Question

In the following exercises, translate to a system of equations and solve.

Hattie had $3,000 to invest and wants to earn 10.6% interest per year. She will put some of the money into an account that earns 12% per year and the rest into an account that earns 10% per year. How much money should she put into each account?

Short answer

The amount invested in first account is $900 and amount invested in second account is $2100.

Step-by-step solution

Step 1. Given Information  

The given data is that Hattie had $3,000 to invest and wants to earn 10.6% interest per year. She will put some of the money into an account that earns 12% per year and the rest into an account that earns 10% per year.

Step 2. Explanation     

Let the amount invested in first account be x and amount invested in second account be y.

The total amount invested is $3000 that is $x+y=3000--\left(1\right)$

The interest wish to earn from investment is $10.6\%$. Thus, the amount received is $3000\times \frac{10.6}{100}=318$

The total interest earned from each investment is$0.12x+0.1y=318--\left(2\right)$

Step 3. Calculation  

Multiply equation (1) with number $0.12$and write the revised equation.

$$\begin{gathered} 0.12(x+y)=0.12\left(3000\right) \\ 0.12x+0.12y=360--\left(3\right) \end{gathered}$$

Solve the equation (3) and (2) by subtracting equation (2) from equation (3).

$$\begin{gathered} 0.12x+0.12y-0.12x-0.1y=360-318 \\ 0.02y=42 \\ y=2100 \end{gathered}$$

Substitute the value of y in equation (1) to find the value of x.

$$\begin{gathered} x+y=3000 \\ x=3000-2100 \\ x=900 \end{gathered}$$