1. Graphing Linear Equations Using a Table of Values
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Chapter 2
1.
Graphing Linear Equations Using a Table of Values
This lesson delves into the methods of graphing linear equations by utilizing tables of values. It provides a step-by-step approach, starting with choosing at least two x-values for the table. The relationship between variables is visualized through these tables, aiding in the graphing process. Real-life scenarios, such as Maya's book buying habits and Ramsha's earnings at a bakery, are used to illustrate the application of these methods. The essence of the lesson is to equip learners with the skills to visualize relationships between two variables and graph them accurately.
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Lesson Settings & Tools
| | 10 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Graphing Linear Equations Using a Table of Values
Slide of 10
Graphs of linear equations are useful for describing relationships between variables that change at a constant rate. They often outperform words or equations. Keeping in mind the definition of a linear equation in one variable, it will now be extended to two variables. This lesson will also cover the most basic method of graphing linear equations in two variables.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Plotting Points Corresponding to Linear Equations
Consider the following linear equations.
lc3=2x+1 & (I) 5=2x+1 & (II) 7=2x+1 & (III)
The right-hand sides of these equations are all the same. Now, each equation will be solved using the Properties of Equality.
The solutions of Equations (I), (II), and (III) are x=1, x=2, and x=3, respectively. The points corresponding to each equation will be plotted on one coordinate plane. First, the points must be identified. Their x-coordinates will be the solutions and their y-coordinates will be the left-hand sides of the original related equations.
Now the points can be plotted on a coordinate plane.
What can be noted about the points? What is the equation of the line that passes through them? Why?
l3=2x+1 5=2x+1 7=2x+1
▼
Solve for x
(I), (II), (III): LHS-1=RHS-1
l2=2x 4=2x 6=2x
(I), (II), (III): .LHS /2.=.RHS /2.
l1=x 2=x 3=x
(I), (II), (III): Rearrange equation
lx=1 x=2 x=3
| Equation | Solution | Point |
|---|---|---|
| 3=2x+1 | x= 1 | ( 1, 3) |
| 5=2x+1 | x= 2 | ( 2, 5) |
| 7=2x+1 | x= 3 | ( 3, 7) |
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A Table of Values and How to Make It
A table of values is a chart that helps to organize and visualize information. It is frequently used to show the relation between two variables.
| x | y |
|---|---|
| 0 | 0 |
| 1 | 3 |
| 2 | 6 |
| 3 | 9 |
| 4 | 12 |
| 5 | 15 |
2corresponds to the y-value
6.This is usually represented with the notation (2,6).
The relation between the variables of an equation can be shown by making a table of values. This is helpful when graphing any equation, not only linear ones. The following linear equation will be drawn as an example.
6x-3y=- 12
There are four steps to making a table of values for an equation.
1
Isolate One Variable
expand_more If in the given equation none of the variables are isolated on one side, it is convenient to do so before making the table of values. This will simplify the rest of the process. A variable can be isolated using the Properties of Equality. If the variables presented in the equation are x and y, usually the latter one will be isolated.
Often the non-isolated variable is called the input and the isolated variable is called the output.
6x-3y=- 12
SubEqn
LHS-6x=RHS-6x
- 3y=- 6x-12
DivEqn
.LHS /(- 3).=.RHS /(- 3).
y=- 6x-12/- 3
▼
Simplify right-hand side
WriteSumFrac
Write as a sum of fractions
y=- 6x/- 3+- 12/- 3
DivNegNeg
- a/- b=a/b
y=6x/3+12/3
CalcQuot
Calculate quotient
y=2x+4
2
Choose the Values for the Non-Isolated Variable(s)
expand_more A table of values will usually have as many columns as there are variables plus one. In this case, the first column is for the values of the input. The second column is for substituting values from the first column into the rewritten equation. The third column shows the values of the output corresponding to its input.
| x | 2x+4 | y=2x+4 |
|---|
Next, the values of the non-isolated input variable x should be chosen. It can be done arbitrarily. However, the values should always belong to the domain of the function represented by the given equation. Recall that the domain of a linear function are all real numbers. In this case - 2, 0, 0.5, and 1 will be used.
| x | 2x+4 | y=2x+4 |
|---|---|---|
| - 2 | ||
| 0 | ||
| 0.5 | ||
| 1 |
3
Substitute the Values Into the Rewritten Equation
expand_more In this step the input values will be substituted into the rewritten equation where the output variable is isolated on the left-hand side.
| x | 2x+4 | y=2x+4 |
|---|---|---|
| - 2 | 2( - 2)+4 | |
| 0 | 2( 0)+4 | |
| 0.5 | 2( 0.5)+4 | |
| 1 | 2( 1)+4 |
4
Calculate the Values of the Isolated Variable
expand_more The last step is to simplify the expressions in the second column and write the result in the the last column. These are the outputs corresponding to the input from each row.
| x | 2x+4 | y=2x+4 |
|---|---|---|
| - 2 | 2( - 2)+4 | 0 |
| 0 | 2( 0)+4 | 4 |
| 0.5 | 2( 0.5)+4 | 5 |
| 1 | 2( 1)+4 | 6 |
If the expressions in the second column are too complicated to calculate mentally, extra columns where the partial results are written can be added as needed. However, the last column should always be the outputs.
| x | 2x+4 | Simplify | y=2x+4 |
|---|---|---|---|
| - 2 | 2( - 2)+4 | - 4+4 | 0 |
| 0 | 2( 0)+4 | 0+4 | 4 |
| 0.5 | 2( 0.5)+4 | 1+4 | 5 |
| 1 | 2( 1)+4 | 2+4 | 6 |
The table of values can be reduced to a table with only two columns — one with inputs and one with outputs. Note that for the output column we write only the variable.
| 6x-3y=- 12 | |
|---|---|
| x | y |
| - 2 | 0 |
| 0 | 4 |
| 0.5 | 5 |
| 1 | 6 |
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Phone Plan
Zain is considering a new phone plan. Currently they spend $15 on their phone each month. The new plan consists of a $9 flat rate for which they get unlimited texts, 300 minutes, and 6GB of data. For any minute above the 300 minutes included in the plan, Zain will have to pay an additional $0.01.
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Hint
Let y represent the total cost of the new phone plan. Let x represent the number of minutes used. Write an equation for y when x is greater than 300.
Solution
Let y represent the total cost of the new phone plan expressed in dollars. Let x represent the number of minutes used. If x is less than or equal to 300, y is equal to the flat rate of $9.
x≤ 300 ⇒ y= 9
If x is greater than 300, the cost of the minutes exceeding the 300 given minutes must be added to the flat rate. There are x-300 minutes that have to be payed for. Each of these minutes costs $0.01.
x> 300 ⇒ y= 9+(x-300) 0.01
The above equation for y can be simplified.
Now, the table of values for x and y will be formed. Let x be 350, 410, and 550. These are the numbers of minutes Zain used in the previous months. Since they never exceeded 600 minutes, this will be the highest considered value of x. Note that all considered values of x are greater than 300, so the expression for y is 0.01x+6.
y=9+(x-300)0.01
y=0.01x+6
| x | y=0.01x+6 | y |
|---|---|---|
| 350 | y=0.01( 350)+6 | 9.5 |
| 410 | y=0.01( 410)+6 | 10.1 |
| 550 | y=0.01( 550)+6 | 11.5 |
| 600 | y=0.01( 600)+6 | 12 |
For the full picture, the values of x=200 and x=300 will also be added to the table. Remember that when x≤ 300, the value of y is always equal to 9.
| Number of Minutes | Total Cost |
|---|---|
| 200 | $ 9 |
| 300 | $ 9 |
| 350 | $ 9.5 |
| 410 | $ 10.1 |
| 550 | $ 11.5 |
| 600 | $ 12 |
Currently Zain spends $15 on their phone plan. It can be noted that all of the prices in the above table are less than $15. Assuming Zain does not change their talking habits and 6GB is enough data for them, they should switch to the new plan.
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Linear Equation and Graph of Its Solutions
A linear equation is an equation with at least one linear term and any number of constants. No other types of terms may be included. Linear equations in one variable have the following form, where a and b are real numbers and a≠ 0.
ax+b=0 or ax = b
Linear equations in two variables have the form below, where a, b, and c are real numbers and a≠ 0 and b≠ 0.
ax+by+c=0 or ax+by = c
A line is a one-dimensional object of infinite length with no width or thickness that never bends or turns. Its graphical representation is a straight line with arrowheads on either end, indicating that it continues indefinitely in both directions.
The graph of a linear equation in two variables is the set of all its solutions plotted on the same coordinate plane, forming a line. Since a line is infinite, this means that a linear equation in two variables has infinitely many solutions.
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Identifying Linear Equations
Consider the given equation in two variables. Determine whether it is a linear equation or not.
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December Pay
Ramsha works in a bakery. She earns $ 10 per hour. Since it is December, she will get a bonus of $ 50 this month.
Answer
Hint
Choose at least two x-values for the table of values. Since x represents the number of hours worked, it must be greater than 0.
Solution
The given linear equation will be graphed using a table of values. y=10x+50 The graph of a linear equation in two variables is a line. There is exactly one line that passes through any two different points. Therefore, at least two different values of x need to be evaluated in the table of values. Also, x must be greater than 0 as it represents the number of hours worked.
| x | y=10x+50 | y |
|---|---|---|
| 35 | y=10( 35)+50 | 400 |
| 95 | y=10( 95)+50 | 1000 |
| 125 | y=10( 125)+50 | 1300 |
Each pair of x and y corresponds to an ( x, y) coordinate pair. Now (35,400), (95,1000), and (125,1300) will be plotted on the same coordinate plane.
Finally, the three points can be connected using a straightedge.
Since in the given case x cannot be negative, the graph includes only one arrowhead and is actually a ray.
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Graphing a Linear Equation in Standard Form
Maya loves reading. She is buying books at a garage sale.
Answer
Graph:
Maximum Number of Books: 15
Hint
Choose at least two x-values for the table of values.
Solution
First, the given equation will be graphed. Then the obtained graph will be used to determine the maximum number of books that Maya can purchase.
Graphing the Equation
Before making the table of values for the given equation, one of the variables has to be isolated. It is more common to isolate y, so let it also be the case here.3x+2y=46
▼
Solve for y
SubEqn
LHS-3x=RHS-3x
2y=- 3x+46
DivEqn
.LHS /2.=.RHS /2.
y=- 3x+46/2
WriteSumFrac
Write as a sum of fractions
y=- 3x/2+46/2
CalcQuot
Calculate quotient
y=- 1.5x+23
| x | y=- 1.5x+23 | y |
|---|---|---|
| 0 | y=- 1.5( 0)+23 | 23 |
| 2 | y=- 1.5( 2)+23 | 20 |
| 4 | y=- 1.5( 4)+23 | 17 |
| 10 | y=- 1.5( 10)+23 | 8 |
Note that in order to simplify the graphing process, the chosen x-values are all even numbers. Each pair of x and y corresponds to a coordinate pair ( x, y). Now (0,23), (2,20), (4,17), and (10,8) will be plotted on the same coordinate plane.
Finally, the four points can be connected using a straightedge.
Since the amount y of money left and the number x of books purchased cannot be negative, the graph is bounded only to the first quadrant. This also means that only part of the line is actually considered, which makes it a segment.
Finding the Maximum Number of Books
Note that the number of books has to be a whole number, as Maya cannot buy part
of a book. Also, she cannot spend more money than she has, which means that the amount of money left has to be non-negative. A point which meets these conditions, the maximum value of the x-coordinate, can be identified on the graph.
The x-coordinate of the point corresponds to the maximum number of books that Maya can buy. As seen on the graph, its value is 15. Therefore, Maya can purchase at most 15 books.
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Matching a Line to a Table of Values
Closure
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Solving a Linear Equation by Graphing
Equations in one variable can be solved by using the properties of equality. Consider the following equation in one variable. 2x+1=5x-5 This type of equation can also be solved by graphing the linear equations in two variables corresponding to the left- and right-hand sides of the given equation. 2x+1=5x-5 ⇒ lcy=2x+1 & (I) y=5x-5 & (II) Equations (I) and (II) will be graphed on one coordinate plane using a table of values. First, the table of values for Equation (I) will be made. Remember that at least two different values of x have to be evaluated in the table.
| x | y=2x+1 | y |
|---|---|---|
| 0 | y=2( 0)+1 | 1 |
| 1 | y=2( 1)+1 | 3 |
The line passing through (0,1) and (1,3) is the graph of y=2x+1. Now the table for Equation (II) will be constructed.
| x | y=5x-5 | y |
|---|---|---|
| 0 | y=5( 0)-5 | - 5 |
| 1 | y=5( 1)-5 | 0 |
The line passing through (0,- 5) and (1,0) is the graph of y=5x-5.
Finally, the point of intersection of y=2x+1 and y=5x-5 can be identified.
The lines intersect at ( 2, 5). This means that both 2x+1 and 5x-5 equal 5 when x= 2.
| Expression | x=2 | Simplify |
|---|---|---|
| 2x+1 | 2( 2)+1 | 5 |
| 5x-5 | 5( 2)-5 | 5 |
Graphing Linear Equations Using a Table of Values
Exercises
Graph each of the linear equations using a table of values.
y=- 8x Which of the graphs corresponds to the equation?
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3x-1=y Which of the following graph corresponds to the given equation?
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x=10-y Which of the graphs corresponds to the equation?
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We are asked to graph the linear equation by making a table of values. y=- 8x The y-variable is already isolated on the left-hand side of the equation, so we can arbitrarily choose the values for x. The graph of a linear equation in two variables is a line. There is exactly one line that passes through any two different points. This is why we need to evaluate at least two different values of x in the table.
| x | y=-8 x | y |
|---|---|---|
| -1 | y=-8( -1) | 8 |
| 0 | y=-8( 0) | 0 |
| 1 | y=-8( 1) | -8 |
In this case, there are no restrictions on the values of x, so they can be any real numbers. Now, by using the x- and y-values as ( x, y) coordinate pairs, we can plot the points. Connecting these points using a straightedge will give us the graph of y=- 8x.
This graph corresponds to option C.
We will graph the given linear equation using a table of values.
3x-1=y
As in Part A, the y-variable is already isolated. However, this time it is on the right-hand side of the equation. For simplicity, let's rearrange our equation so that y is on the left-hand side.
3x-1=y ⇔ y=3x-1
Now, we will follow the same process as in Part A. Let's make a table of values.
| x | y=3x-1 | y |
|---|---|---|
| -1 | y=3( -1)-1 | - 4 |
| 0 | y=3( 0)-1 | - 1 |
| 1 | y=3( 1)-1 | 2 |
We can plot (-1,- 4), (0,- 1), and (1,2) on one coordinate plane and connect them using a straightedge. This will give us the graph of 3x-1=y.
This graph corresponds to option A.
Again, we will graph the given linear equation by making a table of values.
x=10-y
The x-variable is already isolated on the left-hand side. We are used to y being isolated but in this case it does not matter. The procedure is the same — the only thing that changes is the name of the variable. Let's arbitrarily choose the y-values and find the corresponding x-values.
| y | x=10-y | x |
|---|---|---|
| 12 | x=10- 12 | - 2 |
| 10 | x=10- 10 | 0 |
| 8 | x=10- 9 | 2 |
The first column of the table contains the y-values and the last column contains the x-values. Therefore, the points that we will plot are (- 2,12), (0,10), and (2,8). Connecting these three points using a straightedge will give us the graph of x=10-y.
This graph corresponds to option D.
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Graph each of the linear equations with one variable using a table of values.
x=- 2 Which of the graphs corresponds to the equation?
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y=4 Which of the graphs corresponds to the equation?
{"type":"choice","form":{"alts":["A","B","C","D"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":2}
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We are asked to graph a linear equation with only the x-variable present. This means that the coefficient of the y-variable is 0. Since the coefficient is 0, the term 0y can be added to the left- or right-hand side of the given equation. x=- 2+ 0y In this equation, x is already isolated, so let's substitute some values for y. In this case, there are no restrictions on the values of y, so they can be any real numbers. Also, remember to choose at least two different values of y.
| y | x=- 2+0y | x |
|---|---|---|
| 0 | x=- 2+0( 0) | - 2 |
| 1 | x=- 2+0( 1) | - 2 |
| 2 | x=- 2+0( 2) | - 2 |
Note that the value of y does not affect the value of x because the coefficient of the y-variable is 0. Each pair of x and y corresponds to an ( x, y) coordinate pair. Now we will plot (- 2,0), (- 2,1), and (- 2,2) on the same coordinate plane. Then, the three points will be connected using a straightedge.
Since the value of x is the same for all values of y, the obtained line is vertical. This graph corresponds to option D.
We are asked to graph a linear equation with only the y-variable present. This means that the coefficient of the x-variable is 0. Since the coefficient is 0, the term 0x can be added to the left- or to the right-hand side of the given equation.
y=4+ 0x
Here, y is already isolated on the left-hand side of the equation. Let's use this equation and substitute some values for x. Similarly as in Part A, there should be at least two different values of x in the table and they can be chosen from all real numbers.
| x | y=4+0x | y |
|---|---|---|
| 0 | y=4+0( 0) | 4 |
| 2 | y=4+0( 2) | 4 |
| 4 | y=4+0( 4) | 4 |
This time the value of x does not affect the value of y because the coefficient of the x-variable is 0. Each pair of x and y corresponds to an ( x, y) coordinate pair. Let's plot (0,4), (2,4), and (4,4) on the same coordinate plane and connect the points using a straightedge.
Since the value of y is the same for all values of x, the obtained line is horizontal. This graph corresponds to option C.
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Graph each of the linear equations using a table of values.
y-7=- x Which of the graphs corresponds to the equation?
{"type":"choice","form":{"alts":["A","B","C","D"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":0}
4x+5=2y-1 Which of the graphs corresponds to the equation?
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We are asked to graph the linear equation by making a table of values. y-7=- x First, we will isolate the y-variable by adding 7 to both sides of the equation. y-7=- x ⇔ y=- x+7 Now we will arbitrarily choose the x-values and substitute them into the equation to find the corresponding y-values. Remember to choose at least two values and to make sure they are from the domain of the function. In this case, the domain is all real numbers.
| x | y=- x+7 | y |
|---|---|---|
| - 2 | y=-( - 2)+7 | 9 |
| 0 | y=-( 0)+7 | 7 |
Using the x and y values as ( x, y) coordinate pairs, we can plot the points. Connecting these points using a straightedge will give us the graph of y-7=- x.
This graph corresponds to option A.
Again, we need to graph the linear equation by making a table of values.
4x+5=2y-1
Let's start by isolating y on the left-hand side of the equation by using the Properties of Equality.
Next, we will arbitrarily choose the x-values and substitute them into the equation to find the corresponding y-values.
| x | y=2x+3 | y |
|---|---|---|
| 0 | y=2( 0)+3 | 3 |
| 1 | y=2( 1)+3 | 5 |
Finally, we can plot the points from the table and connect them using a straightedge, which will give us the graph of 4x+5=2y-1.
This graph corresponds to option C.
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Solve the following equation by graphing.
- 3x+6=0
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We are asked to solve the given equation by graphing. - 3x+6=0 Each side of this equation corresponds to a linear equation. - 3x+6=0 ⇒ lcy=- 3x+6 & (I) y=0 & (II) Let's graph Equations (I) and (II) on one coordinate plane by making a table of values. First, we will make the table of values for Equation (I). Remember that at least two different values of x have to be evaluated in the table.
| x | y=- 3x+6 | y |
|---|---|---|
| 0 | y=- 3( 0)+6 | 6 |
| 1 | y=- 3( 1)+6 | 3 |
The line passing through (0,6) and (1,3) is the graph of y=- 3x+6. Now we will construct the table for Equation (II). Since y is equal to a constant, for any x-value the corresponding y-value will be 0.
| x | y |
|---|---|
| 0 | 0 |
| 1 | 0 |
The line passing through (0,0) and (1,0) is the graph of y=0, which is the y-axis.
Looking at the above graph, we can identify the point of intersection of y=- 3x+6 and y=0.
The lines intersect at ( 2, 0). This means that - 3x+6 is equal to 0 when x= 2. Since the right-hand side of the equation is always equal to 0, x=2 is the solution to the original equation. We can verify our solution by substituting 2 for x into the original equation and simplifying.
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Solve the following equation by graphing.
2/3(x+3)=1/2(x+5)
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We are asked to solve the given equation by graphing. 2/3(x+3)=1/2(x+5) Each side of this equation corresponds to a linear equation. 2/3(x+3)=1/2(x+5) ⇓ lcy= 23(x+3) & (I) y= 12(x+5) & (II) Let's graph Equations (I) and (II) on one coordinate plane by making a table of values. First, we will make the table of values for Equation (I) by choosing at least two different values of x. Also, we will choose such x-values that the y-values are integers. This will make plotting the corresponding points simpler.
| x | y=2/3(x+3) | Simplify | y |
|---|---|---|---|
| - 3 | y=2/3( - 3+3) | y=2/3(0) | 0 |
| 0 | y=2/3( 0+3) | y=2/3(3) | 2 |
The line passing through (- 3,0) and (0,2) is the graph of y= 23(x+3). Now, we will construct the table for Equation (II). Again, we will choose such x-values that the y-values are integers.
| x | y=1/2(x+5) | Simplify | y |
|---|---|---|---|
| - 1 | y=1/2( - 1+5) | y=1/2(4) | 2 |
| 1 | y=1/2( 1+5) | y=1/2(6) | 3 |
The line passing through (- 1,2) and (1,3) is the graph of y= 12(x+5).
Looking at the above graph we can identify the point of intersection of y= 23(x+3) and y= 12(x+5).
The lines intersect at ( 3, 4). This means that both 23(x+3) and 12(x+5) equal 4 when x= 3. Therefore, x=3 is the solution to the original equation. We can verify our solution by substituting 3 for x into the original equation and simplifying.
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Graphing Linear Equations Using a Table of Values
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