SOLUTION: Mr. and Mrs. Jackson invested some money at 6% simple interest and some money at 10% simple interest. In the second investment they put $1000.00 more than they put in the first. If

Algebra ->  Finance -> SOLUTION: Mr. and Mrs. Jackson invested some money at 6% simple interest and some money at 10% simple interest. In the second investment they put $1000.00 more than they put in the first. If      Log On


   

Question 1194955: Mr. and Mrs. Jackson invested some money at 6% simple interest and some money at 10% simple interest. In the second investment they put $1000.00 more than they put in the first. If the income from both investments for 1 year was $340.00, then how much did they invest at each rate?
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
x = amount invested at 6%.
x + 1000 = amount invested at 10%.
.06x + .1 * (x + 1000) = 340
simplify to get:
.06x + .1x + 100 = 340
subtract 100 from both sides of the equation and combine like terms to get:
.16x = 240
solve for x to get:
x = 240/.16 = 24000/16 = 3000 / 2 = 1500.
.06 * 1500 + .1 * (1500 + 1000) = .06 * 1500 + .1 * 2500 = 90 + 250 = 340.
this confirms the value of x = 1500 is correct.
solution is they invested 1500 at 6% and 2500 at 10%.