Question 554767
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 = {{{120/.02}}}
y = $6000 invested at 12%
:
You should be able to find x, check your solutions in both equations