I need help solving this question to find x,y, and z

Question

I am having trouble solving this one problem. I am in college algebra and I am having difficulties figuring out how to get started.

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

Answer 1

400=0.12x+0.13y+0.14z

x+y=z

x+y+z=3000

z+z=3000

2z=3000

z=1500

x+y=1500 also

0.14*1500=210

400-210=190

190=0.12x+0.13y

x+y=1500---->x=1500-y

190=0.12(1500-y)+0.13y

190=180-0.12y+0.13y

190-180=0.01y

10=0.01y

1000=y

if z=$1500 and y=$1000, then x=$500  (answer)

check: 0.12*500+0.13*1000+0.14*1500=60+130+210=400