Q 4.55 - Intermediate Algebra

Question

Translate to a system of equations and solve:

Leon had $\$50,000$ to invest and hopes to earn $6.2\%$ interest per year. He will put some of the money into a stock

fund that earns $7\%$ per year and the rest into a savings account that earns $2\%$ per year. How much money should

he put into each fund?

Step-by-Step Solution

Answer

Leon should invest $\$42000$ into a stock and $\$8000$ into a saving account to get $6.2$ interest per year.

Step 1 . Given information

• Leon invested $\$50000$ hopes to earn $6.2\%$ interest per year.
• Stocks will give $7\%$ interest per year.
• Saving account will give $2\%$ interest per year.

Step 2. Form the system of equations.

Let

$$\begin{gathered} s = the amount invested in stocks. \\ a = the amount invested in saving account \end{gathered}$$

Total money invested is $\$50000$.

So,

$s+a=50000\_\_\_\left(1\right)$

And, 

The interest he will get at $6.2\%$ is:

$$\begin{gathered} =6.2\%\times 50000 \\ =0.062\times 50000 \\ =3100 \end{gathered}$$

So, according to the question,

$$\begin{gathered} 7\%\times s+2\%\times a=3100 \\ 0.07s+0.02a=3100\_\_\_\left(2\right) \end{gathered}$$

Step 3. Solve the system of equations by substitution method.

Multiplying the equation $\left(1\right)$ by $0.07$, we get:

$$\begin{gathered} 0.07(s+a)=0.07\times 50000 \\ 0.07s+0.07a=3500\_\_\_\left(3\right) \end{gathered}$$

Subtracting equation $\left(3\right)$ from equation $\left(2\right)$, we get:

$$\begin{gathered} 0.07s+0.02a-(0.07s+0.07a)=3100-3500 \\ 0.02a-0.07a=-400 \\ 0.05a=400 \end{gathered}$$

Dividing both sides by $0.05$, we get:

Step 4. Find the value of s

Substituting the value of $a$ in equation $\left(1\right)$, we get:

$$\begin{gathered} s+a=50000 \\ s+8000=50000 \\ s+8000-8000=50000-8000 \\ s=42000 \end{gathered}$$

So, Leon should invest $\$42000$ into a stock and $\$8000$ into a saving account in order to get desired interest.