Video: Solving a System of Linear Equations Involving Percentages

One year ago, you inherited one hundred thousand dollars, and the following bank accounts were available. Account A, which pays 4% compounded annually. Account B, which pays 3% compounded annually. Account C, which pays 2% compounded annually. You invested all of the money in these three accounts and put five times as much in account A than in account B. Given that you earned $3650 in interest in a year, how much did you invest in each account?

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Video Transcript

One year ago, you inherited 100000 dollars, and the following bank accounts were available: account A, which pays four percent compounded annually; account B, which pays three percent compounded annually; account C, which pays two percent compounded annually. You invested all of the money in these three accounts and put five times as much in account A than in account B. Given that you earned 3650 dollars in interest in a year, how much did you invest in each account?

So in this question, we have the three unknown variables of how much we put in each account. We also have compound interest to our account. We can use the compound interest formula 𝐴 equals 𝑃 times one plus π‘Ÿ over 𝑛 to the power 𝑛𝑑, where 𝑃 represents the principal or starting amount. π‘Ÿ represents the rate of interest. 𝑛 represents the number of times the interest is compounded. 𝑑 represents the number of years the amount is deposited. And 𝐴 represents the amount at the end. It’s important to note that the amount at the end is the principal plus the interest and not just the interest on its own. So let’s see if we can use the compound interest formula to write an equation for each account.

Starting with account A, if we write 𝐴 sub 𝐴 as the amount in account 𝐴 and 𝑃 sub 𝐴 as the principal in account A, then our multiplier in parentheses will be written as one plus 0.04 over one. Since our interest rate is four percent compounded annually, that’s the same as 0.04 as a decimal. The one on the denominator represents the fact that the interest rate is only compounded once in a year. And the exponent will be one times one, since we have 𝑛 is equal to one and 𝑑 is equal to one since it’s just invested for one year. We can, therefore, simplify this equation as 𝐴 sub 𝐴 equals 1.04𝑃 sub 𝐴.

In the same way, if we use 𝐴 sub 𝐡 to represent the amount in account B and 𝑃 sub 𝐡 to represent the principal in account B, then with the interest rate of three percent, we would have the equation that 𝐴 sub 𝐡 equals 𝑃 sub 𝐡 times one plus 0.03 over one to the power of one times one, which can be simplified to 1.03𝑃 sub 𝐡. For account C then, we have the equation that 𝐴 sub 𝐢 equals 1.02𝑃 sub 𝐢, with 𝐴 sub 𝐢 being the amount in account C and 𝑃 sub 𝐢 being the principal in account C.

So now, we find a relationship for each account between the principal β€” that’s the amount at the start β€” and the amount at the end. Let’s see if we can write an equation relating the principal amount in each account and the total amount invested. We’re told in this rather delightful situation that we had 100000 dollars and that we invested all of the money in the three accounts. Since our 𝑃 values relate the principal amounts, then we can write the equation that 𝑃 sub 𝐴 plus 𝑃 sub 𝐡 plus 𝑃 sub 𝐢 equals 100000.

Next, let’s see if we can write an equation that represents the amounts in the accounts at the end of this one-year period. The amount in each account will be represented by the 𝐴 values. And we’re told that we earned 3650 dollars after one year. However, we can’t simply write that the sum of the amounts is equal to 3650 since we know that the amount at the end also includes the principal. So we must have 𝐴 sub 𝐴 plus 𝐴 sub 𝐡 plus 𝐴 sub 𝐢 equals 103650. It will be good at this stage if we could write this second equation in terms of the principal amounts rather than in terms of the amount in each account at the end. We have already established that 𝐴 sub 𝐴 is equal to 1.04𝑃 sub 𝐴. So we can substitute this into the equation. We can also do the same for our other values of 𝐴 sub 𝐡 and 𝐴 sub 𝐢.

So now, we have two equations representing our three unknown values. But to solve for three unknown values, we need three equations. So let’s see if we can find another one from the information that we’re given. We’re told that we put five times as much in account 𝐴 than in account 𝐡, which means that we have a relationship between 𝑃 sub 𝐴 and 𝑃 sub 𝐡. This can be written as 𝑃 sub 𝐴 equals five times 𝑃 sub 𝐡, since the principal amount in account 𝐴 is five times that which is in account 𝐡. So now, we have three equations for our three unknown values. It’s time to set about solving to find the value of each one.

Noticing that equation three just has 𝑃 sub 𝐴 and 𝑃 sub 𝐡, then our first step will be to eliminate the other variable 𝑃 sub 𝐢 from equations one and two. We can begin by getting rid of the decimals in equation two. And we can do this by multiplying all the values by 100. This will give us 104𝑃 sub 𝐴 plus 103𝑃 sub 𝐡 plus 102𝑃 sub 𝐢 equals 10365000. In order to eliminate our variable 𝑃 sub 𝐢, it needs to have the same coefficient. We can achieve this by multiplying everything in equation one by 102. So we have 102𝑃 sub 𝐴 plus 102𝑃 sub 𝐡 plus 102𝑃 sub 𝐢 equals 10200000. And now we have the same coefficient of 𝑃 sub 𝐢, then we can eliminate this by subtraction.

Therefore, when we calculate equation two subtract equation one, we’ll have 104𝑃 sub 𝐴 subtract 102𝑃 sub 𝐴, giving us two 𝑃 sub 𝐴. 103𝑃 sub 𝐡 subtract 102𝑃 sub 𝐡 will give us one 𝑃 sub 𝐡 or just 𝑃 sub 𝐡. 102𝑃 sub 𝐢 subtract 102𝑃 sub 𝐢 does give us zero. So we have eliminated that. And on the right-hand side, 10365000 subtract 10200000 gives us 165000. And so, we have now formed a fourth equation. We now have two equations relating 𝑃 sub 𝐴 and 𝑃 sub 𝐡. We can solve these using elimination, like we’ve just seen, or by substitution, which I’ll do here.

If we take the fact that 𝑃 sub 𝐴 equals five 𝑃 sub 𝐡, then we can substitute this into equation four. This means that the value of 𝑃 sub 𝐴 in equation four will be replaced with the value of five 𝑃 sub 𝐡. This gives us the equation that two times five 𝑃 sub 𝐡 plus 𝑃 sub 𝐡 equals 165000. Simplifying our left-hand side will give us that 11𝑃 sub 𝐡 equals 165000. To find the value of 𝑃 sub 𝐡, we divide both sides of our equation by 11. And so, our first missing value of 𝑃 sub 𝐡 is equal to 15000. In other words, the principal amount in account 𝐡 is 15000 dollars.

Now we found our first unknown value, we can substitute this into our other equations to find the other two unknown values. We can use any of the equations to do this, but let’s start with the simplest one: 𝑃 sub 𝐴 equals five 𝑃 sub 𝐡. Substituting our value of 𝑃 sub 𝐡 into this equation will give us that 𝑃 sub 𝐴 equals five times 15000. So 𝑃 sub 𝐴 is equal to 75000. To find the remaining value of 𝑃 sub 𝐢, we can substitute our values of 𝑃 sub 𝐴 and 𝑃 sub 𝐡 into equation one. This gives us 75000 plus 15000 plus 𝑃 sub 𝐢 equals 100000. Simplifying the left-hand side, we have 90000 plus 𝑃 sub 𝐢 equals 100000. And then subtracting 90000 from both sides of our equation, we have 𝑃 sub 𝐢 equals 10000.

So now we have found the three values of 𝑃 sub 𝐴, 𝑃 sub 𝐡, and 𝑃 sub 𝐢, we need to check. The question is asking us, How much did you invest? Since this means that 𝑃 values are not the 𝐴 values, then our final answer will be the ones we’ve just calculated. So we invested in account A 75000 dollars, in account B 15000 dollars, and in account C 10000 dollars.

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