SOLUTION: Jane invested some amount at the rate of 12% simple interest and some other amount at the rate of 10% simple interest. She received yearly interest of $130. Randy also invested in

Algebra ->  Linear-equations -> SOLUTION: Jane invested some amount at the rate of 12% simple interest and some other amount at the rate of 10% simple interest. She received yearly interest of $130. Randy also invested in       Log On


   



Question 132062: Jane invested some amount at the rate of 12% simple interest and some other amount at the rate of 10% simple interest. She received yearly interest of $130. Randy also invested in the same scheme, but he interchanged the amounts invested, and received $4 more as interest. How much amount did each of them invest at different rates?
Please help me solve this problem.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Jane invested
$a at 12%
$b at 10%
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Randy invested
$b at 12%
$a at 10%
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For Jane:
(1) .12a+%2B+.1b+=+130
For Randy:
(2) .12b+%2B+.1a+=+134
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Rewrite (2) as:
.12b+%2B+.1a+=+130+%2B+4
(3) .12b+%2B+.1a+-+4+=+130
Set the left side of (3) equal to the left
side of (1)
.12b+%2B+.1a+-+4+=+.12a+%2B+.1b
.02b+-+.02a+=+4
divide both sides by .02
b+-+a+=+200
b+=+200+%2B+a
Substitute this in (1)
.12a+%2B+.1%28200+%2B+a%29+=+130
.12a+%2B+20+%2B+.1a+=+130
.22a+=+110
a+=+500
b+=+200+%2B+a
b+=+700
Jane invested
$500 at 12%
$700 at 10%
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Randy invested
$700 at 12%
$500 at 10%
check answers:
(1) .12a+%2B+.1b+=+130
.12%2A500+%2B+.1%2A700+=+130
60+%2B+70+=+130
130+=+130
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(2) .12b+%2B+.1a+=+134
.12%2A700+%2B+.1%2A500+=+134
84+%2B+50+=+134
134+=+134
OK