Question 144613
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%
J's total interest = $130
Also
x = amt R invested at 10%
y = amt R invested at 12%
R's total interest = $134
:
Two equations
.12x + .10y = 130 (J's investments)
.10x + .12y = 134 (R's investments)
:
Multiply the 1st equation by 1.2 and subtract the 2nd equation, find x
.144x + .12y = 156
.10x  + .12y = 134
---------------------subtracting eliminates y
.044x = 22
x = {{{22/.044}}}
x = $500; J has invested at 12% and R has invested at 10%
:
Find y using (.12x + .10y = 130)
.12(500) + .10y = 130
60 = .10y = 130
.10y = 130 - 60
.1y = 70
y = {{{70/.1}}}
y = $700; J has invested at 10% and R has invested at 12%
:
:
Check solution using (.10x + .12y = 134)
.10(500) + .12(700) =
50 + 84 = 134; confirms our solutions