In this explainer, we will learn how to solve systems of linear equations using substitution.

When we are asked to solve a system of equations, this means we are looking for a set of values for the variables that satisfy every equation. For example, consider the system of equations

We want to find a value for and a value for such that both equations hold true. In other words, we are looking for two values whose sum is 3 and whose difference is . We could do this by trial and error; however, this will not work for more complicated systems.

Instead, we will use the fact that we can solve any linear equation in one variable. This means if we can find a linear equation in either variable, we can solve for that value. To do this, we can note that both equations must hold true, so we can rearrange one equation to make one variable the subject. For example, we can rearrange the first equation by subtracting from both sides to get

Therefore, if and are solutions to the system of equations, they must also satisfy this equation. Since we have now written in terms of , and both equations must hold true, we can substitute this expression into the other equation to get

Distributing the negative over the parentheses and simplifying yields

Dividing the equation through by 2 gives

We can then substitute this value for into any of these equations to find the value of . Substituting into the first equation gives

Therefore, and solves the system of equations. We can verify that these values solve this system of equations by substituting the values into both equations.

Substituting and into the left-hand side of the first equation gives which is equal to the right-hand side.

Substituting and into the left-hand side of the second equation gives which is equal to the right-hand side. Since both equations hold true, we have confirmed this is a solution to the system of equations.

This method of solving equations is called substitution, since we substitute a rearrangement of one equation into the other. It is worth noting that our choice of substitution does not matter; we could have rearranged the first equation to make the subject or rearranged the second equation for or . In all of these cases, we would arrive at the same solution.

We can generalize this method to attempt to solve any system of two linear equations in two unknowns.

### How To: Solving a System of Linear Equations by Substitution

To solve a system of linear equations using substitution, we use the following method:

- Rearrange one of the equations to make one of the unknowns the subject.
- Substitute this into the other equation and solve the resulting linear equation in one unknown.
- Substitute the value of this unknown into one of the equations and solve for the remaining unknown.
- Verify the answer by checking that the values of both unknowns satisfy the other equation.

Letβs see an example of applying this process to solve a system of two linear equations in two unknowns.

### Example 1: Finding the Value of One Variable in a System of Linear Equations

Find given and .

### Answer

We are asked to find the value of that solves two linear equations in two unknowns. We recall we can do this by substitution. Usually, we would start by rearranging an equation to make a variable the subject; however, we can note that the second equation, , is already in this form. We can now substitute this expression for into the first equation to get

Simplifying, we get

Then, we divide the equation through by to get

In our next example, we will need to solve a system of two linear equations in two unknowns where we must first rearrange one of the equations to make a variable the subject.

### Example 2: Solving Systems of Linear Equations Using Substitution

Solve the following system of equations:

### Answer

We are asked to solve a system of two linear equations in two unknowns and we recall that we can do this by using substitution. We first need to rearrange one of the equations so that a variable is the subject. We can note that none of the equations are already in this form, so we can choose any equation and variable we wish; we will rearrange the second equation for .

We subtract from both sides of the equation to get

Then, we divide the equation through by 4 to get

Now, we can substitute this expression for into the first equation to construct an equation entirely in terms of . We have

We now distribute 5 over the parentheses to get

Simplifying gives

Dividing the equation through by yields

We can now determine the value of by substituting into the first equation; we get

We then add 2 to both sides of the equation, getting

Finally, we divide the equation through by 5 to get

So, and is the solution to this system of equations.

We can verify this solution by substituting both values into the two equations to check if they hold true.

Substituting and into the left-hand side of the first equation gives

This is equal to the right-hand side of the equation, so the solution satisfies the first equation.

Substituting and into the left-hand side of the second equation gives

This is equal to the right-hand side of the equation, so the solution satisfies the second equation.

This confirms that , is the solution to this system of equations.

### Example 3: Solving Simultaneous Equations by Substitution

Use substitution to solve the simultaneous equations

### Answer

To use substitution to solve a system of equations, we first need to rearrange one of the equations to make a variable the subject. In this case, we can notice that the first equation already has as the subject, so we solve the system of equations by substituting this expression for into the second equation. This gives

Distributing over the parentheses gives us

Collecting like terms and rearranging then gives

We can then divide the equation through by to get

We can then determine the value of by substituting into the first equation; we get

We can verify this solution by substituting into the second equation; we get

Subtracting 12 from both sides of the equation gives

Dividing both sides of the equation by yields

Since this agrees with the other value of , we have confirmed this is a solution to the system of equations.

Hence, the solution to the equations is and .

### Example 4: Writing and Solving a System of Linear Equations in Two Unknowns

A manβs age is 9 more than 2 times his sonβs age. Given that the sum of their ages is 57, find each of their ages.

### Answer

Letβs start by converting the information we are given into equations. Letβs call the age of the man and the age of his son . We are told that the manβs age is 9 more than 2 times his sonβs age, so if we double the sonβs age and add 9, we must have the manβs age. We can write this as the equation

We are also told that the sum of their ages is 57, so

This is a system of two linear equations in two unknowns, so we can attempt to solve this by substitution. We will substitute into the second equation to get

We can then simplify to get

We can subtract 9 from both sides of the equation to yield

Finally, we divide through by 3, giving us

Hence, the son is 16 years old. We can determine the age of the man by using either equation. We substitute into the first equation and evaluate to get

So, their ages are 16 years and 41 years.

Thus far, all of our systems of equations have had a unique solution. However, this will not always be the case. In fact, there are two other possibilities for these systems of two linear equations in two unknowns.

First, it is possible that the system will not have any solutions; when this happens, we call the system inconsistent. To see an example of this, letβs consider the system of equations

We can immediately notice there is a problem with this system since we are looking for two numbers that add to give 1 and add to give 2, which is not possible. However, letβs try solving this by substitution to see what occurs.

We can rearrange the first equation to make the subject:

We can then substitute this expression for into the second equation to get

Simplifying then yields

Of course, we know that 1 is not equal to 2. In fact, this means that our original assumption is wrong: there are no values of and that solve both equations, since we cannot choose values of and to make 1 equal to 2.

Second, it is possible that the system will have an infinite number of solutions. To see an example of this, letβs consider the system of equations

Once again, letβs attempt to solve this system by using the substitution method. We can rearrange the first equation to make the subject by adding to both sides, giving us

We can then substitute this expression for into the second equation to get

Distributing the factor of 2 over the parentheses gives

This simplifies to give

At first glance, we can see that this equation is trivial; we know that this is a true statement. However, we can also say that this equation is true for any value of ; this tells us that any value of can be a solution to this system of equations. We can confirm this by choosing a few values of .

Letβs consider ; we have

So, and is a solution to this system.

Letβs also consider ; we have

So, and is a solution to this system.

We can note that any solution to the equation is a solution to the entire system of equations. We can show why this is true by taking a factor of 2 out of the second equation:

In other words, we have shown that the second equation is a scalar multiple of the first equation; we call systems in this form dependent systems.

Letβs now see an example where we need to either solve a system of two linear equations in two unknowns or show that there are no solutions.

### Example 5: Solving a System of Linear Equations Using Substitution If the Solution Exists

Solve the following system of equations if possible:

### Answer

We are asked to solve a system of two linear equations in two unknowns and we recall that we can attempt to do this by using substitution. We first need to rearrange one of the equations so that a variable is the subject. We can note that both equations are already in this form, so we can just equate the two expressions for ; we have

We subtract from both sides of the equation to get

There are no values of or that can make this equation true, so we can conclude that there are no solutions to this system of equations.

In our next example, we will determine the number of solutions to a system of two linear equations in two unknowns.

### Example 6: Finding the Number of Solutions to a System of Equations

Find the number of solutions to the following system of equations:

### Answer

We are asked to solve a system of two linear equations in two unknowns and we recall that we can attempt to do this by using substitution. We first need to rearrange one of the equations so that a variable is the subject. Since the coefficient of in the first equation is 1, we will rearrange the first equation to make the subject.

We subtract from both sides of the first equation to get

We can then substitute this expression for into the second equation:

Distributing the factor of 2 over the parentheses gives

Simplifying then yields

Since this equation is true for any value of , we have that any value of gives a solution to this system of equations.

Hence, there are infinite solutions to this system of equation.

Letβs finish by recapping some of the important points of this explainer.

### Key Points

- To solve a system of linear equations using substitution, we use the following method:
- Rearrange one of the equations to make one of the unknowns the subject.
- Substitute this into the other equation and solve for the unknown.
- Substitute the value of this unknown into one of the equations and solve for the remaining unknown.
- Check that the values of both unknowns satisfy the other equation.

- Solving a system of equations by using substitution gives us exact solutions.
- We can verify our solutions by substituting them back into the system of equations to check that the equations hold.
- Not all systems of equations will have solutions. If we apply the substitution method and end up with an equation that is not true, then there are no solutions to the system of equations and we call this an inconsistent system of equations.
- Systems of two equations in two unknowns can also have an infinite number of solutions. If we apply the substitution method and end up with an equation that is always true, then there are an infinite number of solutions to the equation; these occur when the equations are scalar multiples of each other. We call these dependent systems of equations.