Math Core

One linear equation in two variables has infinitely many solutions: every point on its line. But what if two things have to be true at the same time? Then you need a point that sits on both lines, and a graph shows you exactly where that is.

What is a system?

Two or more equations that use the same variables form a system of equations. You usually write them stacked together:

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

Definition

Solution of a system

A solution of a system of two linear equations is an ordered pair $(x, y)$ that makes both equations true. On a graph, it is a point where the two lines intersect.

Checking a solution

You don't need a graph to test a point. Substitute it into each equation. It has to work in both.

Worked example: Is it a solution?

Is $(1, 4)$ a solution of the system $y = 3x + 1$ and $x + y = 5$ ?

First equation: $3(1) + 1 = 4$ , and $y = 4$ . ✓
Second equation: $1 + 4 = 5$ . ✓

The point works in both equations, so $(1, 4)$ is a solution.

What about $(2, 3)$ ? In the second equation, $2 + 3 = 5$ ✓. But in the first, $3(2) + 1 = 7$ , not $3$ . ✗ One miss is enough: $(2, 3)$ is not a solution.

Solving by graphing

To solve a system by graphing:

Graph the first line. Slope-intercept form $y = mx + b$ makes this quick: start at $b$ on the $y$ -axis and use the slope $m$ .
Graph the second line on the same grid.
Find the point where the lines cross. Read its coordinates.
Check the point in both equations.

Worked example: Reading the intersection

Solve the system by graphing.

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

The first line starts at $(0, 1)$ and rises $1$ for every $1$ to the right. The second starts at $(0, 5)$ and falls $1$ for every $1$ to the right.

The lines y = x + 1 and y = -x + 5 cross at (2, 3).Open in grapher →

The lines cross at $(2, 3)$ .

Check: $3 = 2 + 1$ ✓ and $3 = -2 + 5$ ✓. The solution is $(2, 3)$ .

Worked example: A steeper line

Solve the system by graphing.

\begin{aligned} y &= 2x - 3 \\ y &= -x + 3 \end{aligned}

The first line starts at $(0, -3)$ and rises $2$ for every $1$ to the right. The second starts at $(0, 3)$ and falls $1$ for every $1$ to the right.

The lines y = 2x - 3 and y = -x + 3 cross at (2, 1).Open in grapher →

The lines cross at $(2, 1)$ .

Check: $2(2) - 3 = 1$ ✓ and $-2 + 3 = 1$ ✓. The solution is $(2, 1)$ .

Tip

If an equation isn't in slope-intercept form, solve it for $y$ first. For example, $x + y = 6$ becomes $y = -x + 6$ .

No solution or infinitely many

Two lines don't always cross at exactly one point.

Parallel lines have the same slope but different $y$ -intercepts. They never meet, so the system has no solution.
The same line written two ways has the same slope and the same $y$ -intercept. Every point on it works, so the system has infinitely many solutions.

Worked example: Parallel lines

How many solutions does the system $y = 2x + 1$ and $y = 2x - 3$ have?

Both lines have slope 2. They are parallel and never cross.Open in grapher →

Both lines have slope $2$ , but their $y$ -intercepts are $1$ and $-3$ . The lines are parallel, so there is no solution.

Now try $y = 3x - 2$ and $6x - 2y = 4$ . Solve the second for $y$ : $-2y = -6x + 4$ , so $y = 3x - 2$ . It's the same line, so the system has infinitely many solutions.

Compare slopes and intercepts

Different slopes: the lines cross once, so there is exactly one solution.
Same slope, different $y$ -intercepts: parallel lines, no solution.
Same slope, same $y$ -intercept: the same line, infinitely many solutions.

Common mistake

A graph can fool you when the crossing point isn't on a grid corner. If a point looks like $(2, 3)$ , always check it in both equations. If it fails, the real answer is probably a fraction, and you'll need an algebra method from the next lessons.

Practice

Practice 1

Which point is a solution of the system $y = x + 2$ and $y = -2x + 8$ ?

Practice 2

The graph shows the system $y = x - 1$ and $y = -x + 3$ . What is the solution?

y = x - 1y = -x + 3Open in grapher →

Enter a point like (2, -3)

Practice 3

Solve the system by graphing: $x + y = 9$ and $y = 2x$ .

Enter a point like (2, -3)

Practice 4

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

Enter a point like (2, -3)

Practice 5

How many solutions does the system $y = 4x - 1$ and $y = 4x + 3$ have?

Practice 6

Solve the system by graphing: $y = 3x + 2$ and $y = x - 2$ .

Enter a point like (2, -3)

Practice 7

How many solutions does the system $y = -x + 2$ and $2x + 2y = 4$ have?

Practice 8

Solve the system by graphing: $y = \dfrac{1}{2}x + 1$ and $y = -x + 7$ .

Enter a point like (2, -3)