subtract the second equation from the first...
3r-3r-5s-3s=-14+?56
I am not certain on your second equation, an equals is missing.
3r-5s=-14 3r+3s-56
2 answers
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!
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!