Lesson Objectives

- Demonstrate an understanding of how to solve a system of linear equations in two variables using graphing
- Learn how to solve a system of linear equations in two variables using substitution
- Learn how to identify a system of linear equations in two variables with no solution
- Learn how to identify a system of linear equations in two variables with infinitely many solutions

## How to Solve a Linear System using Substitution

In our last lesson, we introduced the topic of solving linear systems in two variables. We learned how to solve a linear system using graphing. 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.

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 ✔

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".

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 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

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".

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