Question

If $10y=7x-4$ and $12x+18y=1$; find the values of $4x+6y$ and $8y-x$.

Answer 1

Given equations $10y=7x-4$

$-7x+10y=-4$...(1)

$12x+18y=1$...(2)

Multiplying equation no. (1) by $12$ and (2) by $7$ and then adding,

$-84x+120y=-48$
$\underset{–}{84x+126y=7}$
$\Rightarrow 246y=-41$
$\Rightarrow y=\frac{-1}{6}$

From (1)
$\Rightarrow -7x+10\left(\frac{-1}{6}\right)=-4$

$\Rightarrow -7x=-4+\frac{5}{3}$
$\Rightarrow -7x=-\frac{7}{3}$
$\Rightarrow x=\frac{1}{3}$

Now lets evaluate $(4x+6y)$ and $(8y-x)$,

$4x+6y=4\left(\frac{1}{3}\right)+6\left(\frac{-1}{6}\right)$

$=\frac{4}{3}-1$

$=\frac{4-3}{3}$

$=\frac{1}{3}$

And $8y-x=8\left(\frac{-1}{6}\right)-\frac{1}{3}$

$=\frac{-4}{3}-\frac{1}{3}$

$=\frac{-4-1}{3}$

$=\frac{-5}{3}$

Hence, the value of $4x+6y=\frac{1}{3}$ and $8y-x=\frac{-5}{3}$