# algebra

Janet invested \$26,000, part at 6% and part at 3%. If the total interest at the end of the year is \$1,080, how much did she invest at 6%

1. 👍 0
2. 👎 0
3. 👁 146
1. let x = the portion invested at 6%
let y = the portion invested at 3%

x + y = 26000
0.06x + 0.03y = 1080

Solve for x.

1. 👍 1
2. 👎 0

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