Question 401355
First multiply by 10 both equations
3x+2y=40
2x-3y=50/23
then multiply the first one by 2 and the second by -3
6x+4y=80
-6x+9y=-150/23

Then add and keep the first one for next steps
3x+2y=40
13y=80-150/23
so
3x+2y=40
13y=(1840-150)/23
so
3x+2y=40
13y=1690/23
We divide the second equation by 13 (because 169=13*13)
3x+2y=40
13y/13=1690/13*23
so 
3x+2y=40
y=130/23
then we replace the y value in the first equation.
3x+2*130/23=40
y=130/23
so 
3x=40-260/23
y=130/23
so 
3x=(920-260)/23
y=130/23
so 
3x=660/23
y=130/23
We divide the first equation by 3
x=220/23
y=130/23