Answer:
The amounts invested at 4%, 11% and 13% are $1200, $2500 and $4300 respectively.
Step-by-step explanation:
Let the amount invested at 4% be x and the amount invested at 11% be y.
Total amount of investment is $8000, so the amount invested at 13% is (8000-x-y).
The total annual income from these investments was $882.
[tex]\frac{4}{100}\times x+\frac{11}{100}\times y +\frac{13}{100}\times (8000-x-y)=882[/tex]
[tex]\frac{4x}{100}+\frac{11y}{100}+\frac{10400-13x-13y}{100}=882[/tex]
[tex]-9x-2y=-15800[/tex]
[tex]9x+2y=15800[/tex] ..... (1)
The amount of money invested at 13% was $600 more than the invested at 4% and 11% combined.
[tex]8000-x-y=x+y+600[/tex]
[tex]2x+2y=7400[/tex]
[tex]x+y=3700[/tex] ..... (2)
Solve equation (1) and (2) by using elimination method.
[tex]x=1200[/tex]
[tex]y=2500[/tex]
[tex]8000-x-y=8000-1200-2500=4300[/tex]
Therefore amounts invested at 4%, 11% and 13% are $1200, $2500 and $4300 respectively.
