Q. 152 - Intermediate Algebra

Question

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

Sam invested \($48,000$, some at $6$% interest and the rest at $10$%. How much did he invest at each rate if he received \)$4000$ in interest in one year?

Step-by-Step Solution

Answer

Sam should invest $\$20,000$ in the account with $6\%$ interest and $\$28,000$ in the account with $10\%$ interest.

Step 1. Given Information

We are given that Sam invested $\$48,000$ in two account with $6\%$ and $10\%$ interest.

Step 2. Assumptions and Formation of equations.

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

Total amount invested is equal to $48000$, so

$x+y=48000 -\left(1\right)$

He received $\$4000$ every year, so the other equation is given by,

$$\begin{gathered} 6\%x+10\%y=4000 \\ 0.06x+0.1y=4000 -\left(2\right) \end{gathered}$$

Step 3. Solving the equations.

The first equation can be written as,

$x=48000-y -\left(3\right)$

Now, putting the value of $x$ in the second equation, we get

$$\begin{gathered} 0.06x+0.1y=4000 \\ 0.06(48000-y)+0.1y=4000 \\ 2880-0.06y+0.1y=4000 \\ 0.04y=1120 \end{gathered}$$

Dividing by $0.04$, we get

$$\begin{gathered} y=\frac{1120}{0.04} \\ y=28000 \end{gathered}$$

Now, putting the value of $x$ in the third equation, we get

$$\begin{gathered} x=48000-28000 \\ x=20,000 \end{gathered}$$

Hence he invested $\$20,000$ in the first account with $6\%$ and $\$28,000$ in the second account with $10\%$ interest.

Step 4. Checking the solution

Checking the solution by putting the value of $x,y$ in the equations, we get

$$\begin{gathered} x+y=48000 \\ 20000+28000=48000 \\ 48000=48000 \\ 0.06x+0.1y=4000 \\ 0.06\left(20000\right)+0.1\left(28000\right)=4000 \\ 1200+2800=4000 \\ 4000=4000 \end{gathered}$$

This is true, hence the solution is correct.