Question

Solve the following pairs of equations:

$4x+\frac{6}{y}=15,6x-\frac{8}{y}=14;y\ne 0$

Answer 1

Put $\frac{1}{y}=Y$
Therefore, the given equations becomes
$4x+6Y=15$ ...(i)
$6x-8Y=14$ ...(ii)
On multiplying (i) by 3 and (ii) by 2, we get
$12x+18Y=45$
$\underset{–}{\underset{-}{1}2x\underset{+}{-}16Y=\underset{-}{2}8}$
$34Y=17$
$\therefore Y=\frac{1}{2}$
On putting $Y=\frac{1}{2}$ in (i), we get
$4x+6\times \frac{1}{2}=15\Rightarrow 4x=12\Rightarrow x=3$
$\therefore x=3,y=\frac{1}{Y}=2$