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Question 164032: I do not understand how to set this problem up.Please help.Also, how can I check my answer.Thanks in advance.
Walt made an extra $9000 last year from a part-time job. He invested part of the money at 9% and the rest at 8%. He made a total of $770 in interest. How much was invested at 8%?
Found 2 solutions by Earlsdon, Alan3354:
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! Consider this!
You want to find how much of Walt's $9,000.00 was invested at 8%.
Call this amount x dollars.
The interest earned on this amount would then be 8% of x, or 0.08x.
The remainder of Walt's $9,000.00, that's ($9,000-x) was invested at 9%.
The interest earned on this amount would be 9% of ($9000.00-x) or 0.09($9000.00-x).
The sum of these two amounts of interest is $770.00, so we can write the equation to find x.
0.08x+0.09($9000.00-x) = $770.00 Simplify this and solve for x.
0.08x+$810.00-0.09x = $770.00 Combine like-terms.
-0.01x+$810.00 = $770.00 Subtract $810.00 from both sides.
-0.01x = - $40.00 Divide both sides by 0.01
x = $4000.00
Walt invested $4000.00 at 8% interest.
Check:
0.08($4000.00)+0.09($5000.00) = $320.00+$450.00 = $770.00
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! I do not understand how to set this problem up.Please help.Also, how can I check my answer.Thanks in advance.
Walt made an extra $9000 last year from a part-time job. He invested part of the money at 9% and the rest at 8%. He made a total of $770 in interest. How much was invested at 8%?
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He invested some at 9%, call that N dollars. The rest at 8%, call the E dollars.
N + E = 9000
N*.09 + E*.08 = 770
From eqn 1, E = 9000-N
Substitute that into eqn 2
N*.09 + (9000-N)*.08 = 770
.09N + 720 - .08N = 770
.01N = 50
N = 5000
E is 5000-N, or $4000
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To check:
5000*.09 + 4000*.08 ?? 770
450 + 320 = 770, so it's correct.
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