Lesson Objectives
  • Learn how to solve a linear system using graphing
  • Learn how to solve a linear system using substitution
  • Learn how to solve a linear system using elimination

How to Solve a System of Linear Equations Using Graphing, Substitution, and Elimination


Solving Linear Systems by Graphing

Up to this point, we have only dealt with a single linear equation in two variables. Let's suppose we saw an equation such as:
x - y = -3
We can obtain the slope-intercept form of the line by solving for y:
y = x + 3
Now we can graph our equation by plotting the y-intercept (0,3) and using our slope (m = 1) to find additional points: Graphing y=x + 3 In some cases, we will need to work with two linear equations in two variables at the same time. Let's suppose we also had the equation:
5x - y = 1
Again, we can obtain the slope-intercept form of the line by solving for y:
y = 5x - 1
Now we can graph our equation by plotting the y-intercept (0,-1) and using our slope (m = 5) to find additional points: Graphing y=5x - 1 We can combine these two equations into what is known as a "system of linear equations" or a "linear system". The solution for the system is any point (x,y), that satisfies both of the equations. When we look at the graph of any linear equation in two variables, each point on the line is a solution. To find the solution for a system using the graphing method, we graph each equation and look for the point (x,y) of intersection. This point of intersection will lie on both lines and, therefore, be a solution to our system. Let's graph our two equations together and look for the point of intersection.
x - y = -3
5x - y = 1 Graphing x - y=-3 and 5x - y=1 We can see the point of intersection occurs at (1,4). We can verify our solution by plugging in for x and y in each equation of the system.
x - y = -3
(1) - (4) = -3
-3 = -3
5x - y = 1
5(1) - (4) = 1
5 - 4 = 1
1 = 1
We can see that our point (1,4) works as a solution for both equations, therefore, it is the solution for the system.

Solving a Linear System using Graphing

  • Graph each equation of the system
  • Identify the point of intersection
  • Check the solution by plugging into each equation of the system
Let's look at some examples.
Example 1: Solve each linear system using graphing
7x + 3y = 9
2x + 3y = -6
Let's begin by placing each equation in slope-intercept form: $$y=-\frac{7}{3}x + 3$$ $$y=-\frac{2}{3}x - 2$$ Now, we can graph each equation and look for the point of intersection: Graphing 7x + 3y=9, 2x + 3y=-6 We can see that the point of intersection occurs at (3,-4). Let's check the solution for the system in each original equation:
7(3) + 3(-4) = 9
21 - 12 = 9
9 = 9
2(3) + 3(-4) = -6
6 - 12 = -6
-6 = -6
Example 2: Solve each linear system using graphing
2x - 3y = 9
5x + 3y = 12
Let's begin by placing each equation in slope-intercept form: $$y=\frac{2}{3}x - 3$$ $$y=-\frac{5}{3}x + 4$$ Now we can graph each equation and look for the point of intersection: Graphing 2x - 3y=9, 5x + 3y=12 We can see that the point of intersection occurs at (3,-1). Let's check the solution for the system in each original equation:
2(3) - 3(-1) = 9
6 + 3 = 9
9 = 9
5(3) + 3(-1) = 12
15 + (-3) = 12
12 = 12
Example 3: Solve each linear system using graphing
4x + 3y = 6
x + 3y = -3
Let's begin by placing each equation in slope-intercept form: $$y=-\frac{4}{3}x + 2$$ $$y=-\frac{1}{3}x - 1$$ Now we can graph each equation and look for the point of intersection: Graphing 4x - y=3, 2x + y=3 We can see that the point of intersection occurs at (3,-2). Let's check the solution for the system in each original equation:
4(3) + 3(-2) = 6
12 + (-6) = 6
6 = 6
(3) + 3(-2) = -3
3 + (-6) = -3
-3 = -3

Systems of Linear Equations with No Solution

Up to this point, we have only seen linear systems with exactly one solution. When this occurs, we say our system is consistent, and the equations are independent. In some cases, we will not be able to obtain a solution to our linear system. This happens when the two lines are parallel. Recall that parallel lines have the same slope and will never intersect. Since there isn't a point that exists on both lines, there is no solution. When this occurs, our system is said to be "inconsistent". We can state our answer as "no solution" or use the empty set symbol "∅". Let's look at an example.
Example 4: Solve each linear system using graphing
3x + 4y = 8
6x + 8y = -24
Let's begin by placing each equation in slope-intercept form: $$y=-\frac{3}{4}x + 2$$ $$y=-\frac{3}{4}x - 3$$ We can see that these two lines are parallel. They have the same slope (m = -3/4) and different y-intercepts. This means we will never be able to find a point of intersection or a solution for the system. Let's take a look at this graphically: Graphing 3x + 4y=8, 6x + 8y=-24 We can clearly see that the two lines are parallel and will never intersect. We will state that there is no solution to our system.

Systems of Equations with Infinitely Many Solutions

Another special case scenario occurs when we see a system of equations that has an infinite number of solutions. When this happens, we will see the same equation algebraically manipulated to look different. Since the two equations are the same, what works as a solution to one, also works for the other. Therefore, there will be an infinite number of solutions, and the equations are said to be "dependent". We can state our answer as "infinitely many solutions". Let's look at an example.
-7x + 9y = -13
14x - 18y = 26
Let's begin by placing each equation in slope-intercept form: $$y=\frac{7}{9}x - \frac{13}{9}$$ $$y=\frac{7}{9}x - \frac{13}{9}$$ There is no need to graph anything here, as we can see that both equations of the system are the same. Again, when this occurs we have an infinite number of solutions.

Solving Linear Systems by Substitution

Although we can obtain a solution using graphing, the method is not very practical. In this lesson, we will focus on another method to solve a linear system known as "substitution". The substitution method is most useful when one of the coefficients for one of the variables is either 1 or -1.

Solving a Linear System using the Substitution Method

  • Solve either equation for one of the variables
    • Look for a variable with a coefficient of 1 or -1
  • Substitute in for the variable in the other equation
    • The result will be a linear equation in one variable
  • Solve the linear equation in one variable
      This will give us one of our unknowns
  • Plug in for the known variable in either original equation, then solve for the other unknown
  • Check the result
    • Plug in for x and y in each original equation
Let's look at a few examples.
Example 1: Solve each linear system using substitution
x - 2y = 2
2x - y = -5
First, let's label our equations as equation 1 and equation 2:
1) x - 2y = 2
2) 2x - y = -5
Step 1) Solve either equation for one of the variables, we want to look for a variable with a coefficient of 1 or -1. In this case, we have 1x that appears in equation 1 and -1y that appears in equation 2. Since its easier, let's solve equation 1 for x.
x - 2y = 2
x = 2y + 2
Step 2) Substitute in for the variable in the other equation.
The key here is to understand the meaning of an equality "=". Since we solved equation 1 for x, we can say that x is equal to or x is the same as the quantity (2y + 2). This means we can plug in (2y + 2) for x in equation 2.
2x - y = -5
2(2y + 2) - y = -5
4y + 4 - y = -5
3y = -9
Step 3) Solve the linear equation in one variable
3y = -9
y = -3
Step 4) Plug in for the known variable in either original equation. At this point, we know that y is -3. We can plug in a -3 for y in either equation 1 or 2. Let's use equation 1 since it is simpler.
x - 2y = 2
x - 2(-3) = 2
x + 6 = 2
x = -4
Since x is -4 and y is -3, we can write our solution as the ordered pair (-4,-3).
Step 5) Check:
Plug in a -4 for each x and a -3 for each y in the original equations.
x - 2y = 2
-4 - 2(-3) = 2
-4 + 6 = 2
2 = 2
2x - y = -5
2(-4) - (-3) = -5
-8 + 3 = -5
-5 = -5
Example 2: Solve each linear system using substitution
-8x + y = -4
-4x - 5y = 20
First, let's label our equations as equation 1 and equation 2:
1) -8x + y = -4
2) -4x - 5y = 20
Step 1) Solve either equation for one of the variables, we want to look for a variable with a coefficient of 1 or -1. In this case, we have 1y that appears in equation 1. We will solve equation 1 for y.
-8x + y = -4
y = 8x - 4
Step 2) Substitute in for the variable in the other equation.
The key here is to understand the meaning of an equality "=". Since we solved equation 1 for y, we can say that y is equal to or y is the same as the quantity (8x - 4). This means we can plug in (8x - 4) for y in equation 2.
-4x - 5y = 20
-4x - 5(8x - 4) = 20
-4x - 40x + 20 = 20
-44x = 0
Step 3) Solve the linear equation in one variable
-44x = 0
x = 0
Step 4) Plug in for the known variable in either original equation. At this point, we know that x is 0. We can plug in a 0 for x in either equation 1 or 2. Let's use equation 1 since it is simpler.
-8x + y = -4
-8(0) + y = -4
0 + y = -4
y = -4
Since x is 0 and y is -4, we can write our solution as the ordered pair (0,-4).
Step 5) Check:
Plug in a 0 for each x and a -4 for each y in the original equations.
-8x + y = -4
-8(0) + (-4) = -4
-4 = -4
-4x - 5y = 20
-4(0) - 5(-4) = 20
20 = 20
Example 3: Solve each linear system using substitution
8x - 3y = -1
-2x - 5y = -17
First, let's label our equations as equation 1 and equation 2:
1) 8x - 3y = -1
2) -2x - 5y = -17
Step 1) Solve either equation for one of the variables, we want to look for a variable with a coefficient of 1 or -1. In this case, neither equation has a variable with a coefficient of 1 or -1. Let's solve equation 2 for x.
-2x - 5y = -17
-2x = 5y - 17
x = (-5/2)y + 17/2
Step 2) Substitute in for the variable in the other equation.
The key here is to understand the meaning of an equality "=". Since we solved equation 2 for x, we can say that x is equal to or x is the same as the quantity ([-5/2]y + 17/2). This means we can plug in ([-5/2]y + 17/2) for x in equation 1.
8x - 3y = -1
8([-5/2]y + 17/2) - 3y = -1
-20y + 68 - 3y = -1
-23y = -69
Step 3) Solve the linear equation in one variable
-23y = -69
y = 3
Step 4) Plug in for the known variable in either original equation. At this point, we know that y is 3. We can plug in a 3 for y in either equation 1 or 2. Let's use equation 1 since it is simpler.
8x - 3y = -1
8x - 3(3) = -1
8x - 9 = -1
8x = 8
x = 1
Since x is 1 and y is 3, we can write our solution as the ordered pair (1,3).
Step 5) Check:
Plug in a 1 for each x and a 3 for each y in the original equations.
8x - 3y = -1
8(1) - 3(3) = -1
-1 = -1
-2x - 5y = -17
-2(1) - 5(3) = -17
-17 = -17

Special Case Linear Systems

If both variables drop out when solving a system of linear equations:
  • There is no solution when the remaining statement is false
  • There are infinitely many solutions when the remaining statement is true

Linear Systems with No Solution

In some cases, we will not have a solution for our linear system. This will occur when we have two parallel lines. This type of system is known as an "inconsistent system". If we are solving our system using the substitution method, we will notice that our variables disappear and we are left with a false statement. Let's look at an example.
Example 4: Solve each linear system using substitution
2x + 7y = -2
-6x - 21y = -6
First, let's label our equations as equation 1 and equation 2:
1) 2x + 7y = -2
2) -6x - 21y = -6
We should notice that the second equation can be made more simple by dividing each side by -3:
2) 2x + 7y = 2
This should immediately flag a problem since the left side of equation 1 and equation 2 are identical and the right sides are not. Let's go through our normal steps.
Step 1) Solve either equation for one of the variables, we want to look for a variable with a coefficient of 1 or -1. In this case, neither equation has a variable with a coefficient of 1 or -1. Let's solve equation 1 for x.
2x + 7y = -2
2x = -7y - 2
x = [-7/2]y - 1
Step 2) Substitute in for the variable in the other equation.
The key here is to understand the meaning of an equality "=". Since we solved equation 1 for x, we can say that x is equal to or x is the same as the quantity ([-7/2]y - 1). This means we can plug in ([-7/2]y - 1) for x in equation 2.
2x + 7y = 2
2([-7/2]y - 1) + 7y = 2
-7y - 2 + 7y = 2
-2 = 2 (false)
Our variable has dropped out and we have a false statement, this tells us we have an "inconsistent system". We will state our answer as "no solution".

Linear Systems with Infinitely Many Solutions

Another special case scenario occurs when the same equation is presented twice as a system of equations. In this case, what works as a solution for one equation works as a solution to the other. These equations are known as "dependent equations". Let's look at an example.
Example 5: Solve each linear system using substitution
6x - 2y = 34
3x - y = 17
First, let's label our equations as equation 1 and equation 2:
1) 6x - 2y = 34
2) 3x - y = 17
We should be able to notice that multiplying equation 2 by 2 yields equation 1. This means we have the same equation. When you notice this, stop and give the answer of "infinitely many solutions". To see this using the substitution technique, let's use our normal steps.
Step 1) Solve either equation for one of the variables, we want to look for a variable with a coefficient of 1 or -1. In this case, we have -1y that appears in equation 2. We will solve equation 2 for y.
3x - y = 17
-y = -3x + 17
y = 3x - 17
Step 2) Substitute in for the variable in the other equation.
The key here is to understand the meaning of an equality "=". Since we solved equation 2 for y, we can say that y is equal to or y is the same as the quantity (3x - 17). This means we can plug in (3x - 17) for y in equation 1.
6x - 2y = 34
6x - 2(3x - 17) = 34
6x - 6x + 34 = 34
34 = 34 (true)
Our variable has dropped out and we have a true statement, this tells us we have "dependent equations". We will state our answer as "infinitely many solutions".

Solving Linear Systems Using Elimination

The goal of the elimination method is to eliminate one of the variables and obtain a linear equation in one variable. This allows us to get a solution for one of the variables. We can then use substitution to find the other unknown. The elimination method works best when one pair of variable terms are opposites, this means they will have opposite coefficients.

Solving a Linear System using the Elimination Method

  • Place both equations in standard form
    • ax + by = c
  • Transform one or both equations in such a way that one pair of variable terms are opposites
    • This means the variables will have opposite coefficients
  • Add the left sides of the equations and set this equal to the sum of the right sides, the result is a linear equation in one variable
    • This is legal since we are adding equal quantities to both sides of an equation
    • If a = b and c = d, then a + c = b + d
  • Solve the equation and find the other unknown using substitution
  • Check
      Plug in for x and y in both original equations
Let's look at a few examples.
Example 1: Solve each linear system using elimination
-5x + 3y = 18
5x - 2y = -12
Step 1) Place both equations in standard form.
In this case, both equations are in standard form. We will label our equations as 1 and 2.
1) -5x + 3y = 18
2) 5x - 2y = -12
Step 2) Transform one or both equations in such a way that one pair of variable terms are opposites.
In this case, we have -5x and 5x, which are opposites.
Step 3) Add the left sides of the equations and set this equal to the sum of the right sides. $$-5x + 3y=-12$$ $$\underline{+5x - 2y=-12}$$ The -5x and 5x will cancel: $$\require{cancel}\cancel{-5x}+ 3y=18$$ $$\underline{\cancel{+5x}- 2y=-12}$$ $$\hspace{4em}y=-24$$ On the left, we are left with (3y) + (-2y), which is just (y). On the right, we have (18) + (-12), which is (6).
Step 4) Solve the equation and find the other unknown.
In this case, we already know that y is 6. Let's plug this in for y in one of the original equations. Let's choose to plug in for y in equation 1.
-5x + 3y = 18
-5x + 3(6) = 18
-5x + 18 = 18
-5x = 0
x = 0
Our solution is the ordered pair (0,6).
Step 5) Check.
-5x + 3y = 18
-5(0) + 3(6) = 18
18 = 18
5x - 2y = -12
5(0) - 2(6) = -12
-12 = -12
Example 2: Solve each linear system using elimination
4x + 10y = -22
5y = 5x - 25
Step 1) Place both equations in standard form.
We will also label our equations as 1 and 2.
1) 4x + 10y = -22
2) -5x + 5y = -25
Step 2) Transform one or both equations in such a way that one pair of variable terms are opposites.
In this case, we can multiply both sides of equation 2 by (-2). This will give us (10y) in equation 1 and (-10y) in equation 2.
2) 10x - 10y = 50
Step 3) Add the left sides of the equations and set this equal to the sum of the right sides. $$\hspace{.5em}4x + 10y=-22$$ $$\underline{10x - 10y=50}$$ The 10y and (-10y) will cancel: $$\hspace{.5em}4x + \cancel{10y}=-22$$ $$\underline{10x - \cancel{10y}=50}$$ $$14x \hspace{3em}=28$$ On the left, we are left with (10x) + (6x), which gives us (14x). On the right, we have (-22) + (50), which is (28).
This leaves us with the equation:
14x = 28
Step 4) Solve the equation and find the other unknown.
14x = 28
x = 2
Let's plug a (2) in for x in equation 1.
4x + 10y = -22
4(2) + 10y = -22
(8) + 10y = -22
10y = -22 - 8
10y = -30
y = -3
Our solution is the ordered pair (2,-3).
Step 5) Check.
4x + 10y = -22
4(2) + 10(-3) = -22
8 + (-30) = -22
-22 = -22
5y = 5x - 25
5(-3) = 5(2) - 25
-15 = 10 - 25
-15 = -15
Example 3: Solve each linear system using elimination
-9y = 2x - 16
-7x = 22 + 12y
Step 1) Place both equations in standard form.
We will also label our equations as 1 and 2.
1) -2x - 9y = -16
2) -7x - 12y = 22
Step 2) Transform one or both equations in such a way that one pair of variable terms are opposites.
We can multiply equation 1 by (7) and equation 2 by (-2). This will give us (-14x) in equation 1 and (14x) in equation 2.
1) -14x - 63y = -112
2) 14x + 24y = -44
Step 3) Add the left sides of the equations and set this equal to the sum of the right sides. $$-14x - 63y=-112$$ $$\underline{+14x + 24y=-44}$$ The -14x and 14x will cancel: $$\cancel{-14x}- 63y=-112$$ $$\underline{\cancel{+14x}+ 24y=-44}$$ $$\hspace{3.2em}-39y=-156$$ On the left, we are left with (-63y) + (24y), which is (-39y). On the right, we have (-112) + (-44), which is (-156).
This leaves us with the equation:
-39y = -156
Step 4) Solve the equation and find the other unknown.
-39y = -156
y = 4
Let's plug a 4 in for y in equation 1.
-2x - 9y = -16
-2x - 9(4) = -16
-2x - 36 = -16
-2x = 20
x = -10
Our solution is the ordered pair (-10,4).
Step 5) Check.
-9y = 2x - 16
-9(4) = 2(-10) - 16
-36 = -20 - 16
-36 = -36
-7x = 22 + 12y
-7(-10) = 22 + 12(4)
70 = 22 + 48
70 = 70

Special Case Linear Systems

At this point, we have run across special case scenarios while using graphing and substitution methods. The majority of our problems with linear systems will have exactly one solution. This type of system is said to be "consistent", with equations that are "independent". When we have a system that contains two parallel lines, there will not be a solution. This type of system is said to be "inconsistent". Additionally, we will see systems that contain the same equation. These equations are said to be "dependent". For this type of system, there are infinitely many solutions.

Linear Systems with No Solution

When using the elimination method, if both are variables are eliminated and we are left with a false statement, we know we have an inconsistent system. Our answer will be stated as "no solution" or ∅. Let's look at an example.
Example 4: Solve each linear system using elimination
16 = 55x + 10y
-22x - 4y = -4
Step 1) Place both equations in standard form.
We will also label our equations as 1 and 2.
1) 55x + 10y = 16
2) -22x - 4y = -4
Step 2) Transform one or both equations in such a way that one pair of variable terms are opposites.
We can multiply equation 1 by (2) and equation 2 by (5). This will give us (20y) in equation 1 and (-20y) in equation 2.
1) 110x + 20y = 32
2) -110x - 20y = -20
We should be able to see a problem. Both variables are set up such that they have opposite coefficients. If we add the left sides together and set this equal to the sum of the right sides, we will get a false statement.
Step 3) Add the left sides of the equations and set this equal to the sum of the right sides. $$+110x + 20y=32$$ $$\underline{-110x - 20y=-20}$$ The 110x and (-110x) will cancel and the 20y and (-20y) will also cancel: $$\cancel{+110x}+ \cancel{+20y}=32$$ $$\cancel{-110x}+ \cancel{-20y}=-20$$ $$\hspace{6.65em}0=12$$ On the left, we are left with (0). On the right, we have (32) + (-20), which is (12).
0 = 12 (false)
We can stop and report our answer as "no solution".

Solving Linear Systems with Infinitely Many Solutions

When using the elimination method, if both are variables are eliminated and we are left with a true statement, we know we have dependent equations. Our answer will be stated as "infinitely many solutions". Let's look at an example.
Example 5: Solve each linear system using elimination
15y + 3 = -24x
-150y - 30 = 240x
Step 1) Place both equations in standard form.
We will also label our equations as 1 and 2.
1) 24x + 15y = -3
2) -240x - 150y = 30
At this point, we should be able to tell that multiplying equation 1 by (-10) will give us equation 2. Therefore, both equations of the system are the same and we will have infinitely many solutions. To show the full process, let's continue with the steps.
Step 2) Transform one or both equations in such a way that one pair of variable terms are opposites.
We can multiply equation 1 by (10). This will give us (240x) and (-240x).
1) 240x + 150y = -30
2) -240x - 150y = 30
Step 3) Add the left sides of the equations and set this equal to the sum of the right sides. $$+240x + 150y=-30$$ $$\underline{-240x - 150y=30}$$ The 240x and (-240x) will cancel and the 150y and (-150y) will also cancel: $$\cancel{+240x}+ \cancel{+150y}=-30$$ $$\cancel{-240x}+ \cancel{-150y}=30$$ $$\hspace{7.25em}0=0$$ On the left, we are left with (0). On the right, we have (30) + (-30), which is (0).
0 = 0 (true)
We can stop and report our answer as "infinitely many solutions".

Skills Check:

Example #1

Solve each system. $$4x - 3y=-4$$ $$7x + 5y=-7$$

Please choose the best answer.

A
$$(1,0)$$
B
$$(5,3)$$
C
$$(-1,0)$$
D
$$(0,4)$$
E
$$(2,3)$$

Example #2

Solve each system. $$-8x + 2y=-18$$ $$2x - 8y=-18$$

Please choose the best answer.

A
$$(2,3)$$
B
$$(0,7)$$
C
$$(-1,5)$$
D
$$(3,3)$$
E
$$(4,5)$$

Example #3

Solve each system. $$4x + 3y=-14$$ $$4x + 6y=4$$

Please choose the best answer.

A
$$(8,6)$$
B
$$(-8,6)$$
C
$$(-8,-6)$$
D
$$(6,-8)$$
E
$$(4,5)$$
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