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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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