Model and classify linear systems

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Lesson progressPractice problems 0/4

Difficulty: Intermediate
Estimated time: 38 minutes
Domains: Algebra
Techniques: System-modelingSolution-countIntersectionsProportional-coefficientsUnknown-constants

What you’ll learn

Spot a word problem that needs two equations.
Turn each fact in a story into its own equation, with the units lined up.
Read the point where two lines cross as the answer to both equations.
Tell whether a system has one solution, none or infinitely many.
Find the constant that gives the number of solutions a question asks for.

Prerequisites

Why this matters on the SAT

Two facts, two equations

Lots of SAT word problems give you two numbers you don't know and two facts about them. Your plan: give each unknown its own letter, and turn each fact into its own equation. Together, the two equations form a system.

Solution to the example

Start with the games. The team never lost, so every game was a win or a draw:

w+d=18.

Now the points. Each win is worth $3$ points and each draw is worth $1$ , so

3w+d=42.

That's choice C. The first equation counts games and the second counts points. Choice A has the right pieces in the wrong places: $3w+d$ counts points, so it can't equal $18$ games.

SAT example

A soccer team earned $3$ points for each game it won and $1$ point for each game it drew. The team had no losses in its $18$ games and earned $42$ points. Let $w$ be the number of games the team won and $d$ be the number it drew. Which system represents the situation?

A
$3w+d=18$ and $w+d=42$
B
$w+d=18$ and $w+3d=42$
C
$w+d=18$ and $3w+d=42$
D
$3w+3d=18$ and $w+d=42$

Spot a system question

Two kinds of questions belong here:

A story with two unknowns and two separate facts about them. You'll need one equation for each fact.
A question about how two lines meet. Do they cross once, never, or lie on top of each other? Sometimes a constant like $k$ decides which.

Why two facts? Say $a$ is the number of adult tickets sold and $s$ is the number of student tickets, and all you know is

a+s=120.

That could be $100$ adults and $20$ students, or $60$ and $60$ , or many other pairs. One fact can't pin down two numbers. A second fact, like the total money collected, can.

How you solve depends on what's asked:

In a word problem, name both unknowns and write one equation per fact before you solve anything.
When the numbers are messy, build the system first, then graph it in Desmos and click the point where the lines cross.
When the question asks how many solutions, or for a missing constant, compare the coefficients and constants exactly, by hand.

Turn two facts into two equations

Here's a story to build from. An event sold $280$ passes. Standard passes cost $15, premium passes cost $25, and the sale brought in $5,800. You can turn it into a system in four moves.

1. Name both unknowns

Let $x$ be the number of standard passes and $y$ the number of premium passes.

Write this down. Later, when you see a point like $(120,160)$ , your labels tell you that $120$ counts standard passes and $160$ counts premium ones.

2. Write one equation for each fact

All $280$ passes are standard or premium, so the count fact is

x+y=280.

The money fact adds up what each kind of pass brought in:

15x+25y=5800.

3. Keep each price with its own passes

The term $15x$ is the money from standard passes: $15 per pass times $x$ passes. The term $25y$ is the money from premium passes. Swap the prices and you're describing a different sale.

4. Check the units in each equation

Every term in the first equation counts passes. Every term in the second is dollars:

\frac{\text{dollars}}{\text{pass}}\times\text{passes} = \text{dollars}.

Inside one equation, every term must measure the same kind of thing. The two equations can measure different things, though, because they describe different facts.

Try it yourself:

A farm has $34$ chickens and goats, and the animals have $100$ legs altogether. Let $c$ be the number of chickens and $g$ the number of goats. Write one equation for the animals and one for the legs before you read on.

Here's the system:

\begin{aligned} c+g&=34\\[1.4em] 2c+4g&=100. \end{aligned}

The first equation counts animals. The second counts legs: $2$ for each chicken and $4$ for each goat.

Common mistake:

A common slip is writing $2c+4g=34$ . The left side counts legs, but $34$ counts animals, so the equation mixes two different things. Before you write each equation, say what it counts: $c+g=34$ counts animals, and $2c+4g=100$ counts legs.

Solutions are intersections

Every linear equation graphs as a line, and each point on the line makes that equation true. An intersection, a point where two lines cross, sits on both lines at once. So it makes both equations true, and that's exactly what a solution to the system is.

Here's the idea to remember: two lines can meet once, never, or everywhere. Those are the only three possibilities.

The calculator lists two equations: $2x-y=4$ first and $x+y=5$ second. Leave the first equation alone and change only the second:

With $x+y=5$ , the lines cross at $(3,2)$ . The system has one solution.
Change the second equation to $4x-2y=8$ . The two lines merge into one. They meet everywhere, so the system has infinitely many solutions.
Change it to $4x-2y=12$ . Now the lines are parallel and never touch, so the system has no solution.

In step 2 you'll see only one line, because the two graphs lie exactly on top of each other.

Check your understanding:

Why is $(3,2)$ a solution to both $2x-y=4$ and $x+y=5$ ?

Common mistake:

Seeing one line doesn’t mean one solution. Two equations can sit on top of each other, and then every point on that line is a solution. To catch it, turn the second equation off and on. If the line changes color but doesn’t move, both equations draw the same line. Compare the coefficients too before you decide.

Calculator loads as you approach

Keep the first equation as it is and edit only the second to see one crossing, total overlap and no crossing. Reset brings back the first pair.

Example: Build the model, then graph it

Worked example

A museum sold $180$ passes for a special exhibit. A standard pass cost $8.75, and a premium pass cost $12.50. The museum collected $1,905 from these passes. How many premium passes were sold?

A
$80$
B
$88$
C
$92$
D
$100$

Step 1

Name the unknowns

Let $x$ be the number of standard passes and $y$ the number of premium passes. The question asks for $y$ , so keep an eye on it.

Step 2

Write the count equation

Every pass is either standard or premium, so

x+y=180.

Both sides count passes.

Step 3

Write the money equation

Multiply each price by its own number of passes:

8.75x+12.50y=1905.

Every term is in dollars, so this equation tracks the total money.

Step 4

Let Desmos handle the decimals

Solving this by hand means messy decimal arithmetic. Now that the model is built, graphing is the quicker, safer way. Type each equation on its own line. They cross at

(92,88).

The second coordinate is $y$ , the premium passes, so the museum sold

\boxed{88}

premium passes. The answer is B.

Check it against the story:

92+88=180

and

8.75(92)+12.50(88)=805+1100=1905.

Both facts hold.

Check your understanding:

$92$ is part of the correct intersection. Why is it still the wrong answer?

Common mistake:

If you jump into Desmos before naming the variables, it’s easy to swap the prices or misread the point. Write and label both equations first. After you click $(92,88)$ , look back at your labels, $x$ for standard and $y$ for premium, before you choose.

Calculator loads as you approach

Click the intersection, then match each coordinate to its variable: x is standard, y is premium. Reset brings back the original equations.

Count solutions without graphing

Why did those three pairs behave so differently? You can read the answer straight from the equations. Compare each second equation with $2x-y=4$ and ask: can you multiply $2x-y$ by one number and get the second equation's left side?

$x+y=5$ : No. Turning $2x$ into $x$ takes $\frac12$ , but turning $-y$ into $y$ takes $-1$ . The slopes differ, so the lines cross once.
$4x-2y=8$ : Yes, $2(2x-y)=4x-2y$ . The right side doubles too, since $2\times4=8$ . The whole equation is doubled, so it's the same line.
$4x-2y=12$ : Yes, the number is $2$ again, so the slope is the same. But $2\times4$ is $8$ , not $12$ , so the line sits somewhere else. The lines are parallel.

That one number is the multiplier. Here's a way to remember the test: the $x$ and $y$ coefficients set a line's slope, and the constant sets where the line sits. So check the coefficients first, then the constant. It works for any two lines, once both equations are written as $Ax+By=C$ . Here's the whole test in one place:

Variable coefficients	Constants	Graph relationship	Solution count
No single multiplier matches both	No further comparison needed	Intersecting lines	Exactly one
One multiplier matches both	The same multiplier also matches the constants	Same line	Infinitely many
One multiplier matches both	That multiplier does not match the constants	Distinct parallel lines	No solution

You might wonder why we multiply instead of dividing one coefficient by another. A coefficient can be $0$ , and you can't divide by $0$ .

Check your understanding:

How many solutions does the system $6x+9y=15$ and $-2x-3y=-5$ have? Find the multiplier and check what it does to the whole first equation.

Find a missing constant

Sometimes a letter like $k$ hides in the system, and the question tells you how many solutions there are. These look harder, but it's the same multiplier test. The only new thing is the order: match the slope first, then check where the line sits. Take this one: for what value of $k$ does

\begin{aligned} 2x-y&=4\\[1.4em] 4x+ky&=12 \end{aligned}

have no solution?

Tidy both equations. Get each into the form $Ax+By=C$ . These two already are.
Match the $x$ and $y$ coefficients. Going from $2x$ to $4x$ takes a multiplier of $2$ . The $y$ term has to follow, so $k=2(-1)=-2$ .
Check the constants. The same multiplier turns $4$ into $8$ , but the second equation has $12$ . They don't match.
Put $k$ back in and decide. At $k=-2$ , the second equation is $4x-2y=12$ : same slope as $2x-y=4$ , different spot. That's the parallel pair you graphed earlier, so $k=-2$ gives no solution. If the constants had matched, you'd have the same line instead.

What if a question asks for exactly one solution? Flip the idea. The lines cross once whenever no single multiplier turns one equation's $x$ and $y$ coefficients into the other's. So find the value that makes them match, and rule it out. In the system above, every value of $k$ except $-2$ gives exactly one solution.

Check your understanding:

For which values of $k$ does the system $2x+ky=7$ and $6x+9y=12$ have exactly one solution?

Common mistake:

This one trips up a lot of students: the $x$ and $y$ terms match, so they answer “infinitely many” right away. Matching terms only give the same slope. The lines could still be one line or two parallel ones, so multiply the constant by the same number before you decide.

Example: Make the lines overlap

Worked example

The system

\begin{aligned} 4x-3y&=8\\[1.4em] 12x-9y&=k \end{aligned}

has infinitely many solutions, where $k$ is a constant. What is the value of $k$ ?

A
$8$
B
$16$
C
$24$
D
$32$

Solution

The $x$ and $y$ terms in the second equation are $3$ times those in the first:

3(4x-3y)=12x-9y.

For the two equations to draw the same line, the constant has to be multiplied by $3$ too:

k=3(8)=24.

Now the second equation is the first one tripled, so the lines lie on top of each other and share every point. The answer is C.

Practice problems

Your turn. Do the setup and the coefficient work yourself, and use the graphing calculator beside each problem to check it. And answer exactly what the question asks.

Build a quiz-score system

Practice problem

A quiz has $40$ questions, and a student answered every question. Each correct answer earns $4$ points, and each incorrect answer subtracts $1$ point. The student earned $115$ points. Let $c$ be the number of correct answers and $i$ be the number of incorrect answers. Which system represents the situation?

Calculator loads as you approach

Once you have picked a system, solve it here and see whether it gives 40 questions and 115 points.

Pick a line that crosses once

Practice problem

One equation in a system of two linear equations is

3x+2y=12.

Which choice could be the second equation if the system has exactly one solution?

Calculator loads as you approach

Comparing coefficients is quickest. Graph a choice here if it helps you see its one crossing.

Find k for no solution

Practice problem

In the system

\begin{aligned} 3(x+2y)-2y&=12\\[1.4em] 9x+(2k+4)y&=40, \end{aligned}

$k$ is a constant. If the system has no solution, what is the value of $k$ ?

Your answer

Calculator loads as you approach

Find k exactly first, then check your value here.

Find the hidden multiplier

Practice problem

The system

\begin{aligned} (p+2)x-4y&=3p+1\\[1.4em] (3p+6)x+(p-14)y&=9p+3 \end{aligned}

has infinitely many solutions, where $p$ is a constant. What is the value of $p$ ?

Your answer

Calculator loads as you approach

After you find the exact multiplier, graph both equations here to see them overlap.

Finish the lesson

4 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Two unknowns and two separate facts call for a system: one equation per fact.
Name both variables, and check that every term in an equation measures the same thing.
An intersection sits on both lines, so it makes both equations true.
Two lines meet once, never, or everywhere.
If no single multiplier links the $x$ and $y$ coefficients, the lines cross once. If one does, check the constant: a match means the same line, and a mismatch means parallel lines with no solution.
Build and read the model by hand. Use Desmos when a graph makes the picture clearer or the numbers are messy.

Related lessons

Not quite this kind of question? Try these:

Both equations are given and you only need $(x,y)$ or a value like $x+y$ ? Solve linear systems strategically covers elimination, substitution and Desmos for that.
Only one equation, with two variables? See Read and write two-variable linear equations.
One equation with one unknown, and a question about how many values work? See Count solutions and determine constants.
One of the equations isn't linear? See Solve linear and nonlinear systems.

For more practice typing equations into Desmos and reading where they meet, try Solve systems at intersections.

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