Question

Based on equations reducible to linear equations
Solve for x and y: $\frac{16}{x+3}+\frac{3}{y-2}=5;\frac{8}{x+3}-\frac{1}{y-2}=0$

• A. $x=5,y=3$
• B. $x=7,y=2$
• C. $x=1,y=5$
• D. None of these

Answer 1

The correct option is A $x=5,y=3$

The given equations are
$\Rightarrow \frac{16}{x+3}+\frac{3}{y-2}=\mathrm{5....}eq\left(1\right)$
$\Rightarrow \frac{8}{x+3}-\frac{1}{y-2}=\mathrm{0....}eq\left(2\right)$
$Put\frac{1}{x+3}=uand\frac{1}{y-2}=vineq\left(1\right)and\left(2\right)$
$\Rightarrow 16u+3v=\mathrm{5....}eq\left(3\right)$
$\Rightarrow 8u-v=0....eq\left(4\right)$
$multiplyeq\left(4\right)by3andsubtracteq\left(3\right)andeq\left(4\right)$
$\Rightarrow (16u+3v=5)-(24u-v=0)$
$\Rightarrow 40u=5\Rightarrow u=\frac{1}{8}$
$Putu=\frac{1}{8}ineq\left(3\right)$
$\Rightarrow 16\left(\frac{1}{8}\right)+3v=5\Rightarrow 3v=3\Rightarrow v=1$
$Hence,\frac{1}{x+3}=u=\frac{1}{8}\Rightarrow x+3=8\Rightarrow x=5$
$\frac{1}{y-2}=v=1\Rightarrow y-2=1\Rightarrow y=3$