Question 286132
I'm confused.  In the second equation is it 1/(3y) or (1/3)y.  I will assume that it's the latter, so:

7x-y=10-------------------------eq1
x=(1/3)y-2---------------------eq2

If we get rid of the fraction in eq2, then you may not have to worry about it, so:
Multiply each term in eq2 by 3 and we get
3x=y-6 add 6 to each side
3x+6=y----eq2a
now substitute 3x+6 for y in eq 1
7x-(3x+6)=10
7x-3x-6=10  add 6 to each side
7x-3x-6+6=10+6  collect like terms
4x=16
x=4
substitute x=4 into eq2a and we get:
3*4+6=y
y=18

Hope this helps---ptaylor