Question 1118484: A lake manager wants to stock a small lake with 11,550 fish, some bass and some walleyes. He wants to use 1830 more bass. How many of each kind of fish does he need?
Found 3 solutions by Alan3354, addingup, MathTherapy:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! A lake manager wants to stock a small lake with 11,550 fish, some bass and some walleyes. He wants to use 1830 more bass. How many of each kind of fish does he need?
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B + W = 11550
B - W = 1830
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Answer by addingup(3677)  (Show Source):
You can put this solution on YOUR website! b + w = 11,500 (1)
b = w + 1,830
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In (1), substitute for b:
w + (w + 1830) = 11,500
2w = 9,670
w = 4,835
He needs 4,835 walleyes and 4,835+1,830 = 6,670 bass
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website!
A lake manager wants to stock a small lake with 11,550 fish, some bass and some walleyes. He wants to use 1830 more bass. How many of each kind of fish does he need?
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