Write, Solve and Interpret Systems from Word Problems | Grade 8 Math | LevelKite

Some questions ask for two unknown numbers at once. Picture a school play that sells adult tickets and student tickets. You want to know how many tickets of each kind the school sells on opening night.

One fact, such as "the school sells 100 tickets in all," does not settle it. The numbers 40 and 60 fit, but so do 10 and 90. A second fact about the same two numbers can narrow the choices.

When you write each fact as an equation (a statement that two amounts are equal), you get a system of equations: two or more equations that use the same variables and must all be true at the same time. A variable is a letter that stands for a number. A solution of a system with two variables is a pair of values, one for each variable, that makes every equation in the system true.

In this lesson you will write a system from a story, solve it, and say what the solution means in the story.

Start with the unknowns. Each variable needs a clear meaning: say what it counts and in what unit. Writing " $x$ = adults" is too loose. Writing " $x$ = the number of adult tickets the school sells" is clear.

Say which values are permitted (allowed), too. A number of tickets is a whole number that is 0 or greater, so $x$ cannot be $-3$ or $12.5$ .

Then turn each fact into one equation. In every equation you write in this lesson, both sides must describe the same quantity in the same unit: tickets equal tickets, and dollars equal dollars.

A price needs one extra thought. Suppose every adult ticket costs the same 8 dollars. Then 3 adult tickets cost $8 + 8 + 8 = 3 \times 8 = 24$ dollars. In the same way, $x$ adult tickets cost $8 \times x = 8x$ dollars.

Worked example

Write a system from a story

A school play sells adult tickets for 8 dollars each and student tickets for 5 dollars each. On opening night, the school sells 100 tickets and collects 620 dollars from them. Write a system of equations that describes opening night.

Name the two unknowns and give each one a variable with a unit. The story does not say how many tickets of each kind the school sells, so those are the unknowns. Each one is a number of tickets, so it must be a whole number that is 0 or greater.
$x = \text{number of adult tickets}, \quad y = \text{number of student tickets}$
Write the fact about how many tickets there are. The adult tickets and the student tickets together make the 100 tickets the school sells. Both sides of this equation count tickets.
$x + y = 100$
Find the money from each kind of ticket. Every adult ticket costs the same 8 dollars, so x adult tickets bring in 8 times x dollars. Every student ticket costs the same 5 dollars, so y student tickets bring in 5 times y dollars.
$8x \text{ dollars from adult tickets}, \quad 5y \text{ dollars from student tickets}$
Write the fact about the money. The two amounts together make the 620 dollars the school collects. Both sides of this equation count dollars.
$8x + 5y = 620$
Put the two equations together. Both must be true for the same night, so they form a system.
$\begin{cases} x + y = 100 \\ 8x + 5y = 620 \end{cases}$

Answer

With $x$ as the number of adult tickets and $y$ as the number of student tickets, the system is $x + y = 100$ and $8x + 5y = 620$ .

Watch out: a total matched to the wrong equation. Reading the numbers in order, some students write $8x + 5y = 100$ and $x + y = 620$ . That system has the solution $x = -1000$ , $y = 1620$ : you can check that $-1000 + 1620 = 620$ and $8(-1000) + 5(1620) = -8000 + 8100 = 100$ . But a play cannot sell $-1000$ tickets, so something went wrong before the solving began.

Here is why it fails. Both $8x$ and $5y$ are amounts of money, so $8x + 5y$ counts dollars. But 100 is a number of tickets, and the story never says the money is 100 dollars. Before you solve, read each equation back in words and check that both sides describe the same quantity in the same unit.

What does a solution of the ticket system look like? Each row of the table below is a pair of whole numbers that fits the first equation, because the two numbers add to 100. The last column tests the same pair in the second equation.

Only one of these rows brings in exactly 620 dollars: 40 adult tickets and 60 student tickets. That pair makes both equations true, so it is a solution of the system.

The table tests only five pairs, so it cannot tell you whether some other pair also works. Solving the system with algebra finds the solution without guessing, and for this system it will show that this pair is the only one.

Five pairs that add to 100 tickets, tested against the money fact

Adult tickets (x)	Student tickets (y)	Tickets in all (x + y)	Money in dollars (8x + 5y)
$10$	$90$	$10 + 90 = 100$	$8(10) + 5(90) = 80 + 450 = 530$
$30$	$70$	$30 + 70 = 100$	$8(30) + 5(70) = 240 + 350 = 590$
$40$	$60$	$40 + 60 = 100$	$8(40) + 5(60) = 320 + 300 = 620$
$50$	$50$	$50 + 50 = 100$	$8(50) + 5(50) = 400 + 250 = 650$
$70$	$30$	$70 + 30 = 100$	$8(70) + 5(30) = 560 + 150 = 710$

Key concept

One equation for each fact, one pair that fits both

In this lesson, each story has two unknown quantities and two facts about them. You write each fact as one equation, with both sides in the same unit, and the two equations form a system.

A solution of the system is a pair of values that makes both equations true, so it fits both facts of the story. To interpret it, give each value the meaning and unit of its variable, check that the story permits the values, and use them to answer the question the problem asks.

To solve the system, use a method you already know. Substitution replaces one variable with something that is equal to it. Elimination adds or subtracts the equations so that one variable drops out. Pick whichever fits the equations.

Every step must keep the solutions the same. These moves do that:

Add or subtract the same quantity on both sides of an equation.
Multiply or divide both sides by the same number, as long as that number is not zero. You can never divide by zero.
Replace a variable with something that an equation in the system says is equal to it.
Rewrite one side in a form that has the same value for every permitted value of the variables, such as $3x$ in place of $8x - 5x$ . Two forms like these are called equivalent expressions.

When you have both values, check them in both original equations. Use the originals, not equations you changed along the way, because a slip may already be inside a changed equation. If your pair is wrong, at least one original equation will come out false.

Worked example

Solve the ticket system

In the ticket story, $x$ is the number of adult tickets and $y$ is the number of student tickets. Solve the system $x + y = 100$ and $8x + 5y = 620$ . How many tickets of each kind does the school sell?

Use elimination. Multiply both sides of the first equation by 5, so that both equations contain 5y. This keeps the solutions the same because 5 is not zero. Then multiply each part inside the parentheses by 5, which gives an equivalent expression.
$5(x + y) = 5(100) \quad\text{so}\quad 5x + 5y = 500$
Subtract the new equation from the original second equation. Because 5x + 5y equals 500, you subtract the same amount from both sides, so the solutions stay the same. Subtract the matching parts: 8x minus 5x is 3x, and 5y minus 5y is 0.
$(8x + 5y) - (5x + 5y) = 620 - 500 \quad\text{so}\quad 3x = 120$
Divide both sides by 3. This keeps the solutions the same because 3 is not zero.
$\frac{3x}{3} = \frac{120}{3} \quad\text{so}\quad x = 40$
Replace x with 40 in the original first equation, which is valid because x equals 40. Then subtract 40 from both sides to find y.
$40 + y = 100 \quad\text{so}\quad y = 100 - 40 = 60$
Check the pair in both original equations. Both come out true, so x = 40 and y = 60 is a solution. Every step kept the solutions the same, so it is the only solution of this system.
$40 + 60 = 100, \qquad 8(40) + 5(60) = 320 + 300 = 620$

Answer

The school sells $40$ adult tickets and $60$ student tickets.

Solving gives you numbers. Interpreting means saying what those numbers tell you about the story.

You can write the solution of the ticket system as the ordered pair $(x, y) = (40, 60)$ . An ordered pair lists two values in a fixed order. Here the first value goes with $x$ and the second goes with $y$ . Go back to the definitions: $x$ counts adult tickets and $y$ counts student tickets. So the pair means 40 adult tickets and 60 student tickets, not the other way around.

Then ask three questions.

Does it fit both facts? Go back to the story, not only to your equations, because checking in the equations cannot catch an equation that does not match the story. Here $40 + 60 = 100$ tickets and $320 + 300 = 620$ dollars, so it fits.
Does the story permit these values? Yes: both are whole numbers that are 0 or greater.
Does it answer the question the problem asks? A problem may ask for one value, for both values, or for something you work out from them. The table shows how one solution answers different questions.

One solution of the ticket system, four different questions

Question the problem asks	Work with the solution (40, 60)	Answer in words
How many adult tickets does the school sell?	Read the value of $x$ : $x = 40$	The school sells 40 adult tickets.
How many more student tickets than adult tickets does it sell?	$y - x = 60 - 40 = 20$	It sells 20 more student tickets than adult tickets.
How much money comes from student tickets?	$5y = 5(60) = 300$	Student tickets bring in 300 dollars.
How much money comes from adult tickets?	$8x = 8(40) = 320$	Adult tickets bring in 320 dollars.

Worked example

Interpret a solution

A class of 52 students rides to a museum in 9 vehicles. Some are vans and the rest are cars. Each van carries exactly 8 students and each car carries exactly 4 students. Let $v$ be the number of vans and $c$ be the number of cars, both whole numbers that are 0 or greater. The system $v + c = 9$ and $8v + 4c = 52$ has the solution $(v, c) = (4, 5)$ . What does the solution mean, and how many students ride in vans?

Match each number to its variable. The pair is written in the order (v, c), so the first number is the value of v and the second number is the value of c.
$v = 4, \quad c = 5$
Give each number its meaning and unit from the definitions. The variable v counts vans and the variable c counts cars, so the class uses 4 vans and 5 cars.
Check the pair against both facts by using the original equations. The first equation counts vehicles. The second counts students: each van carries the same 8 students, so 4 vans carry 8 times 4 students, and each car carries the same 4 students, so 5 cars carry 4 times 5 students.
$4 + 5 = 9, \qquad 8(4) + 4(5) = 32 + 20 = 52$
Check that the story permits the values. A number of vehicles must be a whole number that is 0 or greater, and both 4 and 5 are.
Answer the question the problem asks. It asks for the students who ride in vans, which is not the value of v or of c. Multiply the 8 students in each van by the 4 vans.
$8 \times 4 = 32$

Answer

The solution means the class uses $4$ vans and $5$ cars, and $32$ students ride in vans.

Worked example

Write, solve and interpret in one problem

A school store sells notebooks for 4 dollars each and pens for 2 dollars each. In one week it sells 3 times as many pens as notebooks, and it takes in 150 dollars from these two items. How many of these items does the store sell in all that week?

Define the variables with their units. Each one counts items, so it must be a whole number that is 0 or greater.
$n = \text{number of notebooks sold}, \quad p = \text{number of pens sold}$
Write one equation for each fact. There are more pens than notebooks, so the number of pens equals 3 times the number of notebooks; for example, 10 notebooks would mean 30 pens. For the money, n notebooks bring in 4n dollars and p pens bring in 2p dollars, and the total is 150 dollars.
$p = 3n, \qquad 4n + 2p = 150$
Substitute 3n for p in the money equation. This is valid because the first equation says p and 3n are equal. Then multiply 2 by 3n to get 6n, and add 4n and 6n to get 10n. Each rewrite gives an equivalent expression.
$4n + 2(3n) = 150 \quad\text{so}\quad 4n + 6n = 150 \quad\text{so}\quad 10n = 150$
Divide both sides by 10, which is not zero, to find n. Then replace n with 15 in the first equation to find p.
$n = 150 \div 10 = 15, \qquad p = 3(15) = 45$
Check the pair in both original equations. Both come out true.
$45 = 3(15), \qquad 4(15) + 2(45) = 60 + 90 = 150$
Interpret the pair. The store sells 15 notebooks and 45 pens, and both are whole numbers that are 0 or greater. The problem asks for all the items, so add the two values.
$15 + 45 = 60$

Answer

The store sells $60$ of these items that week: $15$ notebooks and $45$ pens.

The big idea. In this lesson, each story had two unknowns and two facts about them. You defined each variable with its meaning and unit, and you wrote one equation for each fact to make a system. You solved it with steps that keep the solutions the same and checked the pair in both original equations. Then you finished in words: what each value means, whether the story permits it, and the answer to the question the problem asks.

Next, you will use a system of equations to compare two plans.