Question

Based on equations reducible to linear equations
Solve for x and y $\frac{2}{x}+\frac{3}{y}=2;\frac{1}{x}-\frac{1}{2y}=\frac{1}{3}$

• A. $x=0,y=1$
• B. $x=0,y=-7$
• C. $x=2,y=3$
• D. $x=2,y=7$

Answer 1

The correct option is C $x=2,y=3$

The given equations are
$\Rightarrow \frac{2}{x}+\frac{3}{y}=\mathrm{2....}eq\left(1\right)$
$\Rightarrow \frac{1}{x}-\frac{1}{2y}=\frac{1}{3}....eq\left(2\right)$
$Put\frac{1}{x}=uand\frac{1}{y}=vineq\left(1\right)and\left(2\right)$
$\Rightarrow 2u+2v=\mathrm{2....}eq\left(3\right)$
$\Rightarrow u-\frac{1}{2}v=\frac{1}{3}\Rightarrow 6u-3v=\mathrm{2.....}eq\left(4\right)$
$Substracteq\left(3\right)andeq\left(4\right)$
$\Rightarrow 8u=4\Rightarrow u=\frac{1}{2}$
$putthevalueofuineq\left(3\right)$
$\Rightarrow 2\times \frac{1}{2}+3v=2\Rightarrow v=\frac{1}{3}$
$Hence,\frac{1}{x}=u=\frac{1}{2}\Rightarrow x=2$
$\frac{1}{y}=v=\frac{1}{2}\Rightarrow y=3$