SOLUTION: Hello, my algebra word problem is this "Use two variables and two equations to solve the problem. Part of $9000 was invested at 10% interest and the rest at 12%. If the annual inc

Question 554767: Hello, my algebra word problem is this "Use two variables and two equations to solve the problem. Part of $9000 was invested at 10% interest and the rest at 12%. If the annual income from these investments was $1020 how much was invested at rate?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Use two variables and two equations to solve the problem.
Part of $9000 was invested at 10% interest and the rest at 12%.
If the annual income from these investments was $1020 how much was invested at rate?
:
Let x = amt invested at 10%
Let y = amt invested at 12%
:
The amt equation
x + y = 9000
Rearrange for substitution
x = (9000-y)
:
the interest equation
.10x + .12y = 1020
substitute (9000-y) for x, solve for y
.10(9000-y) + .12y = 1020
900 - .10y + .12y = 1020
.02y = 1020 - 900
.02y = 120
y =
y = $6000 invested at 12%
:
You should be able to find x, check your solutions in both equations
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