# Substitution Method (Systems of Linear Equations)

When two equations of a line intersect at a single point, we say that it has a unique solution which can be described as a point, [latex]\color{red}\left( {x,y} \right)[/latex], in the **XY-plane**.

The **substitution method** is used to solve systems of linear equations by finding the exact values of [latex]x[/latex] and [latex]y[/latex] which correspond to the point of intersection.

## A Diagram Showing Two Lines Intersecting at a Point

The diagram below illustrates two arbitrary lines showing where they cross path as described by the ordered pair [latex]\left( {x,y} \right)[/latex]. In this lesson, we are interested in manually solving for that common point.

### Examples of How to Solve Systems of Equations by Substitution Method

**Example 1:** Use the method of substitution to solve the system of linear equations below.

The idea is to pick one of the two given equations and solve for either of the variables, [latex]x[/latex] or [latex]y[/latex]. The result from our first step will be substituted into the other equation. The effect will be a single equation with one variable which can be solved as usual.

It totally depends on which equation you think will be much easier to deal with. The choice is yours.

Notice that the top equation contains a variable [latex]x[/latex] that is “alone” – meaning its coefficient is [latex]+1[/latex]. Remember to always look for this characteristic (an “alone” variable) because it will make your life much easier.

Now, I start by solving the top equation for [latex]x[/latex].

Since I know what [latex]x[/latex] is equal to in terms of [latex]y[/latex], I can plug this expression into the other equation. With this, I will end up solving an equation with a single variable.

Hopefully, you get the same value of [latex]y = – \,5[/latex]. Now that I know what the exact value of [latex]y[/latex] is, I will solve for the other variable (in this case, [latex]x[/latex]) by evaluating its value into any of the original two equations. It does not matter which original equation you pick because it will ultimately give the same answer.

However, I must say that the “best” route to solve for [latex]x[/latex] is to use the revised equation that I have previously solved since I have “[latex]x =[/latex] some [latex]y[/latex]”. Right?

Here I get [latex]x = 1[/latex]. In point notation form, the final answer can be written as [latex]\left( {1, – \,5} \right)[/latex]. Remember, this is the point at which the two lines intersect.

It is always a good habit to check those values into the original equations to verify if they are truly the correct answers. I suggest that you check them at all times.

Graphically, the solution looks like this.

**Example 2:** Use the method of substitution to solve the system of linear equations.

The obvious choice here is to pick the bottom equation because the variable [latex]y[/latex] has a coefficient of positive one [latex]\left( { + 1} \right)[/latex]. Now I can easily solve for [latex]y[/latex] in terms of [latex]x[/latex]. To start, I will subtract both sides by [latex]3x[/latex].

After solving for [latex]y[/latex] from the bottom equation, I now turn into the top equation and substitute the expression for [latex]y[/latex] in terms of [latex]x[/latex]. The result will be a multistep equation with a single variable.

Solve this equation by simplifying the parenthesis first. After that, combine like terms in both sides and isolate the variable to the left. Your solution should be similar below.

If you correctly solved for [latex]x[/latex], you should also arrive at the value [latex]x = 3[/latex].

Since the revised bottom equation is already written in the form that I like, I will use it to solve for the exact value of [latex]y[/latex].

With the obtained value, [latex]y = 1[/latex], I can now write the final answer as the ordered pair [latex]\left( {3,1} \right)[/latex].

As I mentioned earlier, always verify the final answers yourself to see if they check using the original equations.

In the graph, the solution is the point of intersection of the two given lines.

**Example 3:** Use the method of substitution to solve the system of equations.

This is a great example because I have two ways to approach the problem. The variables [latex]x[/latex] and [latex]y[/latex] both have positive one [latex]\left( { + 1} \right)[/latex] as their coefficients. This means I can go either way.

For this example, I will solve for [latex]y[/latex]. I can easily do it by subtracting both sides by [latex]x[/latex] and then rearrange.

Next, I will write down the other equation and replace its [latex]y[/latex] by [latex]y = – x + 3[/latex].

After solving the multi-step equation above, I get [latex]x = 5[/latex]. Now, I turn to the transformed version of the top equation to solve for [latex]y[/latex].

Here I get [latex]y = – \,2[/latex]. The final answer then is [latex]\left( {x,y} \right) = \left( {5, – \,2} \right)[/latex].

Indeed, the two lines intersect at the point we calculated!

**Example 4:** Use the method of substitution to solve the system of equations.

I find this problem interesting because I cannot find a situation where the variable is “alone”. Again, our definition of being “alone” is having a coefficient of [latex]+1[/latex]. Remember?

Both the top and bottom equations here contain a variable with a negative symbol. I suggest that whenever you see something like this, **change that negative symbol to **[latex]\textbf{- 1}[/latex]. I am placing a blue arrow right next to it for emphasis (see below).

From here, I can proceed to solve for [latex]y[/latex] using the top equation or for [latex]x[/latex] using the bottom. For this exercise, I will work on the bottom equation.

Notice that to solve for [latex]x[/latex], I divided the entire equation by [latex]- 1[/latex]. You can see here that the look of the equation changed drastically.

I hope you got [latex]y = – \,4[/latex] as well. Otherwise, check and recheck your steps in solving the multi-step equation.

Next, use that value of [latex]y[/latex] and substitute it into the transformed version of the bottom equation to solve for [latex]x[/latex].

So I get [latex]x = – \,2[/latex]. The final answer in ordered pair is [latex]\left( {x,y} \right) = \left( { – \,2, – \,4} \right)[/latex].

The graph agrees with us on where the two lines intersect. Great!

**Example 5: **Use the substitution method to solve the system of linear equations.

The first thing I observed here is that there is no case where the coefficient of the variable is either [latex] + 1[/latex] or [latex] – 1[/latex]. To some, this may look confusing.

In this problem, it is possible to isolate the [latex]y[/latex] on the top equation and do the same thing for [latex]x[/latex] at the bottom equation. Do some scratch work and it should make a lot more sense.

You will realize that either [latex]x[/latex] or [latex]y[/latex] can be solved easily because no fractions are generated in the process. For this exercise, I choose to deal with the top equation to solve for [latex]y[/latex].

As predicted, solving for [latex]y[/latex] came out nicely. Now, I will use this value for [latex]y[/latex] and substitute it into the [latex]y[/latex] of the bottom equation. Then, I will proceed with solving the resulting equation as usual.

If you did it correctly, your answer should come out as [latex]x = 2[/latex]. Plug this value of [latex]x[/latex] into the revised version of the top equation to solve for the exact value of [latex]y[/latex].

Here I got [latex]y = – \,5[/latex]. That makes our final answer as the ordered pair [latex]\left( {2, – \,5} \right)[/latex].

The graph confirms our calculated values for [latex]x[/latex] and [latex]y[/latex].

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Substitution Method Practice Problems with Answers