Question

Based on equations reducible to linear equations
Solve for x and y: $\frac{2}{x-1}+\frac{y-2}{4}=2;\frac{3}{2(x-1)}+\frac{2(y-2)}{5}=\frac{47}{20}$

• A. $x=3,y=6$
• B. $x=1,y=5$
• C. $x=2,y=-7$
• D. $x=5,y=-9$

Answer 1

Answer: (A)

$x=3,y=6$

The given equations are
$\Rightarrow \frac{2}{x-1}+\frac{y-2}{4}=\mathrm{2......}eq\left(1\right)$
$\Rightarrow \frac{3}{2(x-1)}\frac{2(y-2)}{4}=\frac{47}{20}......eq\left(2\right)$
$Put\frac{1}{x-1}=uandy-2=vineq\left(1\right)andeq\left(2\right)$
$\Rightarrow 2u+\frac{v}{4}=2\Rightarrow 8u+v=\mathrm{8....}eq\left(3\right)$
$\Rightarrow \frac{3u}{2}+\frac{2v}{5}=\frac{47}{20}\Rightarrow 30u+8v=\mathrm{47.....}eq\left(4\right)$
$Multiplyeq\left(3\right)by8$
$\Rightarrow 64u+8v=\mathrm{64.....}eq\left(5\right)$
$Subtracteq\left(4\right)andeq\left(5\right)$
$\Rightarrow (64u+8v=64)-(30u+8v+47)$
$\Rightarrow 34u=17\Rightarrow u=\frac{1}{2}$
$putu=\frac{1}{2}ineq\left(3\right)$
$\Rightarrow 8\times \frac{1}{2}+v=8\Rightarrow v=4$
$Hence,u=\frac{1}{x-1}=\frac{1}{2}\Rightarrow x-1=2\Rightarrow x=3$
$v=y-2=4\Rightarrow y=6$