Algebra
posted by A.W. .
How do I solve for elimination method on this problem: 0.3x  0.2y=4 and 0.3x + 0.3y= 5/19. I know that multiply by 10 on both equations and get 3x2y=40 and 3x+3y=50/19. and then I have to multiply 3x3y=40 by 3 and 3x+3y= 50/19 by 3 to get 9x9y=120 and 9x9y=150/19. But from there I get lost. Can someone please help me?

I just did your earlier problem. Please study that. If you can do one you can do them all.

first if you have .3 x in both, why multiply by 10?
.3 x  .2 y = 4
.3 x + .3 y = 5/19
subtract second from first
0 x  .5 y = 71/19
y =  142/19
so
.3 x  .2 (142/19) = 40
3 x = 2(142/19) + 760/19
3 x = 476/19
x = 476/57