$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.)

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

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

solve for x

Apply distributive property right side

Group terms

Combine like terns

therefore

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

hello :

70y^8 + 30y^6 = 2 ( 35 y^8 + 15y^6)

the lenght is : 35 y^8

the width is : 15y^6