Math Word Problem

Question

Juan invests a total of $25,000 in two accounts. The first account earned a rate of return of 13% (after a year). However, the second account suffered a 4% loss in the same time period. At the end of one year, the total amount of money gained was $2,060.00. How much was invested into each account?

Answer 1

This problem can be solved using a system of equations where one of the equations is technically an exponential function but since the time period is 1 year, it can be represented as a linear function. A brief explanation of exponential equations follows:

The general formula for an exponential equation is

y = a(b)^x

Where a represents whatever initial value you are starting from, in this case since the initial values are what we are looking for, we will represent them as variables.

b represents the change in percent. In the case of exponential growth, it would be 1 + % (for example, if something increases by 12% each year, b = 1 + 0.12 = 1.12). For exponential decay, b would be 1 - %. So in this case, since the first account increases by 13% b = 1 + 0.13 = 1.13, while the second account loses 4% so that b = 1 - 0.04 = 0.96

x represents the number of intervals you are calculating the future number for. In this problem the interval is in years but the interval itself is unimportant other than knowing how many intervals you are being asked to calculate for. In this problem however, we essentially have a linear equation because the interval is 1 year and x¹ = x

So going back to the system of equations, our first equation where x is the amount in account 1 and y is the amount in account 2, we know that combined the total amount invested is $25,000 so the equation can be written as:

x + y = 25000

Our second equation will represent the total money earned over the year. This is where the information about exponential functions comes into play. We know account x earned 13% and account y lost 4% and the total earnings for both accounts in the year was $2,060.00, but this equation represents the final value in both accounts so it has to equal the initial investment plus the money gained so we can write the equation as:

x(1.13)¹ + y(0.96)¹ = 27060

Like we said, since it was 1 year, the exponent is essentially useless here so our system of equations is:

x + y = 25000

1.13x + 0.96y = 27060

We can now solve this system of equations using either the substitution or elimination method. I will use the elimination method by multiplying the first equation by 1.13 so that the x values will cancel out:

1.13(x + y = 25000)

1.13x + 0.96y = 27060 Thus,

1.13x + 1.13y = 28250

1.13x + 0.96y = 27060

We can now subtract both equations to get a single equation with one variable:

0.17y = 1190

Solving for y gives us

y = 7000

We now know that account y initially had $7,000 in it and we can use either equation to solve for x by plugging in this value, lets use equation 1 since it doesn't have any decimals and makes it easier:

x + 7000 = 25000

x = 18000

We can see now that account x started with $18,000 and account y started with $7,000. To make sure our answer is correct, we can plug both numbers into the other equation to make sure we get the right answer:

1.13(18000) + 0.96(7000) = 20340 + 6720 = 27060

We now know for a fact that x = $18,000 & y = $7,000 satisfies both equations.