3.1.1: Graphing Linear Equations with Two Variables
By the end of this section, you will be able to:
- Plot points in a rectangular coordinate system
- Graph a linear equation by plotting points
- Graph vertical and horizontal lines
- Find the \(x\)- and \(y\)-intercepts
- Graph a line using the intercepts
Before we get started, take this readiness quiz.
- Evaluate \(5x−4\) when \(x=−1\).
- Evaluate \(3x−2y\) when \(x=4\) and \(y=−3\).
- Solve for \(y\): \(8−3y=20\).
In the previous unit we treated equations with one variable which we hoped to solve. Sometimes there were no solutions, and sometimes there were one or two. In this unit we will treat equations with two variables (often \(x\) and \(y\) but sometimes named other things for convenience). We will deal with equations that are familiar in the sense that if we replace either variable with a number, we are left with an equation that we have discussed previously.
A simple example of an equation with two variables is \(3x+2y=6\). For particular values of \(x\) and \(y\), this statement may be either true or false. For example if \(x=1\) and \(y=2\) this equation says \(3\cdot 1+2\cdot 2=6\) which is false, but if \(x=2\) and \(y=0\) then the equation says \(3\cdot 2+2\cdot 0=6\) which is true. In this case, we say that \(x=2\) and \(y=0\) is a solution. It turns out that unlike the previous unit, our equations have solutions that require two numbers to specify, and there are often infinitely many solutions. We will examine in the context of the first section how to represent the solutions to such an equation.
Linear Expressions and Linear Equations with One Variable
The expressions in the Be Prepared section are examples of what we call linear expressions .
1. An expression that can be written as
\(A x+B\)
with \(A\) and \(B\) real numbers, \(A\neq 0\), is called a linear expression (with one variable) , or more specifically, a linear expression in \(x\).
2. An equation that can be written as
\(A x+B=0\)
with \(A\) and \(B\) real numbers, \(A\neq 0\), is called a linear equation (with one variable) , or more specifically, a linear equation in \(x\).
3. A solution to a linear equation with one variable, say \(x\), is a number, say \(a\), that when substituted in for that variable yields a true statement. In this case we say that \(x=a\) is a solution.
Be Prepared (3) can be written as \(0=3y+12\) by adding \(3y\) and subtracting \(8\) from both sides. So \(8-3y=20\) is a linear equation in \(y\). Notice that substituting \(y=1\), for example, into the equation gives
\(\begin{aligned}8-3(1)&=20 \quad \text{ or }\\\quad 5&=20,\end{aligned}\)
which is not true. However, substituting \(y=-4\) into the equation gives
\(\begin{aligned}8-3(-4)&=20 \quad \text{ or }\\\quad 20&=20,\end{aligned}\)
which is true. So \(y=-4\) is a solution to \(8-3y=20\), and \(y=1\) is not. In solving Be Prepared 3, we found that \(y=-4\) is the only solution.
Introduction to Linear Expressions and Linear Equations with Two Variables
In this section we will be looking at equations that have two variables, \(x\) and \(y\), and more than one solution. We want to represent the solutions in a picture. Consider the equation \(2x-3y=6\). The expression \(2x-3y\) is an example of a linear expression with two variables. We can evaluate the expressions on both sides of the equal sign for any particular choice of \(x\) and \(y\). For example, we can choose \(x=3\) and \(y=0\) and substitute them into the equation to get
\(\begin{aligned}2(3) -3(0) &=6\quad \text{ or }\\\quad 6&=6,\end{aligned}\)
which is a true statement. We can also choose \(x=-3\) and \(y=-4\) and substitute them into the equation to get
\(\begin{aligned}2(-3) -3(-4) &=6\quad \text{ or }\\\quad 6&=6,\end{aligned}\)
which is also true. We say that \(x=3\) and \(y=0\) is one solution, and \(x=-3\) and \(y=-4\) is another solution. We will actually see that equations like \(2x-3y=6\) have infinitely many solutions. Next we introduce what is needed to make a picture of the solutions.
Plotting Points on a Rectangular Coordinate System
Just like maps use a grid system to identify locations, a grid system, or a rectangular coordinate system, is used in algebra to represent ordered pairs of numbers, and ultimately, to show a relationship between two variables. The rectangular coordinate system is also called the \(xy\) -plane or the “ coordinate plane .”
The rectangular coordinate system is formed by two intersecting number lines, one horizontal and one vertical. The horizontal number line is called the \(x\) -axis . The vertical number line is called the \(y\) -axis (note that in the context of an application these may take on different names). These axes divide a plane into four regions, called quadrants . The quadrants are identified by Roman numerals, beginning on the upper right and proceeding counterclockwise.
An ordered pair , \((x,y)\), gives the coordinates of a point in a rectangular coordinate system. The first number is the \(x\) -coordinate . The second number is the \(y\) -coordinate .
The phrase “ordered pair” means that the order is important. For example, \((2,5)\) and \((5,2)\) are different points.
What is the ordered pair of the point where the axes cross? At that point both coordinates are zero, so its ordered pair is \((0,0)\). The point \((0,0)\) has a special name.
The point \((0,0)\) is called the origin . It is the point where the \(x\)-axis and \(y\)-axis intersect.
We use the coordinates to locate a point on the \(xy\)-plane. Let’s plot the point \((1,3)\) as an example. First, locate the \(x\)-coordinate \(1\) on the \(x\)-axis and lightly sketch a vertical line through \(x=1\). Then, locate the \(y\)-coordinate \(3\) on the \(y\)-axis and sketch a horizontal line through \(y=3\). Now, find the point where these two lines meet -- that is the point with coordinates \((1,3)\).
Notice that the vertical line through \(x=1\) and the horizontal line through \(y=3\) are not part of the graph. We just used them to help us locate the point \((1,3)\).
When one of the coordinates is zero, the point lies on one of the axes. The graph below shows that the point \((0,4)\) is on the \(y\)-axis and the point \((−2,0)\) is on the \(x\)-axis.
- Points with a \(y\)-coordinate equal to \(0\) are on the \(x\)-axis, and have the form \((p,0)\), where \(p\) is some real number.
- Points with an \(x\)-coordinate equal to \(0\) are on the \(y\)-axis, and have the form \((0,q)\), where \(q\) is some real number.
Plot each point in the rectangular coordinate system and identify the quadrant in which the point is located:
a. \((−5,4\))
b. \((−3,−4)\)
c. \((2,−3)\)
d. \((0,−1)\)
e. \(\left(3,\dfrac{5}{2}\right)\)
f. \((-2,3)\)
- Solution
-
The first number of the coordinate pair is the \(x\)-coordinate, and the second number is the \(y\)-coordinate. To plot each point, sketch a vertical line through the \(x\)-coordinate and a horizontal line through the \(y\)-coordinate. Their intersection is the point.
a. Since the \(x\)-coordinate is \(−5\), the point is to the left of the \(y\)-axis. Also, since the \(y\)-coordinate is \(4\), the point is above the \(x\)-axis. The point \((−5,4)\) is in Quadrant II.
b. Since the \(x\)-coordinate is \(−3\), the point is to the left of the \(y\)-axis. Also, since the \(y\)-coordinate is \(−4\), the point is below the \(x\)-axis. The point \((−3,−4)\) is in Quadrant III.
c. Since the \(x\)-coordinate is \(2\), the point is to the right of the \(y\)-axis. Since the \(y\)-coordinate is \(−3\), the point is below the \(x\)-axis. The point \((2,−3)\) is in Quadrant IV.
d. Since the \(x\)-coordinate is \(0\), the point whose coordinates are \((0,−1)\) is on the \(y\)-axis.
e. Since the \(x\)-coordinate is \(3\), the point is to the right of the \(y\)-axis. Since the \(y\)-coordinate is \(\dfrac{5}{2}\), the point is above the \(x\)-axis. (It may be helpful to write \(\dfrac{5}{2}\) as a mixed number, \(2\dfrac{1}{2}\), or decimal, \(2.5\), so that we know \(\dfrac{5}{2}\) is between \(2\) and \(3\).) The point \(\left(3,\dfrac{5}{2}\right)\) is in Quadrant I.
f. Since the \(x\)-coordinate is \(-2\), the point is to the left of the \(y\)-axis. Since the \(y\)-coordinate is \(3\), the point is above the \(x\)-axis. The point \((-2,3)\) is in Quadrant II.
Plot each point in a rectangular coordinate system and identify the quadrant in which the point is located:
a. \((−2,1)\)
b. \((−3,−1)\)
c. \((4,−4)\)
d. \((−4,4)\)
e. \(\left(−4,\dfrac{3}{2}\right)\)
- Answer
-
The points \((−2,1)\), \((−4,4)\), and \(\left(−4,\dfrac{3}{2}\right)\) are in Quadrant II.
The point \((−3,−1)\) is in Quadrant III.
The point \((4,−4)\) is in Quadrant IV.
The signs of the \(x\)-coordinate and \(y\)-coordinate affect the location of the points. We may have noticed some patterns as we graphed the points in the previous example. We can summarize sign patterns of the quadrants in this way:
| Quadrant I | Quadrant II | Quadrant III | Quadrant IV |
|---|---|---|---|
| \((x,y)\) | \((x,y)\) | \((x,y)\) | \((x,y)\) |
| \((+,+)\) | \((−,+)\) | \((−,−)\) | \((+,−)\) |
Linear Expressions and Linear Equations with Two Variables
Up to now, all the equations we have solved were equations with just one variable. In almost every case, when we solved the equation we got exactly one solution. But equations can have more than one variable. Equations with two variables may be of the form \(Ax+By=C\). An equation of this form is called a linear equation with two variables .
1. An expression that can be written as
\(A x+B y\)
with \(A\) and \(B\) real numbers, not both zero, is called a linear expression (with two variables) , or more specifically, a linear expression in \(x\) and \(y\) .
2. An equation that can be written as
\(A x+By=C\)
with \(A\) and \(B\) real numbers, not both zero, is called a linear equation (with two variables) , or more specifically, a linear equation in \(x\) and \(y\) .
Here is an example of a linear equation with two variables, \(x\) and \(y\).
\(x+4y=8\)
This equation is in the form \(Ax+By=C\) with \(A=1\), \(B=4\), and \(C=8\). The equation
\(y=−3x+5\)
is also a linear equation with two variables, \(x\) and \(y\), but it does not appear to be in the form \(Ax+By=C\). We can rewrite it in \(Ax+By=C\) form in the following way.
| \(y= -3x+5\) | |
|---|---|
| Add \(3x\) to both sides. | \(y+3x= -3x+5+3x\) |
| Simplify. | \(y+3x= 5\) |
| Use the Commutative Property to put it in \(Ax+By=C\) form. | \(3x+y= 5\) |
By rewriting \(y=−3x+5\) as \(3x+y=5\), we can easily see that it is a linear equation with two variables because it is of the form \(Ax+By=C\) with \(A=3\), \(B=1\), and \(C=5\). When an equation is in the form \(Ax+By=C\), we say it is in standard form of a linear equation .
A linear equation with two variables, \(x\) and \(y\), is in standard form when it is written as \(Ax+By=C\).
Most people prefer to have \(A\), \(B\), and \(C\) be integers and \(A \geq 0\) when writing a linear equation in standard form, although it is not strictly necessary.
Linear equations with two variables have infinitely many solutions. For example, if \(A \neq 0\), for every number that is substituted for \(x\) there is a corresponding \(y\)-value. This pair of values is a solution to the linear equation and is represented by the ordered pair \((x,y)\). When we substitute these values of \(x\) and \(y\) into the equation, the result is a true statement, because the value on the left side is equal to the value on the right side.
An ordered pair \((p,q)\) is a solution of the linear equation \(Ax+By=C\), if the equation is a true statement when the \(x\)- and \(y\)-coordinates of the ordered pair, \(p\) and \(q\), respectively, are substituted into the equation. We can also say in this case that \((x,y)=(p,q)\) is a solution, or \(x=p\) and \(y=q\) is a solution.
We can represent the solutions as points in the rectangular coordinate system. The points will line up perfectly in a straight line. We use a straight-edge to draw this line, and put arrows on the ends of each side of the line to indicate that the line continues in both directions.
A graph is a visual representation of all the solutions of the equation. It is an example of the saying, “A picture is worth a thousand words.” The line (with the arrows) shows us all the solutions to that equation. Every point on the line corresponds a solution of the equation. And, every solution of this equation corresponds to a point on this line. This line is called the graph of the equation. Points not on the line do not correspond to solutions! We may say, as is common, then that the points on the line are the solutions.
The graph of the linear equation \(Ax+By=C\) is the collection of all solutions \((x,y)\).
We can represent the graph on the coordinate plane. The representation is a straight line so that
- every solution of the equation is a point on this line, and
- every point on the line is a solution of the equation.
As universally accepted, a representation is also called a graph of the linear equation .
The graph of \(y=2x−3\) is shown below.
For each ordered pair,
\(A(0,−3),\qquad B(3,3), \qquad C(2,−3),\qquad\text{ and } \qquad D(−1,−5),\)
decide:
a. is the ordered pair a solution to the equation?
b. is the point on the line?
- Solution
-
Substitute the \(x\)- and \(y\)-values into the equation to check if the ordered pair is a solution to the equation.
a.Point \(A(0,−3)\) \(B(3,3)\) \(C(2,−3)\) \(D(−1,−5)\) Write the equation of the line. \(y=3x-3\) \(y=2x-3\) \(y=2x-3\) \(y=2x-3\)
Substitute the \(x\)- and \(y\)-values. \(-3\stackrel{?}{=} 2\cdot 0 -3\) \(3\stackrel{?}{=} 2\cdot 3 -3\) \(-3\stackrel{?}{=} 2\cdot 2 -3\) \(-5\stackrel{?}{=} 2\cdot (-1) -3\) Simplify. \(-3\stackrel{?}{=}-3\) \(3\stackrel{?}{=}3\) \(-3\stackrel{?}{=}1\) \(-5\stackrel{?}{=}5\) True or false? True True False True Answer the question. \((0,-3)\) is a solution \((3,3)\) is a solution \((2,-3)\) is not a solution \((-1,-5)\) is a solution b. Plot the points \((0,−3)\), \((3,3)\), \((2,−3)\), and \((−1,−5)\).
The points \((0,3)\), \((3,−3)\), and \((−1,−5)\) are on the line \(y=2x−3\), and the point \((2,−3)\) is not on the line.
The points that are solutions to \(y=2x−3\) are on the line, but the points that are not solutions are not on the line.
The graph of \(y=3x−1\) is shown below.
For each ordered pair,
\(A(0,−1)\qquad\text{ and } \qquad B(2,5),\)
decide:
a.
is the ordered pair a solution to the equation?
b.
is the point on the line?
- Answer
-
a. Both pairs are solutions.
b. Both pairs are on the line.
The graph of \(y=3x−1\) is shown below.
For each ordered pair,
\(A(3,−1)\qquad\text{ and } \qquad B(−1,−4),\)
decide:
a.
Is the ordered pair a solution to the equation?
b.
Is the point on the line?
- Answer
-
a. \(A\) is a solution.
\(B\) is not a solution.
b. \(A\) is on the line.
\(B\) is not on the line.
Graph a Linear Equation by Plotting Points
There are several methods that can be used to graph a linear equation. The first method we will use is called plotting points . We find three points whose coordinates are solutions to the equation and then plot them in a rectangular coordinate system. By connecting these points in a line, we have the graph of the linear equation. While two points are enough to determine a line, using three points helps us detect errors.
Graph the equation \(y=2x+1\) by plotting points.
- Solution
-
Graph the equation \(y=2x−3\) by plotting points.
- Answer
-
Graph the equation \(y=−2x+4\) by plotting points.
- Answer
-
The steps to take when graphing a linear equation by plotting points are summarized here.
- Find three points whose coordinates are solutions to the equation. Organize them in a table.
- Plot the points in a rectangular coordinate system. Check that the points line up. If they do not, carefully check your work.
- Draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line.
It is true that it only takes two points to determine a line, but it is a good habit to use three points. If we only plot two points and one of them is incorrect, we can still draw a line but it will not represent the solutions to the equation. It will be the wrong line.
If we use three points, and one is incorrect, the points will not line up. This tells us something is wrong and we need to check our work. Look at the difference between these illustrations.
When an equation includes a fraction as the coefficient of \(x\), we can still substitute any numbers for \(x\). But the arithmetic is easier if we make “good” choices for the values of \(x\). This way we will avoid fractional answers, which are hard to plot precisely.
Graph the equation \(y=\dfrac{1}{2}x+3\).
- Solution
-
Find three points that are solutions to the equation. Since this equation has the fraction \(\dfrac{1}{2}\) as a coefficient of \(x\), we will choose values of \(x\) carefully. We will use zero as one choice and multiples of 2 for the other choices. Why are multiples of two a good choice for values of \(x\)? By choosing multiples of 2 the multiplication by \(\dfrac{1}{2}\) simplifies to a whole number.
Choose a value for \(x\) that is a multiple of \(2\). \(x=0\) \(x=2\) \(x=4\) Write the equation of the line. \(y=\dfrac{1}{2}x+3\) \(y=\dfrac{1}{2}x+3\) \(y=\dfrac{1}{2}x+3\) Substitute the \(x\)-value into the equation. \(y=\dfrac{1}{2}(0)+3\) \(y=\dfrac{1}{2}(2)+3\) \(y=\dfrac{1}{2}(4)+3\) Simplify. \(y=0+3\) \(y=1+3\) \(y=2+3\) Find \(y\). \(y=3\) \(y=4\) \(y=5\) We organize the three solutions in a table.
\(y=\dfrac{1}{2}x+3\) \(x\) \(y\) \((x,y)\) \(0\) \(3\) \((0,3)\) \(2\) \(4\) \((2,4)\) \(4\) \(5\) \((4,5)\) Plot the points, check that they line up, and draw the line.
Graph the equation \(y=\dfrac{1}{3}x−1\).
- Answer
-
Graph the equation \(y=\dfrac{1}{4}x+2\).
- Answer
-
Graph Vertical and Horizontal Lines
Some linear equations have only one variable. They may have just \(x\) and no \(y\), or just \(y\) without an \(x\). This changes how we make a table of values to get the points to plot.
Let’s consider the equation \(x=−3\). This equation has only one variable, \(x\). The equation says that the \(x\)-coordinate of any solution is equal to \(−3\), so its value does not depend on the \(y\)-coordinate. No matter what is the value of the \(y\)-coordinate, the value of the \(x\)-coordinate is always \(−3\).
So to make a table of values, write \(−3\) in for all the \(x\)-coordinates. Then choose any values for the \(y\)-coordinate. Since the \(x\)-coordinate does not depend on the \(y\)-coordinate, we can choose any numbers we like. But to fit the points on our coordinate graph, we’ll use 1, 2, and 3 for the \(y\)-coordinates.
| \(x=−3\) | ||
|---|---|---|
| \(x\) | \(y\) | \((x,y)\) |
| \(−3\) | \(1\) | \((−3,1)\) |
| \(−3\) | \(2\) | \((−3,2)\) |
| \(−3\) | \(3\) | \((−3,3)\) |
Plot the points from the table and connect them with a straight line. Notice that we have graphed a vertical line .
What if the equation has \(y\) but no \(x\)? Let’s graph the equation \(y=4\). This time the \(y\)-coordinate of any solution is 4, so for this equation, the \(y\)-coordinate of the solution does not depend on the \(x\)-coordinate. Fill in 4 for all the \(y\)-coordinates in the table and then choose any values for \(x\)-coordinates. We will use 0, 2, and 4 for the \(x\)-coordinates.
| \(y=4\) | ||
|---|---|---|
| \(x\) | \(y\) | \((x,y)\) |
| \(0\) | \(4\) | \((0,4)\) |
| \(2\) | \(4\) | \((2,4)\) |
| \(4\) | \(4\) | \((4,4)\) |
In this figure, we have graphed a horizontal line passing through the \(y\)-axis at 4.
1. A vertical line is the graph of an equation (with two variables \(x\) and \(y\)) of the form \(x=a\).
The line passes through the \(x\)-axis at \((a,0)\).
2. A horizontal line is the graph of an equation (with two variables \(x\) and \(y\)) of the form \(y=b\).
The line passes through the \(y\)-axis at \((0,b)\).
Graph:
a. \(x=2\)
b. \(y=−1\)
- Solution
-
a. The equation has only one variable, \(x\), and \(x\) is always equal to 2. We create a table where the \(x\)-coordinate is always 2 and then put in any values for the \(y\)-coordinate. The graph is a vertical line passing through the \(x\)-axis at 2.
\(x=2\) \(x\) \(y\) \((x,y)\) \(2\) \(1\) \((2,1)\) \(2\) \(2\) \((2,2)\) \(2\) \(3\) \((2,3)\) b. Similarly, the equation \(y=−1\) has only one variable, \(y\). The value of the \(y\)-coordinate of any solution is \(-1\). All the ordered pairs in the next table have the same \(y\)-coordinate. The graph is a horizontal line passing through the \(y\)-axis at \(−1\).
\(y=−1\) \(x\) \(y\) \((x,y)\) \(0\) \(−1\) \((0,−1)\) \(3\) \(−1\) \((3,−1)\) \(−3\) \(−1\) \((−3,−1)\)
G raph:
a. \(x=5\)
b. \(y=−4\)
- Answer
-
a.
b.
G raph:
a. \(x=−2\)
b. \(y=3\)
- Answer
-
a.
b.
What is the difference between the equations \(y=4x\) and \(y=4\)?
The equation \(y=4x\) has both \(x\) and \(y\). The value of the \(y\)-coordinate of a solution depends on the value of the \(x\)-coordinate, so the \(y\)-coordinate changes according to the value of the \(x\)-coordinate. The equation \(y=4\) has only one variable. The value of \(y\)-coordinate of any solution is \(4\), it does not depend on the value of the \(x\)-coordinate.
Notice, in the graph, the equation \(y=4x\) gives a slanted line, while \(y=4\) gives a horizontal line.
Graph \(y=−3x\) and \(y=−3\) in the same rectangular coordinate system.
- Solution
-
We notice that the first equation has the variable \(x\), while the second does not. We make a table of points for each equation and then graph the lines. The two graphs are shown.
Graph \(y=−4x\) and \(y=−4\) in the same rectangular coordinate system.
- Answer
-
Graph \(y=3\) and \(y=3x\) in the same rectangular coordinate system.
- Answer
-
Find \(x\)- and \(y\)-intercepts
Every linear equation can be represented by a line. We have seen that when graphing a line by plotting points, we can use any three solutions to graph. This means that two people graphing the line might use different sets of three points.
At first glance, their two lines might not appear to be the same, since they would have different points labeled. But if all the work was done correctly, the lines should be exactly the same. One way to recognize that they are indeed the same line is to look at where the line intersects the \(x\)-axis and the \(y\)-axis. These points are called the intercepts of a line .
The points where a graph crosses the \(x\)-axis and the \(y\)-axis are called the intercepts of the graph .
Let’s look at the graphs of the lines.
First, notice where each of these lines crosses the \(x\)-axis. Now, let’s look at the points where these lines cross the \(y\)-axis.
| Figure |
The line crosses
the \(x\) -axis at |
Ordered pair
for this point |
The line crosses
the \(y\) - axis at |
Ordered pair
for this point |
|---|---|---|---|---|
| Figure (a) | \(3\) | \((3,0)\) | \(6\) | \((0,6)\) |
| Figure (b) | \(4\) | \((4,0)\) | \(−3\) | \((0,−3)\) |
| Figure (c) | \(5\) | \((5,0)\) | \(−5\) | \((0,-5)\) |
| Figure (d) | \(0\) | \((0,0)\) | \(0\) | \((0,0)\) |
| General Figure | \(a\) | \((a,0)\) | \(b\) | \((0,b)\) |
Is there a pattern?
For each line, the \(y\)-coordinate of the point where the line crosses the \(x\)-axis is zero. The point where the line crosses the \(x\)-axis has the form \((a,0)\) and is called the \(x\) -intercept of the line. The \(x\)-intercept occurs when \(y\) is zero.
In each line, the \(x\)-coordinate of the point where the line crosses the \(y\)-axis is zero. The point where the line crosses the \(y\)-axis has the form \((0,b)\) and is called the \(y\) -intercept of the line. The \(y\)-intercept occurs when \(x\) is zero.
- The \(x\) -intercept of a line is the point \((a,0)\) where the line crosses the \(x\)-axis.
- The \(y\) -intercept of a line is the point \((0,b)\) where the line crosses the \(y\)-axis.
Find the \(x\)- and \(y\)-intercepts on each graph shown.
- Solution
-
a. The graph crosses the \(x\)-axis at the point \((4,0)\). The \(x\)-intercept is \((4,0)\).
The graph crosses the \(y\)-axis at the point \((0,2)\). The \(y\)-intercept is \((0,2)\).
b. The graph crosses the \(x\)-axis at the point \((2,0)\). The \(x\)-intercept is \((2,0)\).
The graph crosses the \(y\)-axis at the point \((0,−6)\). The \(y\)-intercept is \((0,−6)\).
c. The graph crosses the \(x\)-axis at the point \((−5,0)\). The \(x\)-intercept is \((−5,0)\).
The graph crosses the \(y\)-axis at the point \((0,−5)\). The \(y\)-intercept is \((0,−5)\).
Find the \(x\)- and \(y\)-intercepts on the graph.
- Answer
-
The \(x\)-intercept is \((2,0)\).
The \(y\)-intercept is \((0,−2)\).
Find the \(x\)- and \(y\)-intercepts on the graph.
- Answer
-
The \(x\)-intercept is \((3,0)\).
The \(y\)-intercept is \((0,2)\).
Recognizing that the \(x\)- intercept occurs when \(y\) is zero and that the \(y\)- intercept occurs when \(x\) is zero gives us a method to find the intercepts of a line from its equation. To find the \(x\)- intercept, let \(y=0\) and solve for \(x\) . To find the \(y\)- intercept, let \(x=0\) and solve for \(y\) .
To find:
- the \(x\)-intercept of the line, let \(y=0\) and solve for \(x\).
- the \(y\)-intercept of the line, let \(x=0\) and solve for \(y\).
Find the intercepts of the graph of \(2x+y=8\).
- Solution
-
We will let \(y=0\) to find the \(x\)-intercept, and let \(x=0\) to find the \(y\)-intercept. We will fill in a table, which reminds us of what we need to find.
Let's find the \(x\)-intercept first.
\(2x+y=8\) To find the \(x\)-intercept, let \(y=0\). Let \(y=0\). \(2x+0=8\) Simplify. \(2x=8\) \(x=4\) Write the \(x\)-intercept as a point \((x,y)\). \((4,0)\) Now, the \(y\)-intercept.
\(2x+y=8\) To find the \(y\)-intercept, let \(x=0\). Let \(x=0\). \(2\cdot 0+y=8\) Simplify. \(y=8\) Write the \(y\)-intercept as a point \((x,y)\). \((0,8)\) The intercepts are the points \((4,0)\) and \((0,8)\) as shown in the table.
\(2x+y=8\) \(x\) \(y\) \((x,y)\) 4 0 \((4,0)\) 0 8 \((0,8)\) The \(x\)-intercept is \((4,0)\).
The \(y\)-intercept is \((0,8)\).
Find the intercepts of the graph of \(3x+y=12\).
- Answer
-
The \(x\)-intercept is \((4,0)\).
The \(y\)-intercept is \((0,12)\).
Find the intercepts of the graph of \(x+4y=8\).
- Answer
-
The \(x\)-intercept is \((8,0)\).
The \(y\)-intercept is \((0,2)\).
Graph a Line Using the Intercepts
To graph a linear equation by plotting points, we need to find three points whose coordinates are solutions to the equation. We can use the \(x\)- and \(y\)- intercepts as two of our three points. Find the intercepts, and then find a third point to ensure accuracy. Make sure the points line up—then draw the line. This method is often the quickest way to graph a line.
Graph \(–x+2y=6\) using the intercepts.
- Solution
-
Graph \(x-2y=4\) using the intercepts.
- Answer
-
Graph \(–x+3y=6\) using the intercepts.
- Answer
-
When the line passes through the origin, the \(x\)-intercept and the \(y\)-intercept are the same point.
Graph \(y=5x\) using the intercepts.
- Solution
-
\(y=5x\) \(x\)-intercept \(y\)-intercept Let \(y=0\). Let \(x=0\). \(0=5x\) \(y=5\cdot 0\) \(0=x\) \(y=0\) \((0,0)\) \((0,0)\) This line has only one intercept. It is the point \((0,0)\).
To check accuracy, we need to plot three points. Since the \(x\)- and \(y\)-intercepts are the same point, we need two more points to graph the line.\(y=5x\) Let \(x=1\). Let \(x=-1\). \(y=5\cdot 1\) \(y=5\cdot (-1)\) \(y=5\) \(y=-5\) The resulting three points are summarized in the table.
\(y=5x\) \(x\) \(y\) \((x,y)\) 0 0 \((0,0)\) 1 5 \((1,5)\) \(−1\) \(−5\) \((−1,−5)\) Plot the three points, check that they line up, and draw the line.
Graph \(y=4x\) using the intercepts.
- Answer
-
Graph \(y=−x\) using the intercepts.
- Answer
-
- Explain how you would choose three x -values to make a table to graph the line \(y=\dfrac{1}{5}x−2\).
- What is the difference between the equations of a vertical and a horizontal line?
- Do you prefer to use the method of plotting points or the method using the intercepts to graph the equation \(4x+y=−4\)? Why?
- Do you prefer to use the method of plotting points or the method using the intercepts to graph the equation \(y=\dfrac{2}{3}x−2\)? Why?
Key Concepts
-
Points on the Axes
- Points with a \(y\)-coordinate equal to \(0\) are on the \(x\)-axis, and have coordinates \((a,0)\).
- Points with an \(x\)-coordinate equal to \(0\) are on the \(y\)-axis, and have coordinates \((0,b)\).
-
Quadrant
Quadrant I Quadrant II Quadrant III Quadrant IV \((x,y)\) \((x,y)\) \((x,y)\) \((x,y)\) \((+,+)\) \((-,+)\) \((-,-)\) \((+,-)\) -
Graph of a Linear Equation:
The graph of a linear equation \(Ax+By=C\) is a straight line.
Every point on the line is a solution of the equation.
Every solution of this equation is a point on this line. -
How to graph a linear equation by plotting points.
- Find three points whose coordinates are solutions to the equation. Organize them in a table.
- Plot the points in a rectangular coordinate system. Check that the points line up. If they do not, carefully check your work.
- Draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line.
-
\(x\)-intercept and \(y\)-intercept of a Line
- The \(x\)-intercept is the point \((a,0)\) where the line crosses the \(x\)-axis.
- The \(y\)-intercept is the point \((0,b)\) where the line crosses the \(y\)-axis.
-
Find the \(x\)- and \(y\)-intercepts from the Equation of a Line
-
Use the equation of the line. To find:
the \(x\)-intercept of the line, let \(y=0\) and solve for \(x\).
the \(y\)-intercept of the line, let \(x=0\) and solve for \(y\).
-
Use the equation of the line. To find:
-
How to graph a linear equation using the intercepts.
-
Find the \(x\)- and \(y\)-intercepts of the line.
Let \(y=0\) and solve for \(x\) .
Let \(x=0\) and solve for \(y\). - Find a third solution to the equation.
- Plot the three points and check that they line up.
- Draw the line
-
Find the \(x\)- and \(y\)-intercepts of the line.