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.

Let $x be invested at 12%, $y at 13% and $z at 14%, then
x + y + z = 3000 . . . (1)
0.12x + 0.13y + 0.14z = 400 . . . (2)
x + y = z . . . (3)

Putting (3) in (2) gives 2z = 3000
z = 3000/2 = 1500
Substituting for z in (2) and (3), gives
0.12x + 0.13y = 190 . . . (4)
x + y = 1500 . . . (5)

Solving (4) and (5) simultaneusly gives that
x = 500 and y = 1000

Therefore, $500 was invested at 12%, $1,000 at 13% and $1,500 at 14%.

It should be a or d i’m not sure