Posted by Nick on Thursday, March 6, 2008 at 2:53pm.
So I guess I have to:
Solve the system using any algebraic method.
I'm not sure how so here is the problem I'm stuck on.
9x5y=30
x+2y=3

Math(Solve the system)  Rhonda, Thursday, March 6, 2008 at 3:03pm
In the second equation, subtract the 2y to get x=2y3. Then plug you new value for x into the first equation:
9(2y3) 5y = 30. Find the answer for y and then plug it into either equation to get the answer for x. 
Math(Solve the system)  Nick, Thursday, March 6, 2008 at 3:12pm
Ummm....huh?? I am more of a visual learner so maybe if it was worked out completly it would help a little more.

Math(Solve the system)  DLK, Thursday, March 6, 2008 at 3:27pm
make the second equation, x+2y = 3, into x = 32y and substitute 32y into the first equation, 9x5y. then, you will get 9(32y)5y = 30. that should simplify to 27  18y  5y = 30. That should simplify to 27  23y = 30. that simplifies to 23y = 3. Therefore, y = 3/23.
In the second equation, x+2y = 3, substitute 3/23 for y. (x + 2 (3/23))= 3. that simplifies to x + 6/23 = 3. therefore, x = 3 6/23.