SOLUTION: When three pumps A,B, and C are running together, they can pump 3200 gal an hour. When only A and B are running, 1900 gal per hour can be pumped. When only A and C are running, 190

Question 457619: When three pumps A,B, and C are running together, they can pump 3200 gal an hour. When only A and B are running, 1900 gal per hour can be pumped. When only A and C are running, 1900 gal per hour can be pumped. What is the pumping capaciity of each pump? I set it up like this.
x+y+z=3200
x+y=1900
x+z=1900
not sure where to continue im stuck
Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
You've gotten off to a good start,
The three pumps are move water in gallons per hour.
The three pumping rates will be shown as: A, B & C.
(You might as well use the given letters to help keep yourself consistent with the problem.)
.
Running together they pump 3200 gallons per hour.
A + B + C = 3200
.
We also are told:
A + B = 1900
which means
A = 1900 - B
and
B = 1900 - A
.
We also are told:
A + C = 1900
So,
A = 1900 - C
C = 1900 - A
.
Looking back at the first equation, we can substitute for B and C.
A = A
B = 1900-A
C = 1900-A
A +1900-A + 1900-A = 3200
Collecting terms:
A-2A + 3800 = 3200
Simplify and subtract 3800 from both sides
-A = -600
Multiply both sides by -1 (or divide both sides by -1, it doesn't matter which):
A = 600
.
That means
B = 1900-600
B = 1300
and
C = 1900-600
C = 1300
.
A + B + C = 600 + 1300 + 1300
A + B + C = 3200
.
Done.
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