Solve systems of equations using substitution

When solving a system by graphing has several limitations. First, it requires the graph to be perfectly drawn, if the lines are not straight we may arrive at the wrong answer. Second, graphing is not a great method to use if the answer is really large, over 100 for example, or if the answer is a decimal the that graph will not help us find, 3.2134 for example. For these reasons we will rarely use graphing to solve our systems. Instead, an algebraic approach will be used.

The first algebraic approach is called substitution. We will build the concepts of substitution through several example, then end with a five-step process to solve problems using this method.

Also Read : Solve systems of equations by graphing and identifying the point of intersection

Also Read:

Example 1:

Solve the solution

x =5

y =2x -3

Solution :

We already knowx =5, substitute this into the other equation

y =2(5) -3

Evaluate, multiply first

y =10 – 3 = 7

the solution is (5, 7)

When we know what one variable equals we can plug that value (or expression) in for the variable in the other equation. It is very important that when we substitute, the substituted value goes in parenthesis. The reason for this is shown in the next example.

Also Read : Solve systems of equations using the addition/elimination method

Example 2:

Solve the solution

2x – 3y =7

y =3x – 7


We know y =3x -7, substitute this into the other equation

2x – 3(3x – 7) = 7

Solve this equation, distributing -3 first

2x -9x +21 =7

Combine like terms 2x – 9x

-7x + 21 = 7

Subtract 21

-7x = – 14

Divide by -7

x = 2

We now have our x, plug into the y = equation to find y

y = 3(2) – 7

Evaluate, multiply first

y = 6 – 7


y = -1

We now also have y, so the solution is (2,-1).

By using the entire expression 3x – 7 to replace y in the other equation we were able to reduce the system to a single linear equation which we can easily solve for our first variable. However, the lone variable (a variable without a coefficient) is not always alone on one side of the equation. If this happens we can isolate it by solving for the lone variable.

Example 3:


3x + 2y = 1

x – 5y = 6


Lone variable is x, isolate by adding 5y to both sides.

x – 5y = 6

x – 5y + 5y = 6 + 5y

x = 6 + 5y

Substitute this into the untouched equation

3(6 + 5y) + 2y = 1

Solve this equation, distributing 3 first

18 + 15y + 2y = 1

Combine like terms 15y + 2y

18 +17y = 1

Subtract 18 from both sides

17y = -17

Divide both sides by 17

y = -1

We have our y, plug this into the x = equation to find x

x = 6 + 5( -1)

Evaluate, multiply first

x =6 – 5


x = 1

We now also have x, so the solution is (1, -1)

The process in the previous example is how we will solve problems using substitution. This process is described and illustrated in the following table which lists the five steps to solving by substitution.

Sometimes we have several lone variables in a problem. In this case we will have the choice on which lone variable we wish to solve for, either will give the same final result.

Example 4:


x + y =5

x – y =-1

solution :

Find the lone variable: x or y in first, or x in second.

We will chose x in the first

x + y = 5

x = 5 – y

Plug into the untouched equation, the second equation

(5 – y) – y = -1

Solve, parenthesis are not needed here, combine like terms

5 – 2y = -1

Subtract 5 from both sides

-2y = -6

Divide both sides by -2

y = 3

We have our y!. Plug into lone variable equation, evaluate

x =5 – (3) = 2

x = 2

Now we have our x

Our Solution is (2,3).


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