Solving systems of equations by substitution answers

To solve systems using substitution, follow this procedure:

• Select one equation and solve it for one of its variables.
• In the other equation, substitute for the variable just solved.
• Solve the new equation.
• Substitute the value found into any equation involving both variables and solve for the other variable.
• Check the solution in both original equations.

Usually, when using the substitution method, one equation and one of the variables leads to a quick solution more readily than the other. That's illustrated by the selection of x and the second equation in the following example. 

Example 1

Solve this system of equations by using substitution.

Solve for x in the second equation. 

Substitute

Solve this new equation.

Substitute the value found for y into any equation involving both variables. 

Check the solution in both original equations.

The solution is x = 1, y = –2. 

If the substitution method produces a sentence that is always true, such as 0 = 0, then the system is dependent, and either original equation is a solution. If the substitution method produces a sentence that is always false, such as 0 = 5, then the system is inconsistent, and there is no solution.

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 𝑥 +𝑦=3,𝑥−𝑦=−1.

We want to find a value for 𝑥 and a value for 𝑦 such that both equations hold true. In other words, we are looking for two values whose sum is 3 and whose difference is −1. 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 𝑥 from both sides to get 𝑦=3−𝑥.

Therefore, if 𝑥 and 𝑦 are solutions to the system of equations, they must also satisfy this equation. Since we have now written 𝑦 in terms of 𝑥, and both equations must hold true, we can substitute this expression into the other equation to get 𝑥−𝑦=−1𝑥−(3−𝑥)=−1.

Distributing the negative over the parentheses and simplifying yields 𝑥−3+𝑥=−12 𝑥−3=−12𝑥=2.

Dividing the equation through by 2 gives 𝑥=1.

We can then substitute this value for 𝑥 into any of these equations to find the value of 𝑦. Substituting 𝑥=1 into the first equation gives 1+𝑦=3𝑦=2.

Therefore, 𝑥=1 and 𝑦=2 solves the system of equations. We can verify that these values solve this system of equations by substituting the values into both equations.

Substituting 𝑥=1 and 𝑦=2 into the left-hand side of the first equation gives 𝑥+𝑦=1+ 2=3, which is equal to the right-hand side.

Substituting 𝑥=1 and 𝑦=2 into the left-hand side of the second equation gives 𝑥−𝑦=1−2=−1, 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 𝑥 the subject or rearranged the second equation for 𝑥 or 𝑦. 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:

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

Let’s 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 𝑥 given 2𝑥−𝑦=5 and 𝑦=7𝑥.

Answer

We are asked to find the value of 𝑥 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, 𝑦=7𝑥, is already in this form. We can now substitute this expression for 𝑦 into the first equation to get 2𝑥−(7𝑥)=5.

Simplifying, we get −5𝑥=5.

Then, we divide the equation through by −5 to get 𝑥=−1.

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: 5𝑥−2 𝑦=8,4𝑥+3𝑦=11.

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 𝑥.

We subtract 3𝑦 from both sides of the equation to get 4𝑥=11−3𝑦.

Then, we divide the equation through by 4 to get 𝑥=114−34𝑦.

Now, we can substitute this expression for 𝑥 into the first equation to construct an equation entirely in terms of 𝑦. We have 5114−34𝑦−2𝑦=8.

We now distribute 5 over the parentheses to get 554−154 𝑦−2𝑦=8.

Simplifying gives −154𝑦−84𝑦=8−554−234𝑦=8−554−234𝑦=324−554−234𝑦=−234.

Dividing the equation through by −234 yields 𝑦=1.

We can now determine the value of 𝑥 by substituting 𝑦=1 into the first equation; we get 5𝑥−2(1)=85𝑥−2=8.

We then add 2 to both sides of the equation, getting 5𝑥=10.

Finally, we divide the equation through by 5 to get 𝑥=2.

So, 𝑥=2 and 𝑦=1 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 𝑥=2 and 𝑦=1 into the left-hand side of the first equation gives 5𝑥−2𝑦=5(2)−2(1)=10−2=8.

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

Substituting 𝑥=2 and 𝑦=1 into the left-hand side of the second equation gives 4𝑥+3𝑦= 4(2)+3(1)=8+3=11.

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

This confirms that 𝑥=2, 𝑦=1 is the solution to this system of equations.

Example 3: Solving Simultaneous Equations by Substitution

Use substitution to solve the simultaneous equations 13𝑥+23=𝑦,6𝑥+35𝑦=645.

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 𝑦 as the subject, so we solve the system of equations by substituting this expression for 𝑦 into the second equation. This gives 6𝑥+3513𝑥+23=645.

Distributing over the parentheses gives us 6𝑥+15𝑥+25=645.

Collecting like terms and rearranging then gives 31𝑥5=625.

We can then divide the equation through by 315 to get 𝑥=2.

We can then determine the value of 𝑦 by substituting 𝑥=2 into the first equation; we get 13(2)+23=𝑦23+23=𝑦43=𝑦.

We can verify this solution by substituting 𝑥=2 into the second equation; we get 6(2)+35𝑦=64512+35𝑦=645.

Subtracting 12 from both sides of the equation gives 35𝑦=45.

Dividing both sides of the equation by 35 yields 𝑦=43.

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

Hence, the solution to the equations is 𝑥=2 and 𝑦=43.

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

A man’s age is 9 more than 2 times his son’s age. Given that the sum of their ages is 57, find each of their ages.

Answer

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

We are also told that the sum of their ages is 57, so 𝑠+𝑚=57.

This is a system of two linear equations in two unknowns, so we can attempt to solve this by substitution. We will substitute 𝑚=2𝑠+9 into the second equation to get 𝑠+(2𝑠+9)=57.

We can then simplify to get 3𝑠+9=57.

We can subtract 9 from both sides of the equation to yield 3𝑠=57−9=48.

Finally, we divide through by 3, giving us 𝑠=483=16.

Hence, the son is 16 years old. We can determine the age of the man by using either equation. We substitute 𝑠=16 into the first equation and evaluate to get 𝑚=2(16)+9=32+9=41.

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, let’s consider the system of equations 𝑥+𝑦=1𝑥+𝑦=2.

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, let’s try solving this by substitution to see what occurs.

We can rearrange the first equation to make 𝑦 the subject: 𝑦=1−𝑥.

We can then substitute this expression for 𝑦 into the second equation to get 𝑥+(1−𝑥)=2.

Simplifying then yields 1=2.

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 𝑥 and 𝑦 that solve both equations, since we cannot choose values of 𝑥 and 𝑦 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, let’s consider the system of equations 𝑥−𝑦=32𝑥−2𝑦=6.

Once again, let’s attempt to solve this system by using the substitution method. We can rearrange the first equation to make 𝑥 the subject by adding 𝑦 to both sides, giving us 𝑥=3+𝑦.

We can then substitute this expression for 𝑥 into the second equation to get 2(3+𝑦)−2𝑦=6.

Distributing the factor of 2 over the parentheses gives 6+2𝑦−2𝑦=6 .

This simplifies to give 6=6.

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 𝑦; this tells us that any value of 𝑦 can be a solution to this system of equations. We can confirm this by choosing a few values of 𝑦.

Let’s consider 𝑦=0; we have 𝑥−𝑦=3𝑥−0=3𝑥=3.

So, 𝑥=3 and 𝑦=0 is a solution to this system.

Let’s also consider 𝑦=1; we have 𝑥−𝑦=3𝑥−1=3𝑥 =4.

So, 𝑥=4 and 𝑦=1 is a solution to this system.

We can note that any solution to the equation 𝑥−𝑦=3 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: 2𝑥−2𝑦=62(𝑥−𝑦)=2×3𝑥−𝑦=3.

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.

Let’s 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: 𝑦=𝑥+1,𝑦=𝑥−9.

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 𝑦; we have 𝑥+1=𝑥−9.

We subtract 𝑥 from both sides of the equation to get 1=−9.

There are no values of 𝑥 or 𝑦 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: 𝑦+2𝑥=4,2𝑦+4𝑥=8.

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 𝑦 in the first equation is 1, we will rearrange the first equation to make 𝑦 the subject.

We subtract 2𝑥 from both sides of the first equation to get 𝑦=4−2𝑥.

We can then substitute this expression for 𝑦 into the second equation: 2(4−2𝑥)+4𝑥=8.

Distributing the factor of 2 over the parentheses gives 8−4𝑥+4𝑥=8 .

Simplifying then yields 8=8.

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

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

Let’s 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:
  1. Rearrange one of the equations to make one of the unknowns the subject.
  2. Substitute this into the other equation and solve for the unknown.
  3. Substitute the value of this unknown into one of the equations and solve for the remaining unknown.
  4. 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.