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?

Short answer

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

Step-by-step solution

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,

role="math" localid="1644487480595" $$\begin{gathered} 8\%x+6\%y=7.25\%2560 \\ 0.08x+0.06y=0.0725\times 2560 \\ 0.08x+0.06y=185.6-\left(2\right) \end{gathered}$$

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

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

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.