Question

Solve for x and y:

$\frac{1}{2(3x+y)}+\frac{5}{3(3x-y)}=\frac{-3}{2}\frac{5}{4(x+2y)}-\frac{3}{5(3x-2y)}=\frac{61}{60}.$

Answer 1

$\frac{1}{2(x+2y)}+\frac{5}{3(3x-2y)}=\frac{-3}{2}\frac{5}{4(x+2y)}-\frac{3}{5(3x-2y)}=\frac{61}{60}$

$Put\frac{1}{(x+2y)}=u\frac{1}{(3x-2y)}=v$

$\frac{u}{2}+\frac{5v}{3}=\frac{-3}{2}\frac{5u}{4}-\frac{3v}{5}=\frac{61}{60}\frac{3u+10v}{6}=\frac{-3}{2}\frac{25u-12v}{20}=\frac{61}{60}\frac{3u+10v}{3}=\frac{-3}{1}\frac{25u-12v}{1}=\frac{61}{3}$

$3u+10v=-9---\left(1\right)75u-36v=61---\left(2\right)$

$\left(1\right)\times 75225u+750v=-675---\left(3\right)\left(2\right)\times 3225u-108v=183---\left(4\right)$

$\left(3\right)-\left(4\right)225u+750v-(225u-108v)=-675-183225u+750v-225u+108v=-858858v=-858v=-1$

$Putv=-1in\left(1\right)3u+10(-1)=-93u-10=-93u=-9+10=1u=\frac{1}{3}$

$Now\frac{1}{(x+2y)}=\frac{1}{3}x+2y=3---\left(5\right)\frac{1}{(3x-2y)}=-1-3x+2y=13x-2y=-1---\left(6\right)$

$\left(5\right)+\left(6\right)x+2y+3x-2y=3+-14x=2x=\frac{1}{2}$

$x+2y=32y=3-x2y=3-\frac{1}{2}2y=\frac{5}{2}y=\frac{5}{4}$

$x=\frac{1}{2}y=\frac{5}{4}$