maths.free › Algebra › Graphs › Graphing Linear Equations

Graphing Linear Equations

Recognize the relation between the solutions of an equation and its graph

Recognize the Relation Between the Solutions of an Equation and its Graph

In Use the Rectangular Coordinate System, we found a few solutions to the equation $3x+2y=6$. They are listed in the table below. So, the ordered pairs $(0,3)$, $(2,0)$, $(1,\frac{3}{2})$, $(4,-3)$, are some solutions to the equation$3x+2y=6$. We can plot these solutions in the rectangular coordinate system as shown on the graph at right.

Notice how the points line up perfectly? We connect the points with a straight line to get the graph of the equation $3x+2y=6$. Notice the arrows on the ends of each side of the line. These arrows indicate the line continues.

Every point on the line is a solution of the equation. Also, every solution of this equation is a point on this line. Points not on the line are not solutions!

Notice that the point whose coordinates are $(-2,6)$ is on the line If you substitute $x=-2$ and $y=6$ into the equation, you find that it is a solution to the equation.

Example

Try it.

The graph of $y=2x-3$ is shown below.

For each ordered pair decide

ⓐ Is the ordered pair a solution to the equation?
ⓑ Is the point on the line?

(a) $(0,-3)$
(b) $(3,3)$
(c) $(2,-3)$
(d) $(-1,-5)$

Solution

Substitute the $x$- and $y$-values into the equation to check if the ordered pair is a solution to the equation.

ⓐ

ⓑ Plot the points A: $(0,-3)$ B: $(3,3)$ C: $(2,-3)$ and D: $(-1,-5)$.
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 which are solutions to $y=2x-3$ are on the line, but the point which is not a solution is not on the line.

Condensed: the full section is in OpenStax Prealgebra 2e.

Graph a Linear Equation by Plotting Points

There are several methods that can be used to graph a linear equation. The method we used at the start of this section to graph is called plotting points, or the Point-Plotting Method.

Let’s graph the equation $y=2x+1$ by plotting points.

We start by finding three points that are solutions to the equation. We can choose any value for $x$ or $y,$ and then solve for the other variable.

Since $y$ is isolated on the left side of the equation, it is easier to choose values for $x.$ We will use $0,1,$ and $-2$ for $x$ for this example. We substitute each value of $x$ into the equation and solve for $y.$

We can organize the solutions in a table.

$y=2x+1$
$x$	$y$	$(x,y)$
$0$	$1$	$(0,1)$
$1$	$3$	$(1,3)$
$-2$	$-3$	$(-2,-3)$

Now we plot the points on a rectangular coordinate system. Check that the points line up. If they did not line up, it would mean we made a mistake and should double-check all our work.

Draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line. The line is the graph of $y=2x+1.$

Example

Try it.

Graph the equation $y=-3x.$

Solution

Find three points that are solutions to the equation. It’s easier to choose values for $x,$ and solve for $y.$ Do you see why?

List the points in a table.

$y=-3x$
$x$	$y$	$(x,y)$
$0$	$0$	$(0,0)$
$1$	$3$	$(1,-3)$
$-2$	$6$	$(-2,6)$

Plot the points, check that they line up, and draw the line as shown.

Example

Try it.

Graph the equation $y=\frac{1}{2}x+3.$

Solution

Find three points that are solutions to the equation. Since this equation has the fraction $\frac{1}{2}$ 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.

The points are shown in the table.

$y=\frac{1}{2}x+3$
$x$	$y$	$(x,y)$
$0$	$3$	$(0,3)$
$2$	$4$	$(2,4)$
$4$	$5$	$(4,5)$

Plot the points, check that they line up, and draw the line as shown.

\[2x+y=3\]\[y=-2x+3\]

$y=-2x+3$
$x$	$y$	$(x,y)$
$0$	$3$	$(0,3)$
$1$	$1$	$(1,1)$
$-1$	$5$	$(-1,5)$

Condensed: the full section is in OpenStax Prealgebra 2e.

Graph Vertical and Horizontal Lines

Can we graph an equation with only one variable? Just $x$ and no $y,$ or just $y$ without an $x?$ How will we make a table of values to get the points to plot?

Let’s consider the equation $x=-3.$ The equation says that $x$ is always equal to $-3,$ so its value does not depend on $y.$ No matter what $y$ is, the value of $x$ is always $-3.$

To make a table of solutions, we write $-3$ for all the $x$ values. Then choose any values for $y.$ Since $x$ does not depend on $y,$ you can choose any numbers you like. But to fit the size of our coordinate graph, we’ll use $1,2,$ and $3$ for the $y$-coordinates as shown in the table.

$x=-3$
$x$	$y$	$(x,y)$
$-3$	$1$	$(-3,1)$
$-3$	$2$	$(-3,2)$
$-3$	$3$	$(-3,3)$

Then plot the points and connect them with a straight line. Notice in that the graph is a vertical line.

Example

Try it.

Graph the equation $x=2.$ What type of line does it form?

Solution

The equation has only variable, $x,$ and $x$ is always equal to $2.$ We make a table where $x$ is always $2$ and we put in any values for $y.$

$x=2$
$x$	$y$	$(x,y)$
$2$	$1$	$(2,1)$
$2$	$2$	$(2,2)$
$2$	$3$	$(2,3)$

Plot the points and connect them as shown.

The graph is a vertical line passing through the $x$-axis at $2.$

What if the equation has $y$ but no $x$? Let’s graph the equation $y=4.$ This time the $y$-value is a constant, so in this equation $y$ does not depend on $x.$

To make a table of solutions, write $4$ for all the $y$ values and then choose any values for $x.$

We’ll use $0,2,$ and $4$ for the $x$-values.

$y=4$
$x$	$y$	$(x,y)$
$0$	$4$	$(0,4)$
$2$	$4$	$(2,4)$
$4$	$4$	$(4,4)$

Example

Try it.

Graph the equation $y=-1.$

Solution

The equation $y=-1$ has only variable, $y.$ The value of $y$ is constant. All the ordered pairs in the table have the same $y$-coordinate, $-1$. We choose $0,3,$ and $-3$ as values for $x.$

$y=-1$
$x$	$y$	$(x,y)$
$-3$	$-1$	$(-3,-1)$
$0$	$-1$	$(0,-1)$
$3$	$-1$	$(3,-1)$

The graph is a horizontal line passing through the $y$-axis at $-1$ as shown.

Example

Try it.

Graph $y=-3x$ and $y=-3$ in the same rectangular coordinate system.

Solution

Find three solutions for each equation. Notice that the first equation has the variable $x,$ while the second does not. Solutions for both equations are listed.

The graph shows both equations.

Condensed: the full section is in OpenStax Prealgebra 2e.

Key Concepts

Graph a linear equation by plotting points.
1. Find three points whose coordinates are solutions to the equation. Organize them in a table.
2. Plot the points on a rectangular coordinate system. Check that the points line up. If they do not, carefully check your work.
3. Draw the line through the points. Extend the line to fill the grid and put arrows on both ends of the line.
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.

Graphing Linear Equations

Recognize the Relation Between the Solutions of an Equation and its Graph

In each of the following exercises, an equation and its graph is shown. For each ordered pair, decide

ⓐ is the ordered pair a solution to the equation?
ⓑ is the point on the line?

Try it.

$y=x+2$

$(0,2)$
$(1,2)$
$(-1,1)$
$(-3,1)$

Solution

ⓐ yes ⓑ yes
ⓐ no ⓑ no
ⓐ yes ⓑ yes
ⓐ no ⓑ no

Try it.

$y=x-4$

$(0,-4)$
$(3,-1)$
$(2,2)$
$(1,-5)$

Try it.

$y=\frac{1}{2}x-3$

$(0,-3)$
$(2,-2)$
$(-2,-4)$
$(4,1)$

Solution

ⓐ yes ⓑ yes
ⓐ yes ⓑ yes
ⓐ yes ⓑ yes
ⓐ no ⓑ no

Try it.

$y=\frac{1}{3}x+2$

$(0,2)$
$(3,3)$
$(-3,2)$
$(-6,0)$

Graph a Linear Equation by Plotting Points

In the following exercises, graph by plotting points.

Try it.

$y=\frac{1}{2}x+2$

Solution

Try it.

$y=\frac{1}{3}x-1$

Try it.

$y=\frac{4}{3}x-5$

Solution

Try it.

$y=\frac{3}{2}x-3$

Try it.

$y=-\frac{2}{5}x+1$

Solution

Try it.

$y=-\frac{4}{5}x-1$

Try it.

$y=-\frac{3}{2}x+2$

Solution

Try it.

$y=-\frac{5}{3}x+4$

Try it.

$2x+3y=12$

Solution

Try it.

$3x-4y=12$

Try it.

$\frac{1}{3}x+y=2$

Solution

Try it.

$\frac{1}{2}x+y=3$

Graph Vertical and Horizontal lines

In the following exercises, graph the vertical and horizontal lines.

Try it.

$x=\frac{7}{3}$

Solution

Try it.

$x=\frac{5}{4}$

In the following exercises, graph each pair of equations in the same rectangular coordinate system.

Try it.

$y=-\frac{1}{2}x$ and $y=-\frac{1}{2}$

Solution

Try it.

$y=-\frac{1}{3}x$ and $y=-\frac{1}{3}$

Try it.

$y=2x$ and $y=2$

Solution

Try it.

$y=5x$ and $y=5$

Mixed Practice

In the following exercises, graph each equation.

Try it.

$y=4x$

Solution

Try it.

$y=2x$

Try it.

$y=-\frac{1}{2}x+3$

Solution

Try it.

$y=\frac{1}{4}x-2$

Try it.

$y=-x$

Solution

Try it.

$y=x$

Try it.

$x-y=3$

Solution

Try it.

$x+y=-5$

Try it.

$4x+y=2$

Solution

Try it.

$2x+y=6$

Try it.

$y=5$

Try it.

$2x+6y=12$

Solution

Try it.

$5x+2y=10$

Try it.

$x=3$

Solution

Try it.

Motor home cost The Robinsons rented a motor home for one week to go on vacation. It cost them $\text{\$594}$ plus $\text{\$0.32}$ per mile to rent the motor home, so the linear equation $y=594+0.32x$ gives the cost, $y,$ for driving $x$ miles. Calculate the rental cost for driving $400,800,\ \text{and}\ 1,200$ miles, and then graph the line.

Solution

$722, $850, $978

Try it.

Weekly earning At the art gallery where he works, Salvador gets paid $\text{\$200}$ per week plus $\text{15\%}$ of the sales he makes, so the equation $y=200+0.15x$ gives the amount $y$ he earns for selling $x$ dollars of artwork. Calculate the amount Salvador earns for selling $\text{\$900, \$1,600},\ \text{and}\ \text{\$2,000},$ and then graph the line.

Condensed: the full section is in OpenStax Prealgebra 2e.

Teď ty. Žádný kalkulačka neurovná tento, ale jeho části jsou vypočítavé. Zkuste jeden níže, nebo zadejte svůj vlastní.

Praxe (40)

Zkuste každý z nich na papíře první. Odhalte odpověď na kontrolu; ověřené lze otevřít v řešiteli pro každý krok.

Evaluate: $3x+2$ when $x=-1.$

Odhalte odpověď

$-1$
Solve the formula: $5x+2y=20$ for $y.$

Odhalte odpověď

$y=\frac{20-5x}{2}$
Simplify: $\frac{3}{8}(-24)\text{.}$

Odhalte odpověď

$-9$
The graph of $y=2x-3$ is shown below.
For each ordered pair decide
1. ⓐ Is the ordered pair a solution to the equation?
2. ⓑ Is the point on the line?
1. (a) $(0,-3)$
2. (b) $(3,3)$
3. (c) $(2,-3)$
4. (d) $(-1,-5)$
Odhalte odpověď

Substitute the $x$- and $y$-values into the equation to check if the ordered pair is a solution to the equation.
ⓐ
ⓑ Plot the points A: $(0,-3)$ B: $(3,3)$ C: $(2,-3)$ and D: $(-1,-5)$.
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 which are solutions to $y=2x-3$ are on the line, but the point which is not a solution is not on the line.
The graph of $y=3x-1$ is shown.
For each ordered pair, decide
1. ⓐ is the ordered pair a solution to the equation?
2. ⓑ is the point on the line?
1. $(0,-1)$
2. $(2,2)$
3. $(3,-1)$
4. $(-1,-4)$
Odhalte odpověď
1. ⓐ yes ⓑ yes
2. ⓐ no ⓑ no
3. ⓐ no ⓑ no
4. ⓐ yes ⓑ yes
Graph the equation $y=-3x.$

Odhalte odpověď

Find three points that are solutions to the equation. It’s easier to choose values for $x,$ and solve for $y.$ Do you see why?
List the points in a table.
$y=-3x$
$x$ $y$ $(x,y)$
$0$ $0$ $(0,0)$
$1$ $3$ $(1,-3)$
$-2$ $6$ $(-2,6)$
Plot the points, check that they line up, and draw the line as shown.
Graph the equation by plotting points: $y=-4x.$
Graph the equation by plotting points: $y=x.$
Graph the equation $y=\frac{1}{2}x+3.$

Odhalte odpověď

Find three points that are solutions to the equation. Since this equation has the fraction $\frac{1}{2}$ 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.
The points are shown in the table.
$y=\frac{1}{2}x+3$
$x$ $y$ $(x,y)$
$0$ $3$ $(0,3)$
$2$ $4$ $(2,4)$
$4$ $5$ $(4,5)$
Plot the points, check that they line up, and draw the line as shown.
Graph the equation: $y=\frac{1}{3}x-1.$
Graph the equation: $y=\frac{1}{4}x+2.$
Graph the equation $x+y=5.$

Odhalte odpověď

Find three points that are solutions to the equation. Remember, you can start with any value of $x$ or $y.$
We list the points in a table.
$x+y=5$
$x$ $y$ $(x,y)$
$0$ $5$ $(0,5)$
$1$ $4$ $(1,4)$
$4$ $1$ $(4,1)$
Then plot the points, check that they line up, and draw the line.
Graph the equation: $x+y=-2.$
Graph the equation: $x-y=6.$
Graph the equation $3x+y=-1.$

Odhalte odpověď

Find three points that are solutions to the equation.
First, solve the equation for $y.$
$\begin{array}{lll}3x+y & = & -1 \\ y & = & -3x-1\end{array}$
We’ll let $x$ be $0,1,$ and $-1$ to find three points. The ordered pairs are shown in the table. Plot the points, check that they line up, and draw the line.
$y=-3x-1$
$x$ $y$ $(x,y)$
$0$ $-1$ $(0,-1)$
$1$ $-4$ $(1,-4)$
$-1$ $2$ $(-1,2)$
If you can choose any three points to graph a line, how will you know if your graph matches the one shown in the answers in the book? If the points where the graphs cross the $x\text{-}$ and $y$-axes are the same, the graphs match.
Graph each equation: $2x+y=2.$
Graph each equation: $4x+y=-3.$
Graph the equation $x=2.$ What type of line does it form?

Odhalte odpověď

The equation has only variable, $x,$ and $x$ is always equal to $2.$ We make a table where $x$ is always $2$ and we put in any values for $y.$
$x=2$
$x$ $y$ $(x,y)$
$2$ $1$ $(2,1)$
$2$ $2$ $(2,2)$
$2$ $3$ $(2,3)$
Plot the points and connect them as shown.
The graph is a vertical line passing through the $x$-axis at $2.$
Graph the equation: $x=5.$
Graph the equation: $x=-2.$
Graph the equation $y=-1.$

Odhalte odpověď

The equation $y=-1$ has only variable, $y.$ The value of $y$ is constant. All the ordered pairs in the table have the same $y$-coordinate, $-1$. We choose $0,3,$ and $-3$ as values for $x.$
$y=-1$
$x$ $y$ $(x,y)$
$-3$ $-1$ $(-3,-1)$
$0$ $-1$ $(0,-1)$
$3$ $-1$ $(3,-1)$
The graph is a horizontal line passing through the $y$-axis at $-1$ as shown.
Graph the equation: $y=-4.$
Graph the equation: $y=3.$
Graph $y=-3x$ and $y=-3$ in the same rectangular coordinate system.

Odhalte odpověď

Find three solutions for each equation. Notice that the first equation has the variable $x,$ while the second does not. Solutions for both equations are listed.
The graph shows both equations.
Graph the equations in the same rectangular coordinate system: $y=-4x$ and $y=-4.$
Graph the equations in the same rectangular coordinate system: $y=3$ and $y=3x.$
$y=x+2$
1. $(0,2)$
2. $(1,2)$
3. $(-1,1)$
4. $(-3,1)$
Odhalte odpověď
1. ⓐ yes ⓑ yes
2. ⓐ no ⓑ no
3. ⓐ yes ⓑ yes
4. ⓐ no ⓑ no
$y=x-4$
1. $(0,-4)$
2. $(3,-1)$
3. $(2,2)$
4. $(1,-5)$
$y=\frac{1}{2}x-3$
1. $(0,-3)$
2. $(2,-2)$
3. $(-2,-4)$
4. $(4,1)$
Odhalte odpověď
1. ⓐ yes ⓑ yes
2. ⓐ yes ⓑ yes
3. ⓐ yes ⓑ yes
4. ⓐ no ⓑ no
$y=\frac{1}{3}x+2$
1. $(0,2)$
2. $(3,3)$
3. $(-3,2)$
4. $(-6,0)$
$y=\frac{1}{2}x+2$
$y=\frac{1}{3}x-1$
$y=\frac{4}{3}x-5$
$y=\frac{3}{2}x-3$
$y=-\frac{2}{5}x+1$
$y=-\frac{4}{5}x-1$
$y=-\frac{3}{2}x+2$
$y=-\frac{5}{3}x+4$
$2x+3y=12$
$3x-4y=12$

Udržujte si vlastní práci

Volný účet přidává poznámky k každé lekci, záznam o tom, co jste dokončili, vaše vyřešené problémy na jednom místě, a doučování můžete požádat o tuto stránku. Matematika sama je otevřena pro každého, zaregistrované nebo ne.

Zaregistrovat se Přihlášení

Zde použité symboly

Ťukněte na jakýkoliv symbol pro úplnou definici, obrázek a co každé písmeno v něm znamená.

Jak se: Graphing Linear Equations

Recognize the relation between the solutions of an equation and its graph
Graph a linear equation by plotting points
Graph vertical and horizontal lines
Every point on the line is a solution of the equation.
Every solution of this equation is a point on this line.
(a)
(b)
(c)

Otázky, které se lidé ptají

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Části této stránky jsou přizpůsobeny z OpenStax Prealgebra 2e (CC BY-NC-SA 4.0). Zkondenzovaný a znovu-vysvětlený zde; chyby jsou naše.

Více v Algebra

Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value

\(y=2x+1\)
\(x\)	\(y\)	\((x,y)\)
\(0\)	\(1\)	\((0,1)\)
\(1\)	\(3\)	\((1,3)\)
\(-2\)	\(-3\)	\((-2,-3)\)

\(y=-3x\)
\(x\)	\(y\)	\((x,y)\)
\(0\)	\(0\)	\((0,0)\)
\(1\)	\(3\)	\((1,-3)\)
\(-2\)	\(6\)	\((-2,6)\)

\(y=\frac{1}{2}x+3\)
\(x\)	\(y\)	\((x,y)\)
\(0\)	\(3\)	\((0,3)\)
\(2\)	\(4\)	\((2,4)\)
\(4\)	\(5\)	\((4,5)\)

\(y=-2x+3\)
\(x\)	\(y\)	\((x,y)\)
\(0\)	\(3\)	\((0,3)\)
\(1\)	\(1\)	\((1,1)\)
\(-1\)	\(5\)	\((-1,5)\)

\(x=-3\)
\(x\)	\(y\)	\((x,y)\)
\(-3\)	\(1\)	\((-3,1)\)
\(-3\)	\(2\)	\((-3,2)\)
\(-3\)	\(3\)	\((-3,3)\)

\(x=2\)
\(x\)	\(y\)	\((x,y)\)
\(2\)	\(1\)	\((2,1)\)
\(2\)	\(2\)	\((2,2)\)
\(2\)	\(3\)	\((2,3)\)

\(y=4\)
\(x\)	\(y\)	\((x,y)\)
\(0\)	\(4\)	\((0,4)\)
\(2\)	\(4\)	\((2,4)\)
\(4\)	\(4\)	\((4,4)\)

\(y=-1\)
\(x\)	\(y\)	\((x,y)\)
\(-3\)	\(-1\)	\((-3,-1)\)
\(0\)	\(-1\)	\((0,-1)\)
\(3\)	\(-1\)	\((3,-1)\)

\(x+y=5\)
\(x\)	\(y\)	\((x,y)\)
\(0\)	\(5\)	\((0,5)\)
\(1\)	\(4\)	\((1,4)\)
\(4\)	\(1\)	\((4,1)\)

\(y=-3x-1\)
\(x\)	\(y\)	\((x,y)\)
\(0\)	\(-1\)	\((0,-1)\)
\(1\)	\(-4\)	\((1,-4)\)
\(-1\)	\(2\)	\((-1,2)\)