Rightarrow 32u+40v-32u-60v=16-12$

$\Rightarrow -20v=4$

$\Rightarrow v=-\frac{1}{5}$

Put $v=-\frac{1}{5}$ in $\left(3\right)$ we get

$\Rightarrow 4u=2-5v=2-5\times -\frac{1}{5}=2+1=3$

$\Rightarrow 4u=3$ or $u=\frac{3}{4}$

$\therefore u=\frac{3}{4},v=-\frac{1}{5}$

$\Rightarrow \frac{1}{x-1}=\frac{3}{4},\frac{1}{y-1}=-\frac{1}{5}$ since $u=\frac{1}{x-1},v=\frac{1}{y-1}$

$\Rightarrow 3x-3=4,y-1=-5$

$\Rightarrow 3x=4+3=7,y=-5+1=-4$

$\therefore x=\frac{7}{3},y=-4$

$\left(ii\right)\frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}{4}$ .........$\left(1\right)$

$\frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}=-\frac{1}{8}$ ........$\left(2\right)$

Let $u=\frac{1}{3x+y},v=\frac{1}{3x-y}$

Equations $\left(1\right)$ and $\left(2\right)$ becomes

$\left(1\right)\Rightarrow \frac{u}{2}-\frac{v}{2}=-\frac{1}{8}$

$\Rightarrow u-v=-\frac{1}{4}$ ........$\left(3\right)$

$\left(2\right)\Rightarrow u+v=\frac{3}{4}$ ........$\left(4\right)$

$\Rightarrow u-v+u+v=-\frac{1}{4}+\frac{3}{4}$

$\Rightarrow 2u=\frac{2}{4}$

$\Rightarrow u=\frac{1}{2}$

Put $u=\frac{1}{2}$ in $\left(4\right)$ we get

$\Rightarrow v=\frac{3}{4}-u=\frac{3}{4}-\frac{1}{2}=\frac{3-2}{4}=\frac{1}{4}$

$\therefore u=\frac{1}{2},v=\frac{1}{4}$

$\Rightarrow \frac{1}{3x+y}=\frac{1}{2},\frac{1}{3x-y}=\frac{1}{4}$ since $u=\frac{1}{3x+y},v=\frac{1}{3x-y}$

$\Rightarrow 3x+y=2,3x-y=4$

$\Rightarrow 3x+y+3x-y=2+4$

$\Rightarrow 6x=6$ or $x=1$

Substitute $x=1$ in $3x+y=2$ we get

$y=2-3x=2-3=-1$

$\therefore x=1,y=-1$