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 ->  Polynomials-and-rational-expressions -> 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 144619: 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?

Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
.12X+.10Y=130 MULTIPLY THIS EQUATION BY -10
.10X+.12Y=134 MULTIPLY THIS EQUATION BY 12 & THRN ADD THEM.
-1.2X-1Y=-1300
1.2X+1.44Y=1608
-----------------------
.44Y=308
Y=308/.44
Y=$700 INVESTED BY JANE @ 10%.
.12X+.10*700=130
.12X+70=130
.12X=130-70
.12X=60
X=60/.12
X=500 INVESTED BY JANE @ 12%.
PROOF:
.12*500+.10*700=130
60+70=130
130=130
AND:
.10*500+.12*700=134
50+84=134
134=134