Math Core

Many real questions involve two unknowns connected by two facts: how many adult and student tickets were sold, or when two phone plans cost the same. A system of equations turns those facts into math you can solve with graphing, substitution or elimination.

From words to a system

The hardest part of a word problem is usually setting it up. Use these steps.

Solving a word problem with a system

Define two variables. Say exactly what each one stands for.
Write two equations, one for each fact in the problem.
Solve the system with any method.
Answer the question in a sentence with units, and check that it makes sense.

Worked example: Sum and difference

The sum of two numbers is $40$ . Their difference is $12$ . Find the numbers.

Let $x$ be the larger number and $y$ the smaller number.

\begin{aligned} x + y &= 40 \\ x - y &= 12 \end{aligned}

Add the equations: $2x = 52$ , so $x = 26$ . Then $26 + y = 40$ , so $y = 14$ .

The numbers are 26 and 14. Check: $26 + 14 = 40$ ✓ and $26 - 14 = 12$ ✓.

Counting and value problems

A very common type has one equation that counts things and another that adds up their value (cost, points, legs, wheels).

Worked example: Ticket sales

A school play sold $50$ tickets for a total of $310. Adult tickets cost $8 and student tickets cost $5. How many of each kind were sold?

Let $a$ be the number of adult tickets and $s$ the number of student tickets.

Counting tickets: $a + s = 50$
Adding up money: $8a + 5s = 310$

Solve the first equation for $s$ : $s = 50 - a$ . Substitute:

\begin{aligned} 8a + 5(50 - a) &= 310 \\ 8a + 250 - 5a &= 310 \\ 3a &= 60 \\ a &= 20 \end{aligned}

Then $s = 50 - 20 = 30$ .

The play sold 20 adult tickets and 30 student tickets. Check: $20 + 30 = 50$ ✓ and $8(20) + 5(30) = 160 + 150 = 310$ ✓.

Common mistake

Don't mix up the two kinds of equation. The count equation adds numbers of things ( $a + s = 50$ ). The value equation multiplies each count by its price first ( $8a + 5s = 310$ ). Writing $a + s = 310$ adds tickets to dollars, which makes no sense.

Comparing two options

When two plans each have a starting cost and a rate, a system tells you when they cost the same. Each plan is a linear equation $y = mx + b$ , where $b$ is the starting fee and $m$ is the rate.

Worked example: Which gym is cheaper?

Gym A charges a $30 sign-up fee plus $5 per visit. Gym B charges a $10 sign-up fee plus $9 per visit. After how many visits do the gyms cost the same? What is that cost?

Let $x$ be the number of visits and $y$ the total cost in dollars.

\begin{aligned} y &= 30 + 5x \quad \text{(Gym A)} \\ y &= 10 + 9x \quad \text{(Gym B)} \end{aligned}

Both expressions equal $y$ , so set them equal: $30 + 5x = 10 + 9x$ . Subtract $5x$ and $10$ from both sides: $20 = 4x$ , so $x = 5$ . Then $y = 30 + 5(5) = 55$ .

Visits (across) and total cost in dollars (up). The lines cross at (5, 55).Open in grapher →

After 5 visits, both gyms cost $55. The graph shows more: for fewer than 5 visits Gym B's line is lower, so Gym B is cheaper. For more than 5 visits, Gym A is cheaper.

Tip

Before you finish, ask: does my answer make sense? Numbers of tickets, people or coins must be whole numbers that aren't negative. If you get $12.5$ tickets, look for a mistake in your setup.