SOLUTION: a person invested $8000 for 1 year, part at 8%, part at 11%, and the remainder at 12%. The total annual income for these investments was $876. The amount of money invested at 12%
Algebra ->
Customizable Word Problem Solvers
-> Finance
-> SOLUTION: a person invested $8000 for 1 year, part at 8%, part at 11%, and the remainder at 12%. The total annual income for these investments was $876. The amount of money invested at 12%
Log On
Question 887182: a person invested $8000 for 1 year, part at 8%, part at 11%, and the remainder at 12%. The total annual income for these investments was $876. The amount of money invested at 12% was $200 more than the amounts invested at 8% and 11% combined. Find the amount invested at each rate. Answer by Theo(13342) (Show Source):
subtract 200 from both sides of the first equation and subtract 24 from both sides of the second equation to get:
2x + 2y = 7800 and .2x + .23y = 852
multiply both sides of the second equation by 10 to get:
2x + 2y = 7800 and 2x + 2.3y = 8520
subtract the first equation from the second equation to get:
.3y = 720
divide both sides of this equation by .3 to get:
y = 2400
replace y with 2400 in the equation of 2x + 2y = 7800 to get:
2x + 2(2400) = 7800
simplify this to get 2x + 4800 = 7800
subtract 4800 from both sides of this equation to get 2x = 3000
divide both sides of this equation by 2 to get x = 1500
you have x = 1500 and y = 2400
go back to your original equation of x + y + z = 8000 and replace x with 1500 and replace y with 2400 and solve for z to get z = 4100.
now you have:
x = 1500
y = 2400
z = 4100
those should be your solutions.
all that is left is to confirm that they are good solutions.
x + y + z = 8000 becomes 1500 + 2400 + 4100 = 8000 which becomes 8000 = 8000 so that part is good.
.08x + .11y + .12z = 876 becomes .08*1500 + .11*2400 + .12*4100 = 876 which becomes 120 + 264 + 492 = 876 which becomes 876 = 876 so that part is good.
looks like all parts are good so the solution is confirmed as good.
the amount invested at 8% is 1500
the amount invested at 11% is 2400
the amount invested at 12% is 4100