Question

You invested 17,000 into accounts paying 2%and 8% annual interest respectfully if the total interest earned for the year was 1,060 how much was was invested at each rate2%8%

Answer 1

Let's define the following variables.

x = the amount invested at 2%

y = the amount invested at 8%

0.02x = interest at 2% account

0.08y = interest at 8% account

If the total investment in both accounts is 17, 000 then, we can say that:

$x+y=17,000$

If the total interest earning in both accounts is 1,060 then, we can say that:

$0.02x+0.08y=1,060$

Now that we were able to form a system of equation, we can solve for the values of x and y using substitution method. Here are the steps.

1. Equation Equation 1 into y = .

$\begin{gathered} x+y=17,000 \\ y=17,000-x \end{gathered}$

2. Replace the value of y in equation 2 using equation 1.

$\begin{gathered} 0.02x+0.08y=1,060 \\ 0.02x+0.08(17,000-x)=1,060 \end{gathered}$

3. Solve for x.

$\begin{gathered} \text{Distribute 0.08.} \\ 0.02x+1,360-0.08x=1,060 \\ Subtract\text{ 0.02x and 0.08x.} \\ -0.06x+1,360=1,060 \\ \text{Subtract 1,360 on both sides of the equation.} \\ -0.06x=-300 \\ \text{Divide both sides by -0.06.} \\ x=5,000 \end{gathered}$

The value of x is 5,000. Hence, the amount invested at 2% is 5,000.

4. Solve for y using equation 1 and the calculated value of x.

$\begin{gathered} y=17,000-x \\ y=17,000-5,000 \\ y=12,000 \end{gathered}$

The value of y is 12,000. Hence, the amount invested at 8% is 12, 000.