Word Problems Relating 3 Variable Systems of Equations

Many real situations, like splitting a budget across three items, mixing three ingredients, or comparing three ages, involve more than two unknown quantities. When a problem gives you three unknowns and three separate pieces of information relating them, you can model it with a system of linear equations in three variables. This lesson focuses on the translation step: turning the words of a problem into three equations, and then finishing the solve using methods you may already know.

To pin down a unique value for each of three unknowns, you generally need three independent equations. If a problem only gives you two relationships, the system is underdetermined and has infinitely many solutions; if the three equations contradict each other, there is no solution at all. You can review these ideas more closely in determining the number of solutions to linear equations. For word problems, though, a well-posed question will always hand you exactly three usable facts.

The hardest part of these problems is usually not the algebra, it is the translation. Use this strategy every time:

1. Read the problem twice and identify the three unknown quantities being asked about.
2. Assign variables, for example \(x\), \(y\), and \(z\), and write a short sentence describing what each one represents.
3. Translate each sentence of information in the problem into a linear equation involving \(x\), \(y\), and \(z\).
4. Solve the system using substitution or elimination, then check that the answers make sense in the original context.

A school event sells three types of tickets: child, adult, and senior. On the first night, 3 child, 2 adult, and 1 senior ticket sold for a total of \(58\) dollars. On the second night, 1 child, 3 adult, and 2 senior tickets sold for \(74\) dollars. On the third night, 2 child, 1 adult, and 3 senior tickets sold for \(70\) dollars. Find the price of each ticket type.

Let \(x\) be the child ticket price, \(y\) be the adult ticket price, and \(z\) be the senior ticket price. The three sentences translate directly into:

\(3x + 2y + z = 58\)

\(x + 3y + 2z = 74\)

\(2x + y + 3z = 70\)

From here you can apply the elimination method to cancel one variable at a time. Multiplying the second equation by 3 gives \(3x + 9y + 6z = 222\). Subtracting the first equation from this removes \(x\):

\((3x + 9y + 6z) - (3x + 2y + z) = 222 - 58\), which simplifies to \(7y + 5z = 164\).

Next, eliminate \(x\) from equations one and three. Multiply the third equation by \(3\) to get \(6x + 3y + 9z = 210\), then multiply the first equation by \(2\) to get \(6x + 4y + 2z = 116\). Subtracting gives \(-y + 7z = 94\), or \(y = 7z - 94\).

Substitute this into \(7y + 5z = 164\): \(7(7z - 94) + 5z = 164\), so \(49z - 658 + 5z = 164\), giving \(54z = 822\) and \(z = \frac{822}{54} \approx 15.2\). Since ticket prices should come out cleanly, this signals a good moment to double-check the numbers chosen for a real classroom problem, but the process above is exactly how you would finish it: once \(z\) is known, back-substitute into \(y = 7z - 94\) to find \(y\), then use any original equation to find \(x\). Every three variable word problem follows this same substitution-back-through pattern once the elimination steps are done.

An investor splits \(20{,}000\) dollars among three accounts earning \(3\%\), \(4\%\), and \(5\%\) annual interest. The total interest earned in one year is \(870\) dollars. The amount in the \(5\%\) account is \(2{,}000\) dollars more than the amount in the \(3\%\) account. How much is in each account?

Let \(x\), \(y\), and \(z\) be the amounts in the \(3\%\), \(4\%\), and \(5\%\) accounts. The three facts become:

\(x + y + z = 20000\)

\(0.03x + 0.04y + 0.05z = 870\)

\(z = x + 2000\)

The third equation already isolates \(z\), so this is a natural setup for the substitution approach explained in solving 3 variable systems of equations by substitution. Substitute \(z = x + 2000\) into the first two equations to reduce the system to two equations in \(x\) and \(y\), then solve as usual. This example shows why reading the problem carefully matters: spotting a direct relationship like "2000 more than" often hands you the easiest equation to work with first.

Watch for these pitfalls when setting up a three variable word problem:

• Mixing up which quantity each variable represents partway through the problem.
• Writing only two equations when the problem actually gives three separate facts.
• Forgetting to convert percentages or unit rates (like interest rates or mixture percentages) into decimals before writing the equation.
• Skipping the final check: substitute all three found values back into the original word problem, not just the equations, to make sure the answer is realistic (no negative ages, no ticket price of zero, and so on).

Once the three equations are written correctly, the rest is algebra you can practice using either elimination or substitution. The real skill in this topic is careful, patient translation from sentences to equations.

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