Solve Real-World Problems Using Systems of Equations

Students model real-world situations (e.g., comparing cell phone plans, mixing solutions, finding break-even points) with a system of two linear equations, solve algebraically or graphically, and interpret the solution in context, including checking reasonableness.

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Grade 8 · Systems of Linear Equations · question 1 of 3

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20 rungs. A star at the top.

A basketball team makes 11 shots that are all two-pointers or three-pointers and scores 25 points. How many three-pointers did they make?

How it is taught

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Two facts about two unknowns need two equations.

Adding or subtracting those equations can find one unknown.

Read it: Systems in stories

Adult tickets cost $9 and child tickets $5. 8 tickets sell for $52. Let a be adults and c children: a + c = 8 and 9a + 5c = 52. Put c = 8 − a in: 4a + 40 = 52, so a = 3.

Name the two unknowns with letters.
Write one equation for how many there are and one for what they come to.
Use the count to write one letter in terms of the other, put it into the second equation, and solve.

The tickets are of two kinds, so two numbers are unknown, and the story gives two facts about them.

c = 8 − a, so 9a + 5(8 − a) = 52, which is 4a + 40 = 52.

a = 3 and c = 5: 3 + 5 = 8 tickets, and 9 × 3 + 5 × 5 = 52.

A story with two kinds of things: one equation counts them, one adds up their worth; solve the two together.

Pricing all 8 tickets as adult tickets: 9 × 8 = $72.

Some tickets are child tickets at $5. Write c = 8 − a and solve: a = 3.

Tickets: adult $9, child $5. 23 tickets sold for $167. How many adults? — 13. a + c = 23 and 9a + 5c = 167. Put c = 23 − a in: 4a = 52, so a = 13.

Carnival: 100 tickets sold for $380

A school carnival sold one hundred tickets and made three hundred eighty dollars. Adult tickets were five dollars and kid tickets were three dollars. How many of each did they sell?

We have two unknowns, so we'll need two equations. Let a stand for adult tickets and c stand for kid tickets.

Equation one comes from the total tickets. Adults plus kids equals one hundred.

Equation two comes from the money. Five dollars times a, plus three dollars times c, equals three hundred eighty.

Let's solve equation one for c. Subtract a from both sides: c equals one hundred minus a.

Now we substitute that into equation two, replacing c with the expression one hundred minus a.

Distribute the three. Three times one hundred is three hundred, and three times negative a is negative three a.

Combine like terms. Five a minus three a leaves two a, so we have two a plus three hundred equals three hundred eighty.

Subtract three hundred from both sides to get two a equals eighty. Then divide by two: a equals forty.

Now back-substitute to find c. One hundred minus forty equals sixty, so c equals sixty.

Always check it makes sense. Five times forty is two hundred, plus three times sixty is one hundred eighty. Together, three hundred eighty dollars. It works!

The carnival sold forty adult tickets and sixty kid tickets. That's how systems of equations solve real-world problems!

Grade 8 · Modeling and solving with two equations

What you see

While they climb, your page writes one true line to ask about tonight, and a short note each Sunday.

MayaGrade 3 · Mathematics

Did you know?

A chessboard has 64 squares — share them out between the 8 rows and each row holds exactly 8.

Today’s goal: 1 star — one skill mastered, every rung of its ladder climbed
Climbing now: Divide within 100

An example student.

MayaGrade 3 · Mathematics

Maya earned 2 stars this week — Interpret multiplication as equal groups, which took 6 answers over 3 days before its star landed 3 days ago and Multiply by 6, 7, 8 and 9. Next up: Divide within 100.

An example week.

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