Rosa invests $1100 in one account and $1300 in an account paying 3 % higher interest. At the end of one year she had earned $255 in interest. At what rates did she invest?

$1100 invested at %

$1300 invested at %

The amount of interest earned in one year is the product of the interest rate and the amount invested at that rate. The total interest earned is the sum of the amounts of interest earned on each investment.

Let x represent the lower interest rate. Then the higher rate is x+0.03 and the total interest earned is ...

1100x +1300(x+0.03) = 255

2400x +39 = 255 . . . simplify

2400x = 216 . . . . . . subtract 39

x = 216/2400 = 0.09 = 9% . . . . divide by the coefficient of x

Rosa invested $1100 at 9%, and $1300 at 12%.

Rosa invests $1100 in one account and $1300 in an account paying 3 % higher interest. At the end of one year she had earned $255 in interest. At what rates did she invest? $1100 invested at __% $1300 invested at __%

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Rosa invests $1100 in one account and $1300 in an account paying 3 % higher interest. At the end of one year she had earned $255 in interest. At what rates did she invest?

$1100 invested at %

$1300 invested at %