**Introduction**

In this lesson, you will graph (or plot) a linear equation by finding two or more points on the line and then connecting those points with a line. To find points on the line, choose any value for *x*, then plug that number into the function to solve for the unknown variable *y*. A linear equation is simply a straight line.

These videos illustrate the lesson material below. Watching the videos is optional.

- Plotting a Line in Slope-Intercept Form (07:51 mins) | Transcript
- Graphing any Linear Equations (Straight Line) (05:18 mins) | Transcript

**Plotting a Line in Slope-Intercept Form**

Linear equations that are in slope-intercept form are written in this format:

\begin{align*} y = mx + b\end{align*}

**Example 1**

Plot a line using the following linear equation: \(y=\frac{1}{2}x+1\).

Figure 1 shows a table of coordinates to plot this line. The *x* values are on the left and the *y* values are on the right since that's how they will be ordered in coordinate pairs. Four values for *x* are chosen that will be easy to calculate \(x = -1, 0, 1, 5\). It is only necessary to do two points, but doing more will help you check for mistakes.

Figure 1

Solve for the *y* values. Begin by substituting -1 for \(x\).

\begin{align*} & y = \frac{1}{2}x + 1 &\color{red}\small\text{Start with the linear equation}\\\\ & y = \frac{1}{2} (-1) + 1 &\color{red}\small\text{Substitute \(-1\) for \(x\)}\\\\ & y = \frac{-1}{2} + 1 &\color{red}\small\text{Multiply \(\frac{1}{2}\) by \(-1\) is \(\frac{-1}{2}\)}\\\\ & y = \frac{-1}{2} + \frac{2}{2} &\color{red}\small\text{Common denominator is \(2\)}\\\\ & y = \frac{1}{2} &\color{red}\small\text{Simplify the fraction} \end{align*}

Now, substitute 0 for \(x\).

\begin{align*} & y = \frac{1}{2}x + 1 &\color{red}\small\text{Start with the linear equation}\\\\ & y = \frac{1}{2} (0) + 1 &\color{red}\small\text{Substitute \(0\) for \(x\)}\\\\ & y = 0 + 1 &\color{red}\small\text{Multiply \(\frac{1}{2}\) by \(0\) is \(0\)}\\\\ & y = 1 &\color{red}\small\text{Simplify by adding} \end{align*}

Now, substitute 1 for \(x\).

\begin{align*} & y = \frac{1}{2}x + 1 &\color{red}\small\text{Start with the linear equation}\\\\ & y = \frac{1}{2} (1) + 1 &\color{red}\small\text{Substitute \(1\) for \(x\)}\\\\ & y = \frac{1}{2} + 1 &\color{red}\small\text{Multiply \(\frac{1}{2}\) by \(1\) is \(\frac{1}{2}\)}\\\\ & y = \frac{1}{2} + \frac{2}{2} &\color{red}\small\text{Common denominator is \(2\)}\\\\ & y = \frac{3}{2} &\color{red}\small\text{Addition}\\\\& y = 1.5 &\color{red}\small\text{Express as a decimal} \end{align*}

Finally, substitute 5 for \(x\).

\begin{align*} & y = \frac{1}{2}x + 1 &\color{red}\small\text{Start with the linear equation}\\\\ & y = \frac{1}{2} (5) + 1 &\color{red}\small\text{Substitute \(5\) for \(x\)}\\\\ & y = \frac{5}{2} + 1 &\color{red}\small\text{Multiply \(\frac{1}{2}\) by \(5\) is \(\frac{5}{2}\)}\\\\ & y = \frac{5}{2} + \frac{2}{2} &\color{red}\small\text{Common denominator is \(2\)}\\\\ & y = \frac{7}{2} &\color{red}\small\text{Addition}\\\\ & y = 3.5 & \color{red}\small\text{Express as decimal} \end{align*}

With these points (-1, 0.5), (0, 1), (1, 1.5), and (5, 3.5), you can plot the line (graph the equation) by drawing a line that connects the points.

Figure 2

**Graphing Any Linear Equation (Straight Line)**

Not all equations come in slope-intercept form, but you can still graph them. You can graph an equation in standard form, where all of the variables are on the same side of the equation.

Standard form is written in this format:

\begin{align*} Ax + By = c \end{align*}

Similar to the first part of this lesson, you will find two or more points to plot, then connect them to form the line.

**Example 2**

Plot a line using this linear equation in standard form: \(2x+3y=10\).

Start by making a table. The *x* value is on the left, and the *y* value is on the right. Next, find values that will be easy to calculate. In this example, you will see coordinates calculated when *x* is 0 and when *y* is 0.

Figure 3

Solve for the *y* value. Substitute 0 for \(x\).

\begin{align*} & 2x+3y=10 &\color{red}\small\text{Start with the linear equation}\\\\ & 2(0)+3y=10 &\color{red}\small\text{Substitute \(0\) for \(x\)}\\\\ & 0+3y=10 &\color{red}\small\text{Multiply \(2\) by \(0\) is \(0\)}\\\\ & 3y = 10 &\color{red}\small\text{Simplify the left side}\\\\ & \frac{3}{\color{red} 3} y = \frac{10}{\color{red}3} &\color{red}\small\text{Divide both side by \(3\)}\\\\ & \frac{\cancel 3}{\color{red}\cancel 3} y = \frac{10}{\color{red}3} &\color{red}\small\text{Cancel the \(3\)'s on the left side}\\\\ & y = \frac{10}{3} &\color{red}\small\text{The value of y is isolated}\\\\ & y = 3.33 &\color{red}\small\text{Express as a decimal} \end{align*}

Now solve for the *x* value when *y* is 0.

\begin{align*} & 2x+3y=10 &\color{red}\small\text{Start with the linear equation}\\\\ & 2x+3(0)=10 &\color{red}\small\text{Substitute \(0\) for \(y\)}\\\\ & 2x+0=10 &\color{red}\small\text{Multiply \(3\) by \(0\) is \(0\)}\\\\ & 2x = 10 &\color{red}\small\text{Simplify the left side}\\\\ & \frac{2}{\color{red} 2} x = \frac{10}{\color{red}2} &\color{red}\small\text{Divide both side by \(2\)}\\\\ & \frac{\cancel 2}{\color{red}\cancel 2} x = \frac{10}{\color{red}2} &\color{red}\small\text{Cancel the \(2 \)'s on the left side}\\\\ & x = \frac{10}{2} &\color{red}\small\text{The value of x is isolated }\\\\ & x = 5 &\color{red}\small\text{Simplify the right side} \end{align*}

Figure 4 shows these values added to the table.

Figure 4

With these points (0, 3.33), (5, 0), you can plot the line (graph the equation) by drawing a line that connects the points.

Figure 5

This means that any point along this line will make this equation correct. For example, it looks like the point (2,2) where *x* is 2 and *y* is 2 is on this line. To find out if this is correct, plug the point (2,2) into the equation for the value of x and y, and check if you get 10.

\begin{align*} &2x + 3y = 10 &\color{red}\small\text{Start with the linear equation}\\\\ &2 (2) + 3 (2) = 10 &\color{red}\small\text{Check if \((2, 2)\) is true}\\\\ &4 + 6 = 10 &\color{red}\small\text{Multiply the left side}\\\\ & 10 = 10 &\color{red}\small\text{Add the left side}\end{align*}

Since 10=10 is a true statement, the point (2, 2) does lie on the line above. Any other point in this line will also satisfy this equation and make it true.

Figure 6

**Things to Remember**

- To find points on the line, choose any value for
*x*, then plug that number in the function and solve for*y*. - Pick values for
*x*that are easy to calculate for the value of y. - Plot points and draw a line through the points.

### Practice Problems

**1. Match the equation of the line with its graph.**

a. \({\text{y}}=4{\text{x}}-3\) (

b. \({\text{y}}=-2{\text{x}}+3\) (

c. \({\text{y}}= \dfrac{1}{3}{\text{x}}\) (

3.

**2. Create a graph of the line represented by the equation:**

\({\text{y}}=2{\text{x}}-5\)

**(**

**3. Create a graph of the line represented by the equation:**

\({\text{y}}=\dfrac{-1}{2} {\text{x}}+2\) (

**4. Graph the line represented by the equation:**

\(3{\text{x}}+4{\text{y}}=-12\) (

**5. Graph the line represented by the equation:**

\(\dfrac{1}{3}{\text{x}}{-}{\dfrac{1}{2}}{\text{y}}=5\) (

**6. Plot the y-intercept for the line:**

\({\text{y}}=-3{\text{x}}+2\)

**(**

**7. Use the information in the previous question to answer this problem.**

Now, choose an x-value (besides \(x=0\)) and plot the point on the line \(y=−3x+2\) that corresponds to this value. Then, use these two points to graph the line. (

**8. Plot the y-intercept for the line:**

\({\text{y}} = \dfrac{1}{2}{\text{x}} - 3\) (

**9. Use the information in the previous question to answer this problem.**Now, choose an x-value (besides \(x=0\)) and plot the point on the line \({\text{y}} = \dfrac{1}{2}{\text{x}} - 3\) that corresponds to this value. Then, use these two points to graph the line. (

**10. Plot the y-intercept for the line:**

\({\text{y}} = -\dfrac{3}{5}{\text{x}} + 1\)

**(**

**11. Use the information in the previous question to answer this problem.**

Now, choose an x-value (besides \(x=0\)) and plot the point on the line \({\text{y}} = -\dfrac{3}{5}{\text{x}} + 1\) that corresponds to this value. Then, use these two points to graph the line. (

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