Answer:
8,000 at 6%; 12,000 at 8%; 10,000 at 10%
Step-by-step explanation:
Let a, b, c represent the amounts borrowed at 6%, 8%, 10%, respectively. Then the problem statement gives rise to three equations:
... a + b + c = 30,000 . . . . . the total borrowed
... a + b - 2c = 0 . . . . . . . . . . the relationship between the quantities
... .06a +.08b + .10c = 2440 . . . . the total interest charge
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Subtracting the second equation from the first, it becomes
... 3c = 30,000
... c = 10,000 . . . . . divide by 3 . . . [equation 4]
Subtracting 0.06 times the first equation from the third, we have
... .02b +.04c = 640
Multiplying [equation 4] by 0.04 and subtracting from this, we get
... .02b = 240
... b = 12,000 . . . . divide by 0.02 .
Substituting for b and c into the first equation, we find ...
... a + 22000 = 30000
... a = 8000
Then the amounts are ...
- 8000 was borrowed at 6%
- 12000 was borrowed at 8%
- 10000 was borrowed at 10%
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Most graphing calculators are capable of solving matrix problems such as this one. The attachment shows a TI-84 solution.
