Question

Solve the following pair of equations:

$x+y=3.3\frac{0.6}{3x-2y}=-1;3x-2y\ne 0$

Answer 1

Given equations are

$x+y=3.3$

$\frac{0.6}{3x-2y}=-1,3x-2y\ne 0$

Rewriting the given equations, we get
$x+y=3.3$

multiplying both sides by $10$

$\Rightarrow 10x+10y=\mathrm{33...................}\left(1\right)$

and

$\frac{6}{10(3x-2y)}=-1$

$\Rightarrow 30x-20y=-\mathrm{6..................}\left(2\right)$

On multiplying $\left(1\right)$ by $3$ and $\left(2\right)$ by $1$, we get
$30x+30y=99$
$\underset{–}{\underset{-}{3}0x\underset{+}{-}20y=\underset{+}{-}6}$
$50y=105$
$\therefore y=2.1$

On putting $y=2.1$ in $\left(1\right)$, we get
$10x+10\times 2.1=33$

$\Rightarrow 10x+21=33$

$\Rightarrow 10x=33-21$

$\Rightarrow 10x=12$

$\Rightarrow x=1.2$

Hence, the solution is $x=1.2,y=2.1$