Question

Solve the following equations:
$4x-3y=1$,
$12xy+13y^2=25$.

Answer 1

$4x-3y=14x=3y+\mathrm{1.....}\left(i\right)$

$12xy+13y^2=25$

$\left(4x\right)3y+13y^2=25$

Substituting $\left(i\right)$

$(3y+1)3y+13y^2=25$

$9y^2+3y+13y^2=25$

$22y^2+3y-25=0$

$22y^2-22y+25y-25=0$

$22y(y-1)+25(y-1)=0$
$(22y+25)(y-1)=0$

$y=-\frac{25}{22},1$

Substituting $y$ in $\left(i\right)$
$4x=3(-\frac{25}{22})+1$

$\Rightarrow x=-\frac{53}{88}$

$4x=3\left(1\right)+1$

$\Rightarrow x=1$

$\therefore x=-\frac{53}{88},1$