Chapter 2: Linear Functions

2.6.2: The Substitution Method

Learning Outcomes

  • Solve a system of linear equations using the substitution method
  • Write the general solution of a dependent system as an ordered pair in terms of one variable

The Substitution Method

Solving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method. We will consider two more methods of solving a system of linear equations that are more precise than graphing. One such method is solving a system of equations by the substitution method. 

Let’s assume we have a consistent system of two linear equations in two variables. This means that the lines representing this system intersect. At any point of intersection the [latex]x[/latex]– and [latex]y[/latex]– values satisfy both equations. So the [latex]x[/latex]-value in the first equation equals the [latex]x[/latex]-value in the second equation and the [latex]y[/latex]-value in the first equation equals the [latex]y[/latex]-value in the second equation. The assumption that an intersection point exists allows us to solve one of the equations for one variable and then substitute the result into the other equation to end up with a linear equation in one variable that can be solved for the second variable. Once the second variable is known, the first variable can be found by substituting the value of the second variable into either of the original equations in the system.

solving a system of linear equations: the substitution method

  1. Solve one of the two equations for one of the variables in terms of the other.
  2. Substitute the expression for this variable into the other original equation, and then solve for the remaining variable.
  3. Substitute that solution into either of the original equations to find the value of the other variable. If possible, write the solution as an ordered pair.
  4. Check the solution in both equations.

Example 1

Solve the following system of equations using the substitution method:

[latex]\begin{array}{l}-x+y=-5\hfill \\ \text{ }2x - 5y=1\hfill \end{array}[/latex]

Solution

First, we will solve the first equation for [latex]y[/latex] by adding [latex]x[/latex] to both sides of the equation:

[latex]\begin{equation}\begin{aligned}-x+y&=-5\\y&=x-5\end{aligned}\end{equation}[/latex]

Now, we can substitute the expression [latex]x - 5[/latex] for [latex]y[/latex] in the second equation:

[latex]\begin{equation}\begin{aligned}2x - 5\color{blue}{y}&=1 \\ 2x - 5\color{blue}{(x - 5)}&=1\\ 2x - 5x+25&=1 \\ -3x&=-24 \\ x=8 \end{aligned}\end{equation}[/latex]

Now, we substitute [latex]x=8[/latex] into the first equation and solve for [latex]y[/latex]:

[latex]\begin{equation}\begin{aligned}-\color{blue}{x}+y&=-5\\-\color{blue}{8}+y=-5 \\ y=3 \end{aligned}\end{equation}[/latex]

Our solution is [latex]\left(8,3\right)[/latex].

Check the solution by substituting [latex](8, 3)[/latex] into both equations.

[latex]\begin{equation}\begin{aligned}-x+y&=-5\\ -(8)+(3)&=-5 \;\;\;\;\;\;\text{True}\end{aligned}\end{equation}[/latex]

[latex]\begin{equation}\begin{aligned}2x - 5y&=1\\ 2(8)-5(3)&=1\;\;\;\;\;\;\;\text{True}\end{aligned}\end{equation}[/latex]

If we had chosen the other equation to start with in the previous example, we would still determine the same solution. It is really a matter of preference which equation to start with. Had we started with the second equation solving for a variable would have resulted in having to work with fractions, so starting with the first equation was, perhaps, the better choice. The substitution method can be used to solve any linear system in two variables, but the method works best if one of the equations contains a coefficient of 1 or [latex]–1[/latex] so that we do not have to deal with fractions.

The following video shows an example of solving a system of two equations using the substitution method.

Try It 1

Solve the following system of equations using the substitution method:

[latex]\begin{array}{l}x+2y=5\hfill \\ \text{ }x - 2y=1\hfill \end{array}[/latex]

Consider the system of equations:

[latex]\begin{array}{l}x=9 - 2y\hfill \\ x+2y=13\hfill \end{array}[/latex]

The first equation is already solved for [latex]x[/latex] so it makes sense to substitute [latex]9-2y[/latex] for [latex]x[/latex] in the second equation:

[latex]\begin{array}{r}\color{blue}{x}+2y=13\hfill \\ \color{blue}{\left(9 - 2y\right)}+2y=13\hfill \\ 9+0y=13\hfill \\ 9=13\hfill \end{array}[/latex]

Clearly, this statement is a contradiction because [latex]9\ne 13[/latex]. But what does this mean?

When we introduced the substitution method, we made an assumption: that the system had a solution.

SInce we have a contradiction, we are contradicting that assumption, so the system has no solution.

To double check, we rearrange the equations so that they are both in slope-intercept form and we will be able to tell whether we have parallel lines (meaning no solution) or coinciding lines (meaning infinite solutions on the line).

First equation:

[latex]\begin{array}{l}\text{ }x=9 - 2y\hfill \\ 2y=-x+9\hfill \\ \text{ }y=-\dfrac{1}{2}x+\dfrac{9}{2}\hfill \end{array}[/latex]

Second equation:

[latex]\begin{array}{l}x+2y=13\hfill \\ \text{ }2y=-x+13\hfill \\ \text{ }y=-\dfrac{1}{2}x+\dfrac{13}{2}\hfill \end{array}[/latex]

Comparing the equations, we see that the lines they represent have the same slope of [latex]-\frac{1}{2}[/latex] but different [latex]y[/latex]-intercepts. Therefore, the lines are parallel and do not intersect.

Contradictions

When we get a contradiction in an equation using the substitution method, it means the system has no solution.

Example 2

Solve the system of equations using the substitution method:

[latex]\begin{array}{l}y=-x+4\hfill \\ x+y=10\hfill \end{array}[/latex]

Solution

The first equation is already solved for [latex]y[/latex] so it makes sense to substitute [latex]-x+4[/latex] for [latex]y[/latex] in the second equation:

[latex]\begin{array}{r}x+\color{blue}{y}=10\hfill \\ x+\color{blue}{\left(-x+4\right)}=10\hfill \\ 4=10\hfill \end{array}[/latex]

Clearly, this statement is a contradiction because [latex]4\ne 10[/latex], so the system has no solution.

The next video shows another example of using the substitution method to solve a system that has no solution.

Try It 2

Solve the system of equations using the substitution method:

[latex]\begin{array}{l}4x=1 - 2y\hfill \\ 4x+2y=3\hfill \end{array}[/latex]

We have seen that the substitution method results in solving a linear equation in one variable. We have also seen that we can get a single solution when this equation leads to a numerical answer, or no solution when this equation leads to a contradiction (an equation that is never true). The only remaining option occurs when this equation turns into an identity (an equation that is always true) with the result of an infinite number of solutions that lie on the line of the original equation.

Consider the system of equations:

[latex]\begin{array}{l}3x=6 - y\hfill \\ 3x+y=6\hfill \end{array}[/latex]

To solve this system we would first rearrange one of the equations for a single variable. Let’s solve the 2nd equation for [latex]y[/latex] by subtracting [latex]3x[/latex] from both sides:  [latex]y=-3x+6[/latex]

Now we can substitute [latex]-3x+6[/latex] for [latex]y[/latex] in the 1st equation:

[latex]\begin{equation}\begin{aligned}3x&=6-\color{blue}{y}\\3x&=6-\color{blue}{(-3x+6)}\\ 3x&=6+3x-6\\3x&=3x\end{aligned}\end{equation}[/latex]

We have an identity. This means that the lines coincide. Consequently, there are infinite solutions that lie on the line with equation [latex]3x+y=6[/latex].

The next video shows that a system can have an infinite number of solutions.

Try It 3

Solve the system of equations using the substitution method:

[latex]\begin{array}{l}-7x=-5 + 2y\hfill \\ 14x+4y=10\hfill \end{array}[/latex]

TRY IT 4

TRY IT 5

TRY iT 6

The substitution method works well when one of the equations in the system can be solved relatively easily for a single variable. However, it can get a bit tricky if fractions are involved. In the next section, we will consider an alternative method that makes dealing with fractions in the equations a little easier.


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