SOLUTION: A total of $12,000 is invested in two simple interest accounts. On one account the annual simple interest rate is 12%, on the second account the annual simple interest rate is 8%.

Question 850724: A total of $12,000 is invested in two simple interest accounts. On one account the annual simple interest rate is 12%, on the second account the annual simple interest rate is 8%. How much should be invested in each account so that the same interest is earned on both?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
let x = amount invested at 12%.
let y = amount invested at 8%.
x + y = 12000
let i = interest earned.
i = .12x for amount invested at 12%.
i = .08y for amount invested at 8%.
the interest earned on each investment needs to be the same.
this means that:
.12x = .08y
solve for x in this equation to get:
x = .08y / .12 which makes x = 2/3 * y
in the equation of x + y = 12000, replace x with 2/3 * y to get:
2/3 * y + y = 12000
combine like terms to get:
5/3 * y = 12000
multiply both sides of this equation by 3/5 to get:
y = 3/5 * 12000 which makes y = 7200.
since x + y = 12000, this makes x = 4800.
your solution should be that:
x = 4800 and y = 7200
replace x with 4800 and y with 7200 in the equation of .12x = .08y and you get:
.12 * 4800 = .08 * 7200
simplify this equation to get:
576 = 576
the interest is the same for both investments.
the solution is good.

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