Question 152103
1.{{{(3/4)x-(2/5)y=2}}}
2.{{{(1/2)x-(3/5)y=-2}}}
I'm not a big fan of fractions so I usually multiply both equations to get rid of denominators.
Multiply eq. 1 by 20 and eq. 2 by 10.
1.{{{20(3/4)x-20(2/5)y=20(2)}}}
1.{{{15x-8y=40}}}
.
.
.
2.{{{10(1/2)x-10(3/5)y=10(-2)}}}
2.{{{5x-6y=-20}}}
Now use the elimination method to solve. 
If I multiply eq. 2 by (-3) and add to eq. 1, I can get rid of x,
{{{(15x-8y)+(-3)(5x-6y)=40+(-3)(-20)}}}
{{{(15x-8y)+(-15x+18y)=40+60}}}
{{{10y=100}}}
{{{highlight(y=10)}}}
Using eq. 1,
1.{{{15x-8y=40}}}
{{{15x-8(10)=40}}}
{{{15x-80=40}}}
{{{15x=120}}}
{{{highlight(x=8)}}}
.
.
.
Check the answer with the original equations.
1.{{{(3/4)x-(2/5)y=2}}}
{{{(3/4)(8)-(2/5)(10)=2}}}
{{{6-4=2}}}
{{{2=2}}}
True.
I'll leave it to you to verify that eq. 2 works also.