SOLUTION: Part of 8,000 was invested at 10% and the rest at 12%. If the annual income from these investments was 860.00 how much was invested at each rate. I need help with this one.Each ra

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Question 260828: Part of 8,000 was invested at 10% and the rest at 12%. If the annual income from these investments was 860.00 how much was invested at each rate. I need help with this one.Each rate meaning 10% and 12 %
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
total invested is 8,000

let x be one part.
let y be the other part.

you have x + y = 8000 (equation 1)

total annual income is 860.000

x part is invested at 10%
y part is invested at 12%

you have .10*x + .12*y = 860 (equation 2)

you need to solve these 2 equations simultaneously.

solve for y in equation 1 to get y = 8000 - x

substitute for y in equation 2 to get .10*x + .12*(8000-x) = 860

solve for x in equation 2.

equation is:

.10*x + .12*(8000-x) = 860

simplify to get:

.10*x + .12*8000 - .12*x = 860

simplify to get:

-.02*x + 960 = 860

subtract 960 from both sides of this equation to get:

-.02*x = -100

divide both sides of this equation by -.02 to get:

x = 5000

that means y = 3000 because x + y = 8000.

you have x = 5000 and y = 3000.

substitute in equation 2 to get:

.10*5000 + .12*3000 = 860

simplify to get:

500 + 360 = 860 which is true confirming the values for x and y are good.

your answer is:

$5000 was invested at 10% and $3000 was invested at 12%.