Question

Solve the following simultaneous equations. and find $x+y$ for $\frac{3}{2}x+\frac{2}{3}y=13\frac{1}{3};\frac{2}{3}x+\frac{3}{2}y=8\frac{1}{3}$

Answer 1

Given,

$\frac{3}{2}x+\frac{2}{3}y=13\frac{1}{3}$ .....(1)

$\frac{2}{3}x+\frac{3}{2}y=8\frac{1}{3}$ .......(2)

Rewriting above equations we get,
$9x+4y=80$ .......(3)
$4x+9y=50$ ........(4)

Multiplying (3) by 4 and (4) by 9 we get,
$36x+16y=320$ .......(5)
$36x+81y=450$ ........(6)

Subtracting (5) and (6) we get,
$y=2$
Re substituting $y=2$ in (5) we get,
$x=8$

$\therefore$ $x+y=8+2=10$