A person invested $7200 for 1 year, part at 4\%, part at 10\%, and the remainder at 15\%. The total annual income from these investments was $863. The amount of money invested at 15\% was $1400 more than the amounts invested at 4\% and 10\% combined. Find the amount invested at each rate.The person invested $__ at 4\%, $__ at 10\%, and $__ at 15\%.A person invested $7200 for 1 year, part at 4\%, part at 10\%, and the remainder at 15\%. The total annual income from these investments was $863. The amount of money invested at 15\% was $1400 more than the amounts invested at 4\% and 10\% combined. Find the amount invested at each rate.Solution –Let the amount invested at 4\% be x, amount at 10\% be y, then the amount at 15\% should be –4\% ————————– x

10\% ————————– y

15\% ————————– 7200 – (x + y)0.04x + 0.1y + 0.15 (7200 – (x + y)) = 863

0.04x + 0.1y + 0.15 (7200 – x – y) = 863

0.04x + 0.1y + 1080 – 0.15x – 0.15y = 863

0.04x – 0.15x + 0.1y – 0.15y = 863 – 1080

-0.11x – 0.05y = -217

0.11x + 0.05y = 217 — INow since the amount of money invested at 15\% was $1400 more than the amounts invested at 4\% and 10\% combined, we have –(x + y) = 7200 – (x + y) + 1400

x + y = 7200 – x – y + 1400

x + x + y + y = 7200 + 1400

2x + 2y = 8600

x + y = 8600/2

x + y = 4300 — IINow solve I and II using substitution to get the values –From II, x = 4300 – y — IIIPut this value of x in equation I, —

0.11(4300 – y) + 0.05y = 217

473 – 0.11y + 0.05y = 217

-0.06y = 217 – 473

y = -256/-0.06

y = 4266.67Now put this value of y in equation III —

x = 4300 – 4266.67

x = 33.33Amount invested at —

4\% ————————– $33.33

10\% ————————– $4266.67

15\% ————————– 7200 – (33.33 + 4266.67) = $2900

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