Word Problem:
- Part of $3000 is invested at 12%, another part at 13%, and the remainder at 14%. The total yearly income from the three investments is $400. The sum of the amounts invested at 12% and 13% equals the amount invested at 14%. Determine how much is invested at each rate.

Equations:
1) x y z=3000
2) .12x .13y .14z=400
3) x y=z

The equations need to be solved by various methods (elimination, substitution).

Since x+y =z and x+y+z=3000 , we get
x+y+z=z+z=2z=3000 \Rightarrow z=1500
And since x+y= z= 1500 \Rightarrow x=1500-y
So
.12 \cdot x +.13 \cdot y + .14 \cdot z = 400
\Rightarrow .12 \cdot (1500 - y) +.13 \cdot y + .14 \cdot 1500 = 400
\Rightarrow 180 -.12 \cdot y +.13 \cdot y + 210 = 400
\Rightarrow .01 \cdot y + 390 = 400
\Rightarrow .01 \cdot y =10
\Rightarrow y =1000
And since x=1500-y = 1500-1000 = 500
So we have x=500 , y=1000 , and z=1500 .

I hope this helps.

is this equation answered using elimiation or substitution?

thanks