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

Answer:

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

Step-by-step 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