Mathematics Algebra/linear equations

Question

Jolene invests her savings in two bank accounts, one paying 6% and the other paying 11% interest per year. She puts twice as much in the lower-yielding account because it is less risky. If she earned $4669 of total interest for the year, how much was invested in each account?

Answer 1

Let x = the amount of money invested at 11%.

Then 2x is the amount of money invested at 6%.

2x×6% + x×11% = 4669

2x×0.06 + 0.11x = 4669

0.12x + 0.11x = 4669

0.23x = 4669

x = 20300

Check:

20300 × 0.11 = 2,233

20300×2 = 40,600

40,600 × 0.06 = 2,436

2,233 + 2,436 = 4,669

The values check out.

The amount invested at 6% is $40,600, and the amount invested at 11% is $20,300.