Question

Solve the following simultaneous equations :

$\frac{27}{x-2}+\frac{31}{y+3}=85;\frac{31}{x-2}+\frac{27}{y+3}=89$

• A. $x=\frac{3}{4},y=4$
• B. $x=\frac{5}{2},y=-2$
• C. $x=\frac{4}{3},y=3$
• D. $x=\frac{7}{2},y=6$

Answer 1

Answer: (B)

$x=\frac{5}{2},y=-2$

On putting $\frac{1}{x-2}=X$ and $\frac{1}{y+3}=Y$ in given equations, we get

$27X+31Y=85$ ...(i)
$31X+27Y=89$ ...(ii)
On multiplying (i) by 31 and (ii) by 27, we get
$837X+961Y=2635$
$\underset{–}{\underset{-}{8}37X\underset{-}{+}729Y=\underset{-}{2}403}$
$232Y=232$

$\therefore Y=1$
On putting $Y=1$ in (i), we get
$27X+31=85\Rightarrow 27X=54\Rightarrow X=2$

$\therefore x-2=\frac{1}{X}\Rightarrow x=2+\frac{1}{2}=\frac{5}{2}$

and $y+3=\frac{1}{Y}=1-3=-2$

Therefore, $x=\frac{5}{2},y=-2$