Question

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

Sam invested \(48,000, some at 6% interest and the rest at 10%. How much did he invest at each rate if he received \)4,000 in interest in one year?

Short answer

The amount invested in first account is $20,000 and amount invested in second account is $28,000.

Step-by-step solution

Step 1. Given Information  

The given data is that Sam invested $48,000, some at 6% interest and the rest at 10%.

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 $48,000 that is $x+y=48000--\left(1\right)$

The interest earned from investment is $4000.

The total interest earned from each investment isrole="math" localid="1647779119715" $0.06x+0.1y=4000--\left(2\right)$

Step 3. Calculation   

Multiply equation (1) with number $0.06$and write the revised equation.

$$\begin{gathered} 0.06(x+y)=0.06\left(48000\right) \\ 0.06x+0.06y=2880--\left(3\right) \end{gathered}$$

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

$$\begin{gathered} 0.06x+0.1y-0.06x-0.06y=4000-2800 \\ 0.04y=1120 \\ y=28000 \end{gathered}$$

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

$$\begin{gathered} x+y=48000 \\ x=48000-28000 \\ x=20000 \end{gathered}$$