Enez recently invested $5000, part at 6.5% annual interest and the rest at 7%. How much was invested at each rate if her annual income from both investments was $340?
2006-11-21 09:38:03 · 2 answers · asked by mr.murphy76 1 in Science & Mathematics ➔ Mathematics
Pretend that she invested x dollars at 6.5% and y dollars at 7%. You know that x+y=5000 (because she invested $5000 total), and .065x+.07y=340 (the amount of interest from both). From there, you are solving a system.
With a little help from my graphing calculator, I find that x=2000, y=3000.
Let's check. $2000 dollars at 6.5% would give $130, and $3000 at 7% will give $210. $2000+$3000=$5000, and $130+$210=$340.
2006-11-21 09:50:25 · answer #1 · answered by dennismeng90 6 · 1⤊ 0⤋
Let x be amount invested at 6.5 and y be the amount invested at 7.
x + y = 5000
.065x + .07y = 340
Solve the equations simultaneously:
x = 5000 - y
.065 (5000 - y) + .07y = 340
325 -.065y + .07y = 340
.005y = 15
y = 15 / .005
x + (15/.005) = 5000
x = 5000 - (15/.005)
Sorry, I don't have a calculator with me right now.
2006-11-21 17:43:58 · answer #2 · answered by mattmedfet 3 · 0⤊ 0⤋