Solutions of a System and the Intersection Point | Grade 8 Math | LevelKite

You already know how to graph a line and how to check a solution of an equation. Now you will work with two equations at once.

In this lesson, $x$ and $y$ stand for real numbers. An ordered pair $(x, y)$ gives an $x$ -value first and a $y$ -value second. Many ordered pairs make the equation $y = 2x + 1$ true, such as $(0, 1)$ , $(1, 3)$ and $(2, 5)$ . Many ordered pairs make $y = -x + 4$ true as well, such as $(0, 4)$ , $(1, 3)$ and $(4, 0)$ . One pair, $(1, 3)$ , is in both lists.

A system of equations is a set of two or more equations that use the same variables and are considered together. In this lesson, each system has two linear equations in $x$ and $y$ , which are equations whose graphs are straight lines. A solution of the system is an ordered pair that makes every equation in the system true at the same time.

Worked example

Testing an ordered pair in both equations

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

Read the ordered pair. The first number is the x-value and the second number is the y-value.
$x = 1, \quad y = 3$
Substitute both values into the first original equation, y = 2x + 1, so that it reads 3 = 2(1) + 1. Compute the right side to see whether that statement is true. The right side is 3, the same as the left side, so the first equation is true.
$2(1) + 1 = 2 + 1 = 3$
A solution of a system must make every equation true, so substitute the same values into the second original equation, y = -x + 4. It reads 3 = -(1) + 4. The right side is 3, the same as the left side, so the second equation is true.
$-(1) + 4 = -1 + 4 = 3$
Compare the two checks. The same ordered pair turned both equations into the true statement 3 = 3.

Answer

Yes. The ordered pair $(1, 3)$ makes both equations true, so it is a solution of the system.

Now look at the same system as a picture. The graph of an equation in $x$ and $y$ is the set of all points whose coordinates make the equation true.

In the figure below, the horizontal axis shows $x$ and the vertical axis shows $y$ . Each grid step is 1 unit on both axes. The figure shows only the part of the plane where $x$ and $y$ run from $-5$ to $7$ , and the lines continue past its edges. Line A is the graph of $y = 2x + 1$ and Line B is the graph of $y = -x + 4$ .

Point C is at $(1, 3)$ , the pair you just tested. It lies on both lines, which matches the two true checks. Point D is at $(2, 5)$ . It lies on Line A but not on Line B.

Line A is

y = 2x + 1

and Line B is

y = -x + 4

. Point C is at

(1, 3)

, on both lines. Point D is at

(2, 5)

, on Line A only.

Key concept

Why an intersection point solves both equations

For a system of two linear equations in $x$ and $y$ graphed on one coordinate plane:

Each line contains exactly the points whose coordinates make its equation true.
So a point is on both lines exactly when its coordinates make both equations true.

A point of intersection is a point where the lines cross, so it is on both lines. That is why its coordinates are a solution of the system. A point on only one line is not a solution.

A picture can only suggest where lines cross, because a reading from a graph may be approximate. Use the exact coordinates and confirm them by substituting into both original equations.

Now practice giving that explanation yourself with a new system: $y = x + 3$ and $y = -2x - 3$ .

In the figure below, the horizontal axis shows $x$ and the vertical axis shows $y$ , and each grid step is 1 unit on both axes. In this figure, Line A is the graph of $y = x + 3$ and Line B is the graph of $y = -2x - 3$ . The lines cross at Point C, at $(-2, 1)$ .

Two more points are marked where $x = 0$ . Point D is at $(0, 3)$ on Line A, and Point E is at $(0, -3)$ on Line B. They have the same $x$ -value but different $y$ -values, so they are two different points.

Line A is

y = x + 3

and Line B is

y = -2x - 3

. Point C is at

(-2, 1)

, Point D is at

(0, 3)

and Point E is at

(0, -3)

.

Worked example

Explaining why an intersection point works

In the figure above, Line A is the graph of $y = x + 3$ and Line B is the graph of $y = -2x - 3$ . The lines cross at Point C, at $(-2, 1)$ . Explain why $(-2, 1)$ solves both equations.

Point C is on Line A. Line A contains exactly the points that make y = x + 3 true, so the coordinates of Point C must make it true. Confirm it by substituting x = -2 into the right side. The result is 1, which is the y-coordinate of Point C.
$-2 + 3 = 1$
Point C is also on Line B, which contains exactly the points that make y = -2x - 3 true. Confirm it by substituting x = -2. A negative number times a negative number is positive, so -2 times -2 is 4. The result is again 1.
$-2(-2) - 3 = 4 - 3 = 1$
Compare a point on only one line. Point D, at (0, 3), is on Line A because 0 + 3 = 3. On Line B, x = 0 gives y = -3, which is Point E instead. So (0, 3) makes only the first equation true.
$-2(0) - 3 = 0 - 3 = -3$
Put the two facts about Point C together. It is a single point on both lines, so its one ordered pair makes both equations true.

Answer

The pair $(-2, 1)$ solves both equations because Point C lies on both lines, and each line contains exactly the points that make its own equation true. So $(-2, 1)$ is a solution of the system.

Now use both ideas on a system written in a different form: $2x + y = 4$ and $x - 2y = 7$ .

In the figure below, the horizontal axis shows $x$ and the vertical axis shows $y$ , and each grid step is 1 unit on both axes. In this figure, Line A is the graph of $2x + y = 4$ and Line B is the graph of $x - 2y = 7$ . Three points are marked: Point C, Point D and Point E.

The picture can suggest which marked point is on both lines. Substituting exact coordinates into the original equations proves it.

Line A is

2x + y = 4

and Line B is

x - 2y = 7

. Point C is at

(1, 2)

, Point D is at

(3, -2)

and Point E is at

(5, -1)

.

Worked example

Choosing the solution from three marked points

Use the figure above, where Line A is the graph of $2x + y = 4$ and Line B is the graph of $x - 2y = 7$ . Which marked point is a solution of the system: Point C at $(1, 2)$ , Point D at $(3, -2)$ or Point E at $(5, -1)$ ? Explain how the graph agrees with your answer.

Test Point C, at (1, 2), by computing the left side of each original equation. The first left side is 4, so 2x + y = 4 is true. The second left side is -3, not 7, so x - 2y = 7 is false. Point C fails one equation, so it is not a solution. It is on Line A only.
$2(1) + 2 = 2 + 2 = 4 \qquad 1 - 2(2) = 1 - 4 = -3$
Test Point E, at (5, -1), the same way. The first left side is 9, not 4, so the first equation is false. In the second, 2 times -1 is -2, and subtracting -2 is the same as adding 2. That left side is 7, so the second equation is true. Point E is on Line B only, so it is not a solution.
$2(5) + (-1) = 10 - 1 = 9 \qquad 5 - 2(-1) = 5 + 2 = 7$
Test Point D, at (3, -2). In the second equation, 2 times -2 is -4, and subtracting -4 is the same as adding 4. The left sides are 4 and 7, which equal the right sides, so both equations are true.
$2(3) + (-2) = 6 - 2 = 4 \qquad 3 - 2(-2) = 3 + 4 = 7$
Connect the checks to the graph. Of the three marked points, only Point D makes both equations true, so only Point D is on both lines. It is the point where Line A and Line B cross. The picture suggested this, and the exact substitutions prove it.

Answer

Point D, at $(3, -2)$ , is a solution of the system because it makes both $2x + y = 4$ and $x - 2y = 7$ true. On the graph, it is the point where Line A and Line B intersect.

Here is the idea to keep. A solution of a system is an ordered pair that makes every equation in the system true. On a graph, each line contains exactly the points that make its equation true. So a point where the two lines intersect makes both equations true, and a point on only one line does not.

To be sure about a solution, substitute its exact coordinates into both original equations. In this lesson, each pair of lines crossed at one point. Next, you will graph systems yourself to solve them, find how many solutions a system has, and find and correct graphing errors.