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Rebecca invested a total of ${\$12{,}000}$ in two accounts. After a year, one account lost 8.1%, while the other account gained 6.5%. In total, Rebecca lost ${\$388}\text{.}$ How much money did she invest in each account?

Write and solve a system of equations to answer the following questions.

Rebecca invested

in the account with 8.1% loss.

Rebecca invested

in the account with 6.5% gain.

Solution

Assume Rebecca invested $x$ dollars in the account with 8.1% loss, and $y$ dollars in the account with 6.5% gain.

It’s given that the total amount of investment was ${\$12{,}000}\text{,}$ so we have:

$x+y=12000$

Since the first account lost 8.1% for $x$ dollars, the amount of loss was $0.081x$ dollars.

Similarly, since the second account gained 6.5% for $y$ dollars, the amount of gain was $0.065y$ dollars.

It’s given that the total amount of loss was ${\$388}\text{,}$ so we have:

$-0.081x+0.065y= -388$

Now we have a system of equations:

$\begin{cases}x+y=12000 \\ -0.081x+0.065y= -388 \end{cases}$

We will use substitution to solve this problem. First, we solve for $y$ in the first equation:

$\begin{aligned} x+y\amp=12000\\ x+y\subtractright{x}\amp=12000\subtractright{x}\\ y=12000-x \end{aligned}$

Next, we substitute $y$ with $12000-x$ in $-0.081x+0.065y= -388\text{,}$ and we have:

$\begin{aligned} -0.081x+0.065y\amp= -388\\ -0.081x+0.065(\substitute{12000-x})\amp= -388\\ -0.081x+0.065(12000)-0.065x\amp=-388\\ 780-0.146x\amp=388\\ 780-0.146x\subtractright{780}\amp=-388\subtractright{780}\\ -0.146x\amp=-1168\\ x\amp=8000 \end{aligned}$

Finally, we substitute $x=8000$ into $y=12000-x\text{,}$ and we have:

$\begin{aligned} y\amp=12000-x\\ y\amp=12000-\substitute{8000}\\ y\amp=4000 \end{aligned}$

The solution to the system of equations is $x=8000\text{,}$ $y=4000\text{.}$ This implies that Rebecca invested ${\$8{,}000}$ in the account with 8.1% loss and ${\$4{,}000}$ in the account with 6.5% gain.