Asked by Emily
Micah Beasley invested $10,000 for one year, part at 10% annual interest and part at 14% annual interest. Vos total interest for the year was $1,280. How much money did he invest at each rate?
Answers
Answered by
Reiny
let the part invested at 10% be x
let the part invested at 14% be 10000-x
solve for x
.1x + .14(10000-x) = 1280
let me know what you get, so I can check it
let the part invested at 14% be 10000-x
solve for x
.1x + .14(10000-x) = 1280
let me know what you get, so I can check it
Answered by
scott
x at 10%, and y at 14%
x + y = 10000
.10 x + .14 y = 1280 ... multiplying by -10 ... -x - 1.4y = -12800
adding equations (to eliminate x) ... -.4 y = -2800
solve for y , then substitute back to find x
x + y = 10000
.10 x + .14 y = 1280 ... multiplying by -10 ... -x - 1.4y = -12800
adding equations (to eliminate x) ... -.4 y = -2800
solve for y , then substitute back to find x
Answered by
Emily
thank toi!
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