Q. 151 - Intermediate Algebra

Question

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

Carol invested $$2560$ into two accounts. One account paid $8$% interest and the other paid $6$% interest. She earned $7.25$% interest on the total investment. How much money did she put in each account?

Step-by-Step Solution

Answer

Carol put $\$1,600$ in the account with $8$% interest and $\$960$ in account with $6\%$ interest.

Step 1. Given Information

We are given that Carol invested$\$2560$ in two account with interest $8\%$ and $6\%$.

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$\$2560$, so

$x+y=2560 -\left(1\right)$

The other equation is given by,

$$\begin{gathered} 8\%x+6\%y=7.25\%2560 \\ 0.08x+0.06y=0.0725\times 2560 \\ 0.08x+0.06y=185.6 -\left(2\right) \end{gathered}$$

Step 3. Solving the equations.

The first equation can be written as $x=2560-y -\left(3\right)$

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

,$$\begin{gathered} 0.08 x+0.06 y =185.6 \\ 0.08(2,560-y)+0.06 y =185.6 \\ 204.8-0.08 y+0.06 y=185.6 \\ -0.02 y=185.6-204.8 \\ -0.02 y =-19.2 \end{gathered}$$

Dividing by $-0.02$, we get

$y=960$

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

$$\begin{gathered} x=2560-y \\ x=2560-960 \\ x=1600 \end{gathered}$$

Step 4. Checking the solution

Putting the value of $x, y$ in the equations, we get

$$\begin{gathered} x+y=2560 \\ 1600+960=2560 \\ 2560=2560 \\ 0.08x+0.06y=185.6 \\ 0.08\left(1600\right)+0.06\left(960\right)=185.6 \\ 128+57.6=185.6 \\ 185.6=185.6 \end{gathered}$$

This is true, hence the solution is correct.