SOLUTION: I need help with an investment problem: Tom made an extra $5000 last year from a part-time job. He invested part of the money at 6% and the rest at 8%. He made a total of $380 in i

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Question 487208: I need help with an investment problem: Tom made an extra $5000 last year from a part-time job. He invested part of the money at 6% and the rest at 8%. He made a total of $380 in interest. How much was invested at 8%?
Found 3 solutions by jorel1380, josmiceli, ej_03:
Answer by jorel1380(3719) About Me  (Show Source):
You can put this solution on YOUR website!
.06(n)+.08(5000-n)=380
6n+8(5000-n)=38000
6n+40000-8n=38000
2n=2000
n=1000
5000-n=4000
Tom invested $1000 at 6%, and $4000 at 8%. ☺☺☺☺

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +a+ = amount invested @ 6%
Let +b+ = amount invested @ 8%
given:
(1) +a+%2B+b+=+5000+
(2) +.06a+%2B+.08b+=+380+
----------------------
Multiply both sides of (2) by +100+
Then multiply both sides of (1) by +6+
Then subtract (1) from (2)
(2) +6a+%2B+8b+=+38000+
(1) +-6a+-+6b+=+-30000+
+2b+=+8000+
+b+=+4000+
and, since
(1) +a+%2B+b+=+5000+
(1) +a+%2B+4000+=+5000+
(1) +a+=+1000+
$4,000 was invested @ 8%
check:
(2) +.06a+%2B+.08b+=+380+
(2) +.06%2A1000+%2B+.08%2A4000+=+380+
(2) +60+%2B+320+=+380+
(2) +380+=+380+
OK

Answer by ej_03(8) About Me  (Show Source):
You can put this solution on YOUR website!
let x=6%
y=8%
eq1 x+y=5000
y=5000-x
eq2 .06x+.08y=380
perform the indicated operations in the equation:
.06x+.08(5000-x)=380
.06x+400-.08x=380
.06x-.08x=380-400
.02x/.02x=20/.02
x=1000
y=5000-1000=4000
therefore tom invested $4000 at 8%... thank you..