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

Question 79794This question is from textbook Elementary and intermediate algebra
: 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
This question is from textbook Elementary and intermediate algebra

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
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.
:
Let x = amt J invested at 12%
Let y = amt J invested at 10%
:
Randy interchanged the amts so:
x = amt R invested at 10%
y = amt R invested at 12%
:
.12x + .10y = 130; J's earnings
.10x + .12y = 134; R's earnings
:
Multiply 1st equation by 120 and 2nd equation by 100:
14.4x + 12y = 15600
10x + 12y = 13400
--------------------- subtracting eliminates y
4.4x + 0 = 2200
x = 2200/4.4
x = $500 invested at 12% by J, $500 invested at 10% by R
:
Find y:
.12(500) + .10y = 130
60 + .1y = 130
.1y = 130 - 60
y = 70/.1
y = $700 invested at 10% by J, $700 invested at 12% by R
:
:
Check our solutions using; .10x + .12y = 134:
.1(500) + .12(700) =
50 + 84 = 134
:
Make sense to you? Hope so.

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