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 was $340? (I need to see all the work to catch on to the steps)
2006-11-21 07:35:07 · 2 answers · asked by mr.murphy76 1 in Education & Reference ➔ Primary & Secondary Education
Here ya go.
x + y = 5000 so y = 5000 - x
(0.065)x + (0.070)y = 340
(0.065)x + (0.070)(5000 - x) = 340
(0.065)x + (350) - (.070)x = 340
(.005)x = 10
x = $2000
2006-11-21 08:07:50 · answer #1 · answered by teran_realtor 7 · 0⤊ 0⤋
Very simple.
You invest X at 6.5% and (5000-X) at 7% = $340
I assume the $340 is the interest?
so .065X + .07(5000-X)=340 then distribute
.065X + 350 - .07X = 340 add like terms
- 340 -340
.065X + 10 - .07X = 0 put the X's together
10 - .005X = 0 add the X over
10 = .005X divide both sides by .005
2000 = X which was how much was invested at 6.5% and therefore 5000-2000, or 3000 is what was invested at 7%
Hope that helped
2006-11-21 22:00:14 · answer #2 · answered by kaisermojo 2 · 0⤊ 0⤋