This Question: 1 pt 22 of 42 (21 complete)
You invested $16,000 in two accounts paying 5% and 9% annual interest, respectively. If the total interest earned for the year was $880, how much was invested at each rate?

Amount invested in 5% interest = $14,000

Amount invested in 9% interest = $2,000

Step-by-step explanation:

Given:

Total amount invested = $16,000

Total interest earned = $880

Uknowns:

Let:

x = Money invested in 5%(0.05) interest

y = Money invested in 9%(0.09) interest

Equations:

Equation for the total amount invested:

x + y = 16,000 <--- equation 1

Equation for the total interest earned:

0.05x + 0.09y = 880 <--- equation 2

Solution:

Solving for x and y using substitution method,

In equation 1:

x + y = 16,000

y = 16,000 - x <--- equation 3

Substitute equation 3 in equation 2 to solve for x

0.05x + 0.09y = 880

0.05x + 0.09(16,000 - x) = 880

0.05x + 1,440 - 0.09x = 880

0.05x - 0.09x = 880 - 1,440

-0.04x = -560

x = -560/-0.04

x = $14,000 (amount invested in 5% interest)

Substitute the value of x in equation 3 to solve for y

y = 16,000 - x

y = 16,000 - 14,000

y = $2,000 (amount invested in 9% interest)

ian gonna lie that is way to much

y > 1/3x

y < = -x + 4

hope this : )