Question

Solve:

$\frac{x+1}{2}+\frac{y-1}{3}=8$

$\frac{x-1}{3}+\frac{y+1}{2}=9$

Answer 1

Given:

$\frac{x+1}{2}+\frac{y-1}{3}=8$

$\frac{3x+3}{2\times 3}+\frac{2y-2}{3\times 2}=8$

$\frac{3x+3}{6}+\frac{2y-2}{6}=8$

$\frac{3x+3+2y-2}{6}=8$

$3x+2y+1=48$

$3x+2y=47------\left(1\right)$

$\frac{x-1}{3}+\frac{y+1}{2}=9$

$\frac{2x-2}{2\times 3}+\frac{3y+3}{3\times 2}=9$

$\frac{2x-2}{6}+\frac{3y+3}{6}=9$

$\frac{2x-2+3y+3}{6}=9$

$2x+3y+1=54$

$2x+3y=53$ -------- (2)

By solving (1) and (2)

$3x+2y=47-------\times 3$

$2x+3y=53-------\times 2$

$9x+6y=141$

$(-)$

$4x+6y=106$

$5x=35$

$x=7$

Substituting the value of $x$ in $\left(1\right)$

$3\left(7\right)+2y=47$

$21+2y=47$

$2y=47-21$

$2y=26$

$y=13$

So, $x=7$ and $y=13$