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?
thank toi!
To solve this problem, we can set up a system of equations. Let's assume that Micah invested x amount of money at 10% interest and (10000 - x) amount of money at 14% interest.
The interest earned on the investment at 10% is found by multiplying the investment amount by the interest rate (0.10). Therefore, the interest earned on the investment at 10% is 0.10x.
Similarly, the interest earned on the investment at 14% is 0.14 times the amount invested at that rate or 0.14(10000 - x).
According to the problem, the total interest earned is $1,280. So we can write the equation:
0.10x + 0.14(10000 - x) = 1280
To solve the equation, we can start by simplifying it:
0.10x + 1400 - 0.14x = 1280
Next, combine like terms:
-0.04x + 1400 = 1280
Now, isolate x by subtracting 1400 from both sides:
-0.04x = -120
Finally, divide both sides by -0.04 to solve for x:
x = (-120)/(-0.04) = 3000
Therefore, Micah invested $3000 at 10% interest and $7000 (10000 - 3000) at 14% interest.
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
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