Posted by beth on Wednesday, January 5, 2011 at 2:00pm.
3r5s=14 3r+3s56

algebra  bobpursley, Wednesday, January 5, 2011 at 2:02pm
subtract the second equation from the first...
3r3r5s3s=14+?56
I am not certain on your second equation, an equals is missing.

algebra  Kay, Wednesday, January 5, 2011 at 3:57pm
I tried and verified the equation with the following assumption to make it work.
3r  5s = 14
3r +3s = 56
subtract the second equation from first equation:
You will get 8s = 70
s = 70/8
substitute s in any of the original problem equations to get the value of r.
3r = 14 + 5s
3r = 14 + 5 (70/8)
= (112 + 350)/8
= 238/8
3r = 119/4
r = 119/12
To verify the values  apply the values to one of the equations to hold true.
3r 5s should be equal to 14
= 3 (119/12)  5 (70/8)
= 119/4  350/8
= (238  350) / 8
= 112/8
= 14
So we are good!
Additional note regarding posting of your question:
So, your problem statement should be
3r + 3s = 56
Hope this clarifies!
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