SOLUTION: An engineer invests some money at 5% and $3000 more than twice as much at 8%. The total annual income from the investments is $1710. Find the amount invested at 5%. Use elimination

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Question 1136424: An engineer invests some money at 5% and $3000 more than twice as much at 8%. The total annual income from the investments is $1710. Find the amount invested at 5%. Use elimination method.
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Solving by elimination is absurd. The problem is perfectly suited for solving using a single variable:

.05%28x%29%2B.08%282x%2B3000%29+=+1710

However, given the instruction to use elimination....

(1) .05%28x%29%2B.08%28y%29+=+1710
(2) y+=+2x%2B3000 --> -2x%2By+=+3000

Multiply the first equation by 100 (to get rid of decimals) and multiply the second by -8, then add; that eliminates y.

5x%2B8y+=+171000
16x-8y+=+-24000
21x+=+147000
x+=+7000

ANSWER: $7000 was invested at 5%

CHECK: .05(7000) + .08(17000) = 350+1360 = 1710