Question

$23,803 is invested, part at 8% and the rest at 5%. If the interest earned from the amount invested at 8% exceeds the interest earned from the amount invested at 5% by $357.89, how much is invested at each rate? (Round to two decimal places if necessary.)

Answer 1

The amount invested at 8% is $11,908 and the amount invested at 5% is $11,895

Explanation:

Let

x ----> the amount invested at 8%

(23,803-x) ----> the amount invested at 5%

Remember that

`8\%=8\100=0.08`

`5\%=5\100=0.05`

we know that

The interest earned from the amount invested at 8% is equal to the interest earned from the amount invested at 5% plus $357.89

so

The linear equation that represent this situation is

`0.08x=0.05(23,803-x)+357.89`

solve for x

Apply distributive property right side

`0.08x=1,190.15-0.05x+357.89`

Group terms

`0.08x+0.05x=1,190.15+357.89`

Combine like terns

`0.13x=1,548.04`

`x=\$11,908`

`\$23,803-x=\$23,803-\$11,908=\$11,895`

therefore

The amount invested at 8% is $11,908 and the amount invested at 5% is $11,895