Solving Systems of Equations By Substitution

After studying this lesson, you will be able to:

• Solve systems of equations by substitution.

To Solve a System of Equations by Substitution:

1. Solve either equation for a variable. (Hint: Try to solve for a variable that does not have a coefficient.)

2. Substitute on equation into the other equation. This will eliminate one variable.

3. Solve for the remaining variable.

4. Substitute the solution into the other equation and solve for the other variable.

Example 1

Solve x + y = 6, x = y + 2

1 st : The second equation is already solved for a variable so we don't have to do anything in this step.

2 nd : Now we substitute x = y + 2 into the first equation. Since we know that x is the same as y + 2, we can substitute y + 2 for x in the first equation:

( y + 2 ) + y = 6

3 rd : Solve for y which is the remaining variable:

2y + 2 = 6

2y = 4

y = 2

4 th : Now we substitute the solution ( y = 2 ) into the second equation:

x = y + 2

x = (2) + 2

x = 4

The solution is (4, 2)

Example 2

Solve 2x + y = 13, 4x -3y = 11

1 st : We need to solve one equation for a variable. Let's solve the first equation for y since that is the variable without a coefficient:

2x + y = 13

y = 13 - 2x

2 nd : Now we substitute y = 13 - 2x into the second equation. Since we know that y is the same as 13 - 2x, we can substitute 13 - 2x for y in the second equation:

4x - 3y = 11

4x - 3 (13 - 2x ) = 11 (making the substitution)

3 rd : Solve for x which is the remaining variable:

4x - 3 (13 - 2x ) = 11

4x - 39 + 6x = 11

10x -39 = 11

10x = 50

x = 5

4 th : Now we substitute the solution ( x = 5 ) into the first equation:

2x + y = 13

2 (5) + y = 13

10 + y = 13

y = 3

The solution is (5, 3)

Special Cases: If you end up with a system where all variables cancel out, you have what we might call a "special case".

If the statement you're left with is true, the solution will be

If the statement you're left with is false, the solution will be Ø

Example 3

Solve x - 3y = -6, x - 3y = 6

1 st : We need to solve one equation for a variable. Let's solve the first equation for x since that is the variable without a coefficient:

x - 3y = -6

x = -6 + 3y

2 nd : Now we substitute x = -6 + 3y into the second equation.

x - 3y = 6

(-6 + 3y) -3y = 6 (making the substitution)

3 rd : Solve for y which is the remaining variable:

(-6 + 3y) -3y = 6

-6 = 6 All variables cancelled out so we have a special case.

Since the remaining statement is false, the solution is

Example 4

Solve x + 2y = 5, 3x - 15 = -6y

1 st : We need to solve one equation for a variable. Let's solve the first equation for x since that is the variable without a coefficient:

x + 2y = 5

x = 5 - 2y

2 nd : Now we substitute x = 5 - 2y into the second equation.

3x - 15 = -6y

3 (5 - 2y) - 15 = -6y (making the substitution)

3 rd : Solve for y which is the remaining variable:

3 (5 - 2y) - 15 = -6y

15 - 6y -15 = -6y

-6y = -6y

-6 = -6 All variables cancelled out so we have a special case.

Since the remaining statement is true, the solution is Ø