Lesson Objectives
  • Demonstrate an understanding of a System of Linear Equations in Two Variables
  • Demonstrate an understanding of how to Solve a Linear System using Graphing
  • Demonstrate an understanding of how to check the solution for a Linear System
  • Learn how to solve a System of Linear Equations in Two Variables using Substitution
  • Learn how to deal with special case scenarios: dependent equations and inconsistent systems

How to Solve a System of Linear Equations using Substitution


At this point, we should know that solving a system of linear equations in two variables involves finding the point that lies on both lines. In our last lesson, we learned how to solve a system of linear equations using the graphing method. Although the graphing method is a good way to visualize the concepts related to solving a linear system, it is not practical for many reasons. It is very hard to find a solution graphically for non-integer values or values that are very large or very small. In this lesson, we will learn a better algebraic method to solve a linear system in two variables. This method is known as substitution. It gets its name from the process of substituting in for one of the variables. Let’s take a look at the steps for the substitution method.

Substitution Method

  • Solve either equation for one of the variables
    • Look for a variable with a coefficient of 1 or -1
  • Substitute in for that variable in the other equation
  • Solve the linear equation in one variable
  • Substitute the known value into either original equation and solve for the other unknown
  • Check the solution
Let's look at a few examples.
Example 1: Solve each linear system using substitution
2x - 5y = 4
8x + y = 16
First and foremost, we will label our equations so we can refer to them:
1) 2x - 5y = 4
2) 8x + y = 16
Step 1) Solve either equation for one of the variables. Notice how the y variable in equation 2 has a coefficient of 1. This is the easiest case scenario. Let's solve equation 2 for y:
8x + y = 16
y = -8x + 16
Step 2) Substitute in for that variable in the other equation. Since y is equal to the quantity (-8x + 16), we can plug in (-8x + 16) for y in equation 1:
2x - 5y = 4
2x - 5(-8x + 16) = 4
Step 3) Notice how we have a linear equation in one variable. We can now solve this equation and find out one of our unknowns.
2x - 5(-8x + 16) = 4
2x + 40x - 80 = 4
42x = 84
x = 2
Step 4) Now that we know one of our unknowns (x = 2), we can plug in for x in either original equation. This will allow us to solve for the other unknown.
2(2) - 5y = 4
4 - 5y = 4
-5y = 0
y = 0
Our solution for the system is (2,0)
Step 5) Check by plugging in a 2 for each x and a 0 for each y in the original equations.
2(2) - 5(0) = 4
4 = 4
8(2) + 0 = 16
16 = 16
We can see that our point (2,0) works as a solution for each equation of the system.
Example 2: Solve each linear system using substitution
5x + 8y = 4
x + 5y = -6
First and foremost, we will label our equations so we can refer to them:
1) 5x + 8y = 4
2) x + 5y = -6
Step 1) Solve either equation for one of the variables. Notice how the x variable in equation 2 has a coefficient of 1. This is the easiest case scenario. Let's solve equation 2 for x:
x + 5y = -6
x = -5y - 6
Step 2) Substitute in for that variable in the other equation. Since x is equal to the quantity (-5y - 6), we can plug in (-5y - 6) for x in equation 1:
5x + 8y = 4
5(-5y - 6) + 8y = 4
Step 3) Notice how we have a linear equation in one variable. We can now solve this equation and find out one of our unknowns.
5(-5y - 6) + 8y = 4
-25y - 30 + 8y = 4
-17y = 34
y = -2
Step 4) Now that we know one of our unknowns (y = -2), we can plug in for y in either original equation. This will allow us to solve for the other unknown.
Let's plug in a (-2) for y in equation 2:
x + 5(-2) = -6
x - 10 = -6
x = 4
Our solution for the system is (4,-2)
Step 5) Check by plugging in a 4 for each x and a -2 for each y in the original equations.
5(4) + 8(-2) = 4
20 - 16 = 4
4 = 4
4 + 5(-2) = -6
4 - 10 = -6
-6 = -6
We can see that our point (4,-2) works as a solution for each equation of the system.
Example 3: Solve each linear system using substitution
4x + 5y = 5
2x + 3y = 1
First and foremost, we will label our equations so we can refer to them:
1) 4x + 5y = 5
2) 2x + 3y = 1
Step 1) Solve either equation for one of the variables. In this case, we don't have any variables where the coefficient is 1 or (-1). Let's solve equation 1 for y:
4x + 5y = 5
5y = -4x + 5
y = -4/5 x + 1
Step 2) Substitute in for that variable in the other equation.
Since y is equal to the quantity (-4/5 x + 1), we can plug in (-4/5 x + 1) for y in equation 2:
2x + 3y = 1
2x + 3(-4/5 x + 1) = 1
Step 3) Notice how we have a linear equation in one variable. We can now solve this equation and find out one of our unknowns.
2x + 3(-4/5 x + 1) = 1
2x - 12/5 x + 3 = 1
10x - 12x + 15 = 5
-2x + 15 = 5
-2x = -10
x = 5
Step 4) Now that we know one of our unknowns (x = 5), we can plug in for y in either original equation. This will allow us to solve for the other unknown.
Let's plug in a 5 for x in equation 2:
2(5) + 3y = 1
10 + 3y = 1
3y = -9
y = -3
Our solution for the system is (5,-3)
Step 5) Check by plugging in a 5 for each x and a -3 for each y in the original equations.
4(5) + 5(-3) = 5
20 - 15 = 5
5 = 5
2(5) + 3(-3) = 1
10 - 9 = 1
1 = 1
We can see that our point (5,-3) works as a solution for each equation of the system.

Solving Inconsistent Systems with Substitution

We know that parallel lines will never intersect. When we run into a linear system with parallel lines, we will not have a solution. We refer to this type of system as an inconsistent system. When we are using the substitution method, we will know we have an inconsistent system when we obtain a false statement. Let's look at an example.
Example 4: Solve each linear system using substitution
3x = 15 - 6y
2x + 4y = 6
First and foremost, we will label our equations so we can refer to them:
1) 3x = 15 - 6y
2) 2x + 4y = 6
Step 1) Solve either equation for one of the variables. Let's solve equation 1 for x:
3x = 15 - 6y
x = 5 - 2y
Step 2) Substitute in for that variable in the other equation. Since x is equal to the quantity (5 - 2y), we can plug in (5 - 2y) for x in equation 2:
2x + 4y = 6
2(5 - 2y) + 4y = 6
Step 3) Notice how we have a linear equation in one variable. We can now solve this equation and find out one of our unknowns.
2(5 - 2y) + 4y = 6
10 - 4y + 4y = 6
10 = 6 (false)
Notice how our variable dropped out and we are left with a false statement. This means we have an inconsistent system. For our answer, we simply state "no solution".

Dependent Equations - Substitution Method

Lastly, we come across systems which have the same equation listed twice. The equation is algebraically manipulated to look different. For these problems, we will see the variable drop out and we will be left with a true statement. Let's look at an example.
Example 5: Solve each linear system using substitution
-18x + 3y = -3
-6x + y = -1
First and foremost, we will label our equations so we can refer to them:
1) -18x + 3y = -3
2) -6x + y = -1
Step 1) Solve either equation for one of the variables. Let's solve equation 2 for y:
-6x + y = -1
y = 6x - 1
Step 2) Substitute in for that variable in the other equation. Since y is equal to the quantity (6x - 1), we can plug in (6x - 1) for y in equation 1:
-18x + 3y = -3
-18x + 3(6x - 1) = -3
Step 3) Notice how we have a linear equation in one variable. We can now solve this equation and find out one of our unknowns.
-18x + 3(6x - 1) = -3
-18x + 18x - 3 = -3
-3 = -3 (true)
In this case, the variable is gone and we are left with a true statement. This means we have dependent equations. For this scenario, we can state our answer as "infintely many solutions" or an "infinite number of solutions".