8.2 Graphing Linear Equations

A linear equation is an algebraic equation with one or two variables (each with one exponent), producing a straight line when plotted on a graph.

Examples of linear equations with one variable in the rectangular coordinate system are:

[latex]3x - 5 = 0[/latex],   [latex]x = 3[/latex],   [latex]y + 2 = 0[/latex],   [latex]\displaystyle{y = -\frac{7}{5}}[/latex]

Finding the value of the variables for which the equation is true is known as solving the equation. A linear equation with two variables has infinitely many pairs of values as solutions; therefore, it is often convenient to represent these solutions by drawing a graph.

Linear equations with two variables are generally represented by the variables x and y and are expressed either in the form of Ax + By = C  (where A, B, and C are integers and A is positive), known as standard form, or in the form of y = mx + b (where m and b are integers or fractions), known as slope-intercept form.

Linear Equations in Standard Form

The ‘standard’ form for a linear equation with two variables, x, and y, is written as Ax + By = C, where A, B, and C are integers, v is positive, and A, B, and C have no common factors other than 1.

For example, consider the following simple linear equation with two variables: [latex]2x - y = -3[/latex]

The equation Ax + By = C, where A = 2, B = −1, and C = −3.

• If a given equation has fractions, multiply each term by the least common denominator (LCD), divide each term by any common factors, and rearrange the equation into standard form.
• If a given equation has decimals, multiply each term by an appropriate power of 10 to eliminate the decimals, divide each term by common factors, and rearrange the equation into standard form.
• If a given equation has no fractions or decimals, divide each term by any common factors and rearrange it into standard form.

Solution

1. [latex]\displaystyle{\frac{2}{3}x + \frac{1}{2}y - 3 = 0}[/latex]
   Multiplying each term by the LCD of 6 and simplifying, [latex]4x + 3y - 18 = 0[/latex]
   Rearranging, [latex]4x + 3y = 18[/latex]
   This is in the form Ax + By = C.
   Therefore, the equation [latex]\displaystyle{\frac{2}{3}x + \frac{1}{2}y - 3 = 0}[/latex], in standard form, is [latex]4x + 3y = 18[/latex]
2. [latex]0.3x = 1.25y - 2[/latex]
   Multiplying each term by 100, [latex]30x = 125y - 200[/latex]
   Dividing each term by the common factor 5 and simplifying, [latex]6x = 25y - 40[/latex]
   Rearranging, [latex]6x - 25y = -40[/latex]
   This is in the form Ax + By = C.
   Therefore, the equation [latex]0.3x = 1.25y - 2[/latex], in standard form, is [latex]6x - 25y = -40[/latex]
3. [latex]\displaystyle{y = \frac{2}{3}x - 5}[/latex]
   Multiplying each term by 3 and simplifying,[latex]3y = 2x - 15[/latex]
   Rearranging,[latex]-2x + 3y = -15[/latex]
   Multiplying each term by –1 to make A positive,[latex]2x - 3y = 15[/latex]
   This is in the form Ax + By = C.
   Therefore, the equation [latex]\displaystyle{y = \frac{2}{3}x - 5}[/latex], in standard form, is [latex]2x - 3y = 15[/latex]

Linear Equations in Slope-Intercept Form

The ‘slope-intercept’ form for a linear equation with two variables, x, and y, is written as y = mx + b, where m and b are either integers or fractions. ‘m’ represents the slope, and ‘b’ represents the y-coordinate of the y-intercept.

The slope is the steepness of the line relative to the X-axis.

The y-intercept is the point at which the line crosses the Y-axis and where the x-coordinate is zero.

For example, consider a simple linear equation such as y = 2x + 3. This equation is in the slope-intercept form of y = mx + b, where m = 2 and b = 3; hence, the slope is 2, and the y-intercept is (0, 3).

Solution

1. [latex]4x + 3y = 18[/latex]
   Rearranging the equation with y on the left, [latex]3y = -4x + 18[/latex]
   Dividing each term by 3 and simplifying,[latex]\displaystyle{y = -\frac{4}{3}x + 6}[/latex]
   This is in the form y = mx + b.
   Therefore, the slope is [latex]\displaystyle{m = -\frac{4}{3}}[/latex] and the y-intercept is the point (0, 6).
2. [latex]\displaystyle{x = \frac{25}{6}y + \frac{20}{3}}[/latex]
   Multiplying each term by the LCD of 6 and simplifying, [latex]6x = 25y + 40[/latex]
   Rearranging the equation with y on the left,[latex]-25y = -6x + 40[/latex]
   Multiplying each term by −1,[latex]25y = 6x - 40[/latex]
   Dividing each term by 25 and simplifying, [latex]\displaystyle{y = \frac{6}{25}x - \frac{8}{5}}[/latex]
   This is in the form y = mx + b.
   Therefore, the slope is [latex]\displaystyle{m = -\frac{6}{25}}[/latex] and the y-intercept is the point [latex]\displaystyle{(0, -\frac{8}{5})}[/latex].

Determining the Solution Set of a Linear Equation

If a linear equation has two variables, x, and y, then there are infinitely many solutions to the equation, and it is not possible to solve the equation for a single value of each variable. However, it is possible to create a set of solutions by replacing one variable (either x or y) with any value and then computing the value of the other variable.

For example, consider the equation: [latex]y = 2x + 3[/latex]

Choosing [latex]x = 1[/latex] and substituting [latex]x = 1[/latex] into the equation:

[latex]y = 2x + 3[/latex]

[latex]y = 2(1) + 3 = 5[/latex]

Therefore, [latex]x = 1[/latex] and [latex]y = 5[/latex] is one of the infinitely many solutions to the equation; i.e., the ordered pair (1, 5) is a point that satisfies the equation.

Choosing [latex]x = 2[/latex] and substituting [latex]x = 2[/latex] into the equation:

[latex]y = 2x + 3[/latex]

[latex]y = 2(2) + 3 = 7[/latex]

Therefore, [latex]x = 2[/latex] and [latex]y = 7[/latex] is another solution to the equation; i.e., the ordered pair (2, 7) is another point that satisfies the equation.

Similarly, we can obtain any number of points that satisfy the equation by choosing different values for x, and computing the corresponding value for y.

We can represent the full solution set by graphing the linear equation. The graph of the linear equation will be a line formed by all the solutions to the linear equation. Conversely, any point on the line is a solution to the linear equation.

Graphing Linear Equations Using a Table of Values

Follow these steps to graph a linear equation using a table of values:

Step 1:   Create a table of values by choosing any value for the variable x (0 is often a good first choice).

Step 2:   Compute the corresponding value for the variable y (this is easiest if the equation is in slope-intercept form).

Step 3:   Form the ordered pair (x, y).

Step 4:   Repeat Steps 1 to 3 at least two more times to create at least three ordered pairs.

Step 5:   Plot the ordered pairs (points) on the coordinate system using an appropriate scale.

Step 6:   Join the points in a straight line, continuing the line indefinitely in both directions using arrows.

Step 7:   Label the graph with the equation of the line.

For example, consider the linear equation [latex]y = 2x + 3[/latex]. We will first determine the coordinates of four ordered pairs on this line by choosing values for x and finding the corresponding values for y; then, we will draw the graph.

• Choosing [latex]x = 0[/latex], [latex]y = 2x + 3 = 2(0) + 3 = 0 + 3 = 3[/latex], (0, 3) is a point on the line.
• Choosing [latex]x = 1[/latex], [latex]y = 2x + 3 = 2(1) + 3 = 2 + 3 = 5[/latex], (1, 5) is a point on the line.
• Choosing [latex]x = 2[/latex], [latex]y = 2x + 3 = 2(2) + 3 = 4 + 3 = 7[/latex], (2, 7) is a point on the line.
• Choosing [latex]x = 3[/latex], [latex]y = 2x + 3 = 2(3) + 3 = 6 + 3 = 9[/latex], (3, 9) is a point on the line.

Since all the points in the Table of Values fall on a line when joined, it verifies that the plotted line represents the equation.

Graphing Linear Equations Using the x-intercept and the y-intercept

Recall that the y-intercept is the point at which the line crosses the Y-axis and where the x-coordinate is zero. Similarly, the x-intercept is the point at which the line crosses the X-axis and where the y-coordinate is zero.

If the intercepts are not at the origin (0, 0), we may use the x-intercept and y-intercept as two points to draw a linear graph and use a third point to test the drawn line.

If the intercepts are at the origin (0, 0), we need to compute another ordered pair to use as the second point (along with the origin) to draw the line and a third point to test the drawn line.

For example, consider a linear equation, [latex]3x - y = -9[/latex], where we will find the x-intercept, y-intercept, and another ordered pair to draw the graph.

Finding the x-intercept:

• Substituting [latex]y = 0[/latex] into the given equation and solving for x,
• [latex]3x - 0 = -9[/latex], thus, [latex]x = -3[/latex]. Therefore, (−3, 0) is the x-intercept.

Finding the y-intercept:

• Substituting [latex]x = 0[/latex] into the given equation and solving for y,
• [latex]3(0) - y = -9[/latex], thus, [latex]y = 9[/latex]. Therefore, (0, 9) is the y-intercept.

Finally, finding another ordered pair as a third point on the line to use as a test point to verify the plotted line joining the x-intercept and y-intercept:

• Choosing [latex]x = -1[/latex], substituting this in the given equation, and solving for y,
• [latex]3(-1) - y = -9[/latex], thus, [latex]y = 6[/latex]. Therefore, (–1, 6) is a point on the line [latex]3x - y = -9[/latex].

Now that we have the x-intercept, the y-intercept, and a test point:

• Plot the ordered pairs on the coordinate system using an appropriate scale.
• Draw a line to join the x-intercept and y-intercept.
• Verify that the test point falls on the graph of the plotted line.

Since the point (–1, 6) falls on the line when plotted on the graph, it verifies that the plotted line represents the linear equation.

Graphing Linear Equations Using the Slope and the y-intercept

Recall that a linear equation in the form of [latex]y = mx + b[/latex] is the equation in slope-intercept form, where ‘m’ is the slope and ‘b’ is the y-coordinate of the y-intercept.

If the equation is in the standard form [latex]Ax + By = C[/latex], it can be rearranged to represent the slope-intercept form as follows:

[latex]Ax + By = C[/latex]

[latex]By = -Ax + C[/latex]

[latex]\displaystyle{y = -\frac{A}{B}x + \frac{C}{B}}[/latex]

This is in the form [latex]y = mx + b[/latex], where the slope [latex]\displaystyle{m = -\frac{A}{B}}[/latex], and the y-coordinate of the y-intercept [latex]\displaystyle{b = \frac{C}{B}}[/latex].

The Slope and y-intercept of a Line

The slope (m) is the steepness of the line relative to the X-axis. It is the ratio of the change in the value of y (called the ‘rise’ and denoted [latex]\Delta y[/latex]) to the corresponding change in the value of x (called the ‘run’ and denoted [latex]\Delta x[/latex]).

If P ([latex]x_1, y_1[/latex]) and Q ([latex]x_2, y_2[/latex]) are two different points on a line, then the slope of the line PQ between the two points is:

[latex]\boldsymbol{\displaystyle{m = \frac{Rise}{Run} = \frac{Change~in~y~value}{Change~in~x~value} = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}}}[/latex]

This is illustrated in Exhibits 8.2-c and 8.2-d:

Solution

[latex]4x + 3y = 18[/latex]

Rearranging the equation with y on the left,

[latex]3y = -4x + 18[/latex]

Dividing each term by 3 and simplifying,

[latex]\displaystyle{y = -\frac{4}{3}x + 6}[/latex]

This is in the form y = mx + b.

Therefore, the slope is [latex]\displaystyle{m = -\frac{4}{3}}[/latex] and the y-intercept is the point (0, 6).

Therefore, (0, 4) is a point on the line and the slope, [latex]\displaystyle{m = \frac{Rise}{Run} = \frac{Change~in~y~value}{Change~in~x~value} = \frac{2}{3}}[/latex], represents an increase of 2 in the vertical direction (‘rise’) for every increase of 3 in the horizontal direction (‘run’).

Representing this on a graph:

1. First, plot the y-intercept (0, 4).
2. From this point, move 3 units to the right and then move 2 units up to locate the new point (3, 6).
   Note: This is the same as moving 2 units up and then 3 units to the right.
3. Similarly, from the point (3, 6), move 3 units to the right and 2 units up to locate another point (6, 8).
4. Draw the line through these points to graph the equation.

Or,

1. First, plot the y-intercept (0, 4).
2. From this point, move 3 units to the left and then move 2 units down to locate the new point (−3, 2).
3. Similarly, from the point (−3, 2), move 3 units to the left and 2 units down to locate another point (−6, 0).
4. Draw the line through these points to graph the equation.

Note: All the points must lie on the same line, or a mistake has been made.

Solution

The equation is [latex]y = mx + b[/latex].

Therefore, [latex]\displaystyle{m = -\frac{3}{4}}[/latex] and [latex]b = -2[/latex], which means the y-intercept is the point (0, –2) and the slope is

[latex]\displaystyle{m = \frac{Rise}{Run} = \frac{Change~in~y~value}{Change~in~x~value} = \frac{-3}{4}}[/latex] or [latex]\displaystyle{\frac{3}{-4}}[/latex]

First, plot the point (0, –2). Then, using the slope, [latex]\displaystyle{m = \frac{-3}{4}}[/latex], from the point (0, –2), move 4 units to the right and 3 units down to locate the new point, (4, –5).

Alternatively, using the slope, [latex]\displaystyle{m = \frac{3}{-4}}[/latex], from the point (0, –2), move 4 units to the left and 3 units up to locate another point on the line, (–4, 1).

Draw a line through these points to graph the equation.

Slope and Direction of a Line

A line’s slope, m, is a number that describes the line’s direction and steepness.

The direction of the line either a. slopes upwards to the right, b. slopes downwards to the right, c. is horizontal, or d is vertical, as discussed below:

If the sign of ‘m‘ is positive, the line slopes upwards to the right (i.e., the line rises from left to right), as illustrated in Exhibit 8.2-e.
If the sign of ‘m‘ is negative, the line slopes downwards to the right (i.e., the line falls from left to right), as illustrated in Exhibit 8.2-f.

If ‘m’ is zero, the line is horizontal (parallel to the X-axis). A slope of zero means that when the x-coordinate increases or decreases, the y-coordinate does not change (i.e., ‘rise’ = 0). This is a special case of the linear equation [latex]Ax + By = C[/latex], where the value [latex]A = 0[/latex]. Therefore, the equation of a horizontal line will be in the form y = b, and the y-intercept of the line is (0, b). For example, in the equation, [latex]y = 3[/latex], (i.e., [latex]y = 0x + 3[/latex]), the slope is zero, and the value of the y-coordinate is 3 for all values of x. Therefore, the line is horizontal and passes through the y-intercept (0, 3), as illustrated in Exhibit 8.2-g.

If ‘m’ is undefined, the line is vertical (parallel to the Y-axis). An undefined slope means that when the y-coordinate increases or decreases, the x-coordinate does not change (i.e., ‘run’ = 0). This is a special case of the linear equation [latex]Ax + By = C[/latex], where the value [latex]B = 0[/latex]. An equation in this form cannot be rearranged into slope-intercept form, as it would require dividing by 0. As such, rather than isolating for y, we isolate for the variable x. Therefore, the equation of a vertical line will be in the form [latex]x = a[/latex], and the x-intercept of the line is ([latex]a, 0[/latex]). For example, in the equation, [latex]\boldsymbol{x = 2}[/latex], the slope is undefined, and the value of the x-coordinate is 2 for all values of y. Therefore, the line is vertical and passes through the x-intercept (2, 0), as illustrated in Exhibit 8.2-h.

Slope and Steepness of a Line

The steepness of the line is also measured by the slope (‘m’) of the line. The farther the coefficient ‘m’ is away from zero, the steeper the slope. The closer the coefficient ‘m’ is to zero, the flatter the slope.

• A greater value for a positive slope indicates a steeper rise.
• A lesser value for a negative slope indicates a steeper fall.
• As discussed earlier, a slope of 0 indicates a horizontal line (no steepness), and an undefined slope indicates a vertical line (infinite steepness).

Lines Passing Through the Origin

A line passing through the origin means the point (0, 0) is on the line. The x- and y-intercept coordinates are (0, 0). Hence, the equation of a line passing through the origin will be in the form [latex]y = mx[/latex], with the exception of two special cases:

• [latex]y = 0[/latex] – the equation of the X-axis (i.e., the horizontal line passing through the origin)
• [latex]x = 0[/latex] – the equation of the Y-axis (i.e., the vertical line passing through the origin)

For example, in the equation, [latex]\boldsymbol{y = 2x}[/latex], [latex]m = 2[/latex], and [latex]b = 0[/latex]; hence the slope is 2, and the y-intercept is the point (0, 0).

Similarly, in the equation [latex]\boldsymbol{y = -x}[/latex], [latex]m = -1[/latex], and [latex]b = 0[/latex]; hence the slope is –1 and the y-intercept is the point (0, 0).

Therefore, both lines above pass through the origin (0, 0), as illustrated in Exhibit 8.2-i.

8.2 Exercises

Answers to the odd-numbered questions are available at the end of the book.

1. For the equation [latex]2x + 3y = 18[/latex], find the missing values in the following ordered pairs:
   a. (3, ?)
   b. (-6, ?)
   c. (0, ?)
   d. (?, 0)
   e. ( ?, -4)
   f. (?, 2)
2. For the equation [latex]x + 5y = 20[/latex], find the missing values in the following ordered pairs:
   a. (0, ?)
   b. (-15, ?)
   c. (5, ?)
   d. (?, 6)
   e. (?, -3)
   f. (?, 0)
3. For the equation [latex]\displaystyle{y = -\frac{2}{3}x + 1}[/latex], find the missing values in the following ordered pairs:
   a. (6, ?)
   b. (–3, ?)
   c. (0, ?)
   d. (?, 0)
   e. (?, −3)
   f. (?, 5)
4. For the equation [latex]\displaystyle{y = -\frac{2}{3}x + 1}[/latex], find the missing values in the following ordered pairs:
   a. (2, ?)
   b. (–3, ?)
   c. (0, ?)
   d. (?, 0)
   e. (?, −3)
   f. (?, 9)

For problems 5 to 10, write the equations in standard form.

5. [latex]\displaystyle{y = \frac{5}{2}x + 1}[/latex]
6. [latex]\displaystyle{y = \frac{2}{5}x - 1}[/latex]
7. [latex]\displaystyle{y = -\frac{3}{4}x - 3}[/latex]
8. [latex]\displaystyle{y = -\frac{4}{3}x + 4}[/latex]
9. [latex]\displaystyle{y = \frac{1}{2}x + \frac{3}{2}}[/latex]
10. [latex]\displaystyle{y = \frac{3}{2}x - \frac{1}{4}}[/latex]

For problems 11 to 16, write the equations in slope-intercept form.

11. [latex]4y + 6x = -3[/latex]
12. [latex]9y + 2x = 18[/latex]
13. [latex]3y - 2x = 15[/latex]
14. [latex]5y - 2x = -20[/latex]
15. [latex]\displaystyle{\frac{x}{2} + \frac{y}{3} = 1}[/latex]
16. [latex]\displaystyle{\frac{x}{4} + \frac{y}{5} = 2}[/latex]

For problems 17 to 24, graph the equations using a table of values.

17. [latex]y = x + 3[/latex]
18. [latex]y = 3x + 2[/latex]
19. [latex]y = -5x + 1[/latex]
20. [latex]y = -2x + 3[/latex]
21. [latex]2x + y + 1 = 0[/latex]
22. [latex]4x + y + 2 = 0[/latex]
23. [latex]2x - y - 3 = 0[/latex]
24. [latex]x - y - 1 = 0[/latex]

For problems 25 to 30, determine the x- and y-intercepts and graph the equations.

25. [latex]3x + y = -2[/latex]
26. [latex]5x + y = -3[/latex]
27. [latex]x + y - 3 = 4[/latex]
28. [latex]x + y - 4 = 7[/latex]
29. [latex]y = 4x + 1[/latex]
30. [latex]4x + y + 2 = 0[/latex]

For problems 31 to 34, determine the slopes and y-intercepts of the equations and graph the equations.

31. [latex]2x - 3y - 18 = 0[/latex]
32. [latex]5x - 2y + 10 = 0[/latex]
33. [latex]-4x + 7y - 21 = 0[/latex]
34. [latex]-7x + 8y - 32 = 0[/latex]

35. Point ‘A’ is in the 3^rd quadrant and Point ‘B’ is in the 1^st quadrant. Determine the sign of the slope of the line AB.
36. Point ‘C’ is in the 4^th quadrant and Point ‘D’ is in the 2^nd quadrant. Determine the sign of the slope of the line CD.

For problems 37 to 40, determine the slopes of the lines passing through:

37. (2, 1) and (6, 1)
38. (−6, 4) and (2, 4)
39. (−5, 4) and (3, −1)
40. (5, 6) and (5, −4)