Systems of Equations: Solve by Graphing

A system of equations is two or more equations that use the same variables. Its solution is the point that makes every equation true at the same time. When you graph a system of two linear equations, the solution is the point where the two lines cross.

Before this lesson: You should know how to graph a line in slope-intercept form and what a solution to a two-variable linear equation is.

Extra practice: The Algebra 1 Unit 5 Practice Pack has printable problems, answer keys and a cheat sheet for this unit.

What Is a System of Equations?

One linear equation with x and y has many solutions. Every point on its line works. A system asks a harder question: which point works for both equations at once?

Here is a system of two equations. The brace shows that they go together.

\begin{cases} y = x + 1 \\ y = -x + 5 \end{cases}

The point (2, 3) is the solution. It is on both lines, so it makes both equations true.

\begin{aligned} 3 &= 2 + 1 && \checkmark \\ 3 &= -2 + 5 && \checkmark \end{aligned}

The lines y = x + 1 and y = −x + 5 cross at the point (2, 3), which is the solution of the system. — The lines y = x + 1 and y = −x + 5 cross at (2, 3), the solution of the system.

How to Check a Solution

Put the x-value and y-value into the first equation.
Put the same values into the second equation.
If both equations are true, the point is a solution. If either one is false, it is not.

How to Solve a System by Graphing

Write each equation in slope-intercept form, $y = mx + b$ .
Graph both lines on the same grid.
Find the point where the lines cross.
Check that point in both equations.

Worked Examples

Example 1: Check a solution that works

Is (3, −1) a solution of this system?

\begin{cases} x + y = 2 \\ 2x - y = 7 \end{cases}

3 + (-1) = 2

Step 1. Substitute $x = 3$ and $y = -1$ into the first equation, $x + y = 2$ .

Step 2. The left side equals 2, so the first equation is true.

2(3) - (-1) = 6 + 1 = 7

Step 3. Substitute into the second equation, $2x - y = 7$ . Multiply first, then subtract a negative by adding (see adding and subtracting integers).

Step 4. The left side equals 7, so the second equation is true.

Answer: Yes. (3, −1) makes both equations true, so it is the solution.

Example 2: Check a point that doesn’t work

Is (1, 4) a solution of $y = 2x + 2$ and $y = -x + 3$ ?

2(1) + 2 = 2 + 2 = 4

Step 1. Substitute $x = 1$ into the first equation and compare with $y = 4$ .

Step 2. The result is 4, which matches y. The first equation is true.

-1 + 3 = 2

Step 3. Substitute $x = 1$ into the second equation.

Step 4. The result is 2, not 4. The second equation is false.

Answer: No. The point is on the first line but not the second, so it is not a solution of the system.

Example 3: Solve by graphing

Solve the system $y = x + 1$ and $y = -x + 5$ by graphing.

Step 1. Graph $y = x + 1$ . Plot the y-intercept (0, 1). The slope is 1, so go up 1 and right 1 to (1, 2). Draw the line. (See slope-intercept form.)

Step 2. Graph $y = -x + 5$ . Plot (0, 5). The slope is $-1$ , so go down 1 and right 1 to (1, 4). Draw the line.

Step 3. Find where the lines cross. They meet at (2, 3), as shown in the graph above.

y = 2 + 1 = 3

Step 4. Check (2, 3) in the first equation.

y = -2 + 5 = 3

Step 5. Check (2, 3) in the second equation.

Answer: (2, 3). It works in both equations.

Example 4: Rewrite first, then graph

Solve the system $2x + y = 4$ and $x - y = -1$ by graphing.

y = -2x + 4

Step 1. Solve the first equation for y. Subtract $2x$ from both sides. (See literal equations.)

-y = -x - 1

Step 2. Solve the second equation for y. Subtract $x$ from both sides.

y = x + 1

Step 3. Divide both sides by $-1$ .

Step 4. Graph both lines. The first starts at (0, 4) with slope −2. The second starts at (0, 1) with slope 1.

The lines y = −2x + 4 and y = x + 1 cross at the point (1, 2). — The lines y = −2x + 4 and y = x + 1 cross at (1, 2).

Step 5. Find where the lines cross. They meet at (1, 2).

Answer: (1, 2)

Check: First equation: $2(1) + 2 = 2 + 2 = 4$ . Second equation: $1 - 2 = -1$ . Both are true.

Example 5: Word problem

Streaming Plan A costs $20 to join plus $5 per month. Plan B has no fee and costs $10 per month. After how many months do the plans cost the same?

\begin{cases} C = 5m + 20 & \text{Plan A} \\ C = 10m & \text{Plan B} \end{cases}

Step 1. Let $m$ be the number of months and $C$ the total cost. Write an equation for each plan.

Step 2. Graph Plan A. Start at (0, 20) and rise $5 per month.

Step 3. Graph Plan B. Start at (0, 0) and rise $10 per month.

Step 4. Find where the lines cross. They meet at (4, 40).

5(4) + 20 = 20 + 20 = 40

Step 5. Check Plan A at 4 months.

10(4) = 40

Step 6. Check Plan B at 4 months.

Answer: After 4 months, both plans cost $40. Before that Plan B is cheaper, and after that Plan A is cheaper.

When Graphing Isn’t Enough

Graphing shows you what a solution means, but it isn’t always exact. If the lines cross at a point like $\left(\frac{7}{3}, \frac{5}{2}\right)$ , you can only estimate it from a graph.

For exact answers, use algebra. The next two lessons show how: the Substitution Method and the Elimination Method.

Sometimes the lines never cross, or they are the same line. You’ll learn those cases in Number of Solutions.

Common Mistakes

Mistake	What it looks like	How to fix it
Checking only one equation	(1, 4) works in $y = 2x + 2$ , so you call it the solution	A solution must work in every equation in the system.
Switching x and y	Reading the crossing point as (3, 2) instead of (2, 3)	Write the x-value first: (x, y).
Graphing before solving for y	Graphing $2x + y = 4$ with slope 2	Solve for y first: $y = -2x + 4$ has slope −2.
Trusting a rough sketch	Guessing (2, 3) when the lines really cross at (2.2, 3.1)	Use a ruler and graph paper, and always check the point in both equations.

Practice Problems

For problems 1–3, decide if the point is a solution of the system.

(2, 5); $y = 3x - 1$ and $y = x + 3$
(−1, 2); $x + y = 1$ and $3x - y = -4$
(0, −3); $y = 4x - 3$ and $2x + y = -3$

For problems 4–8, solve the system by graphing. Check your answer.

$y = x - 2$ and $y = -x + 4$
$y = 2x + 1$ and $y = -x - 2$
$y = \frac{1}{2}x + 1$ and $y = -x + 4$
$x + y = 5$ and $x - y = 1$
$y = 3$ and $x = -2$
Ana has $10 and saves $5 each week. Ben has $30 and saves $3 each week. After how many weeks will they have the same amount? How much will each have?

Show answers

3(2) - 1 = 6 - 1 = 5

Substitute (2, 5) into the first equation.

2 + 3 = 5

That matches y = 5, so it is true. Substitute into the second equation.

That is true too
Yes, (2, 5) is a solution.

-1 + 2 = 1

Substitute (−1, 2) into the first equation.

3(-1) - 2 = -3 - 2 = -5

That is true. Substitute into the second equation.

The result is −5, not −4, so the second equation is false
No, (−1, 2) is not a solution.

4(0) - 3 = 0 - 3 = -3

Substitute (0, −3) into the first equation.

2(0) + (-3) = 0 + (-3) = -3

That matches y = −3. Substitute into the second equation.

That is true too
Yes, (0, −3) is a solution.

Graph y = x − 2: plot (0, −2), then go up 1 and right 1
Graph y = −x + 4: plot (0, 4), then go down 1 and right 1
The lines cross at (3, 1)
Check the first equation: $3 - 2 = 1$
Check the second equation: $-3 + 4 = 1$
The solution is (3, 1).

Graph y = 2x + 1: plot (0, 1), then go up 2 and right 1
Graph y = −x − 2: plot (0, −2), then go down 1 and right 1
The lines cross at (−1, −1)
Check the first equation: $2(-1) + 1 = -2 + 1 = -1$
Check the second equation: $-(-1) - 2 = 1 - 2 = -1$
The solution is (−1, −1).

Graph y = ½x + 1: plot (0, 1), then go up 1 and right 2
Graph y = −x + 4: plot (0, 4), then go down 1 and right 1
The lines cross at (2, 2)
Check the first equation: $\frac{1}{2}(2) + 1 = 1 + 1 = 2$
Check the second equation: $-2 + 4 = 2$
The solution is (2, 2).

y = -x + 5

Solve the first equation for y. Subtract x from both sides.

-y = -x + 1

Solve the second equation for y. Subtract x from both sides.

y = x - 1

Divide both sides by −1.

Graph both lines. They cross at (3, 2)
Check the first equation: $3 + 2 = 5$
Check the second equation: $3 - 2 = 1$
The solution is (3, 2).

Graph y = 3: a horizontal line through (0, 3)
Graph x = −2: a vertical line through (−2, 0)
The lines cross at (−2, 3). It has x = −2 and y = 3, so it works in both equations
The solution is (−2, 3).

A = 5w + 10

Let w be the number of weeks. Write Ana’s equation.

A = 3w + 30

Write Ben’s equation.

Graph both lines. Ana’s starts at (0, 10) and rises 5 per week. Ben’s starts at (0, 30) and rises 3 per week
The lines cross at (10, 60)
Check Ana: $5(10) + 10 = 50 + 10 = 60$
Check Ben: $3(10) + 30 = 30 + 30 = 60$
After 10 weeks, each has $60.

Want more practice? The Algebra 1 Unit 5 Practice Pack has 55 pages of problems on this and the other lessons in the unit, with step-by-step answers and a cheat sheet.

Systems of Equations FAQ

What is a system of equations?

A system of equations is a group of two or more equations with the same variables. You solve the equations together to find values that make all of them true.

What does the solution of a system mean?

The solution is the point that lies on every line in the system. On a graph, it is where the lines cross.

What are the three ways to solve a system of equations?

The three main methods are graphing, substitution and elimination. Graphing helps you see the answer, while substitution and elimination give exact answers.

Can a system of equations have no solution?

Yes. If the lines are parallel, they never cross, so there is no solution. See Number of Solutions.

Summary

A system of equations is two or more equations with the same variables.
The solution makes every equation true at the same time.
On a graph, the solution is where the lines cross.
To solve by graphing, write each equation as $y = mx + b$ , graph both lines and find where they cross.
Always check the point in every equation.

Previous lesson: Arithmetic Sequences

Next lesson: Substitution Method

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