Two or more equations considered simultaneously form a system of equations. A solution of a system of equations in two variables is an ordered pair (a,b) that satisfies all equations of the system.

One method of solving a system of equations is the substitution method. This method involves converting the system to one equation in one variable by an appropriate substitution. It is a particularly useful method when it is easy to solve one of the equations for one of the variables.

The Substitution Method

1. Solve one of the equations for one variable in terms of the other.

2. Substitute the expression found in step 1 into the other equation.

3. Solve the equation obtained in step 2.

4. Back-substitute the value found in step 3 into the equation from step 1 to find the

value of the remaining variable.

5. Check the solution in both of the system’s given equation.

Example. Solve the system:

2x – y = 9

3x – 4y = -24

Solution.

2x – y = 9 -y = 9 – 2x y = 2x - 9	Solve the first equation for y
3x – 4y = -24 3x – 4(2x – 9) = -24	Substitute the expression 2x – 9 for y in the second equation
3x – 8x + 36 = -24 -5x + 36 = -24 -5x = -60 x = 12	Solve for x
y = 2x – 9 y = 2(12) – 9 y = 24 – 9 y = 15	Back-substitute x = 12 into the expression found in step 1

The solution is (12, 15).

Check: We will substitute the solution into both of the original equations to check:

2x – y = 9 2(12) – 15 = 9 24 – 15 = 9 9 = 9

3x – 4y = -24 3(12) – 4(15) = -24 36 – 60 = -24 -24 = -24

Another method for solving systems of equations is the addition method, sometime called the elimination method.

The Addition/Elimination Method

1. If necessary, rewrite both equations in the form Ax + By = C.

2. If necessary, multiply either equation or both equations by appropriate numbers so that

the coefficients of x and y will be opposites with a sum of 0.

3. Add the equations in step 2. The sum is an equation in one variable.

4. Solve the equation from step 3.

5. Back-substitute the value obtained in step 4 into either of the given equations and

solve for the other variable.

6. Check the solution in both of the original equations.

Example. Solve the system:

3x = 33 – 5y

3y = 4x – 15

Solution. Rewrite both equations in the form Ax + By = C:

3x = 33 – 5y 3x + 5y = 33

3y = 4x – 15 -4x + 3y = -15

To eliminate x, multiply the first equation by 4 and the second equation by 3:

4(3x + 5y) = 4(33) 12x + 20y = 132

3(-4x + 3y) = 3(-15) -12x + 9y = -45

The coefficients of x are now opposites. Add the equations together; then solve for y:

12x + 20y = 132

-12x + 9y = -45

29y = 87

y = 3

Back-substitute y = 3 into the first given equation; then solve for x:

3x = 33 – 5y

3x = 33 – 5(3)

3x = 33 – 15

3x = 18

x = 6

Solution: (6, 3)

Check:

3x = 33 – 5y 3(6) = 33 – 5(3) 18 = 33 – 15 18 = 18

3y = 4x – 15 3(3) = 4(6) – 15 9 = 24 – 15 9 = 9

Inconsistent and Dependent System of Linear Equations

If both variables are eliminated when a system of linear equations is solved by substitution or addition/elimination:

1. There is no solution if the resulting statement is false. The system is inconsistent.

2. There are infinitely many solutions if the resulting statement is true. The system is

dependent.

Example. Solve:

8x – 2y = 8

12x – 3y = 12

Solution. Each equation is already written in the form Ax + By = C. So we will begin by multiplying the first equation by 3 and the second equation by -2:

8x – 2y = 8 3(8x – 2y) = 3(8) 24x – 6y = 24

12x – 3y = 12 -2(12x – 3y) = -2(12) -24x + 6y = -24

We will now add the two resulting equations:

24x – 6y = 24

-24x + 6y = -24

0 = 0

Both variables have been eliminated, and the resulting statement, 0 = 0, is true for all values of x and y. This means that the equations of the given system represent two different ways of writing the equation of the same line. All points on the line have coordinates that satisfy the system. We can use either equation to solve for y:

8x – 2y = 8

-2y = 8 – 8x

y = 4x – 4

Any ordered pair of the form (x, 4x – 4) is a solution of the system.

Note: We could have solved for x instead of y:

8x – 2y = 8

8x = 2y + 8

The solution set could also be represented as any ordered pair of the form (, y).

Example. Solve:

3x – 3y = -2

x – y = 5

Solution. We will use the substitution method, and begin by solving for x in the second equation:

x – y = 5

x = y + 5

We can now substitute this expression for x in the first equation:

3x – 3y = -2

3(y + 5) – 3y = -2

3y + 15 – 3y = -2

15 = -2

The false statement, 15 = -2, indicates that the system has no solution.

Solving a System in Three Variables

The method for solving a system of linear equations in three variables is similar to that used on systems of linear equations in two variables. We use addition to eliminate any variable, reducing the system to two equations in two variables.

A solution of a system of three linear equations in three variables is an ordered triple of real numbers that satisfies all equations of the system.

Example. Solve the system:

x – 4y – z = 6 Equation 1

2x – y + 3z = 0 Equation 2

-3x + 2y – z = -4 Equation 3

Solution. We first choose two equations and use the addition method to eliminate a variable. There are many ways to accomplish this, but we will eliminate x from equations 1 and 2:

x – 4y – z = 6	Equation 1
-2(x – 4y – z) = -2(6) -2x + 8y + 2z = -12	Multiply both sides of Equation 1 by -2
-2x + 8y + 2z = -12 2x - y + 3z = 0 7y + 5z = -12 Equation 4	Add the result from the previous step to Equation 2; we will call the result Equation 4

We must now eliminate the same variable, x, from another pair of equations. This time we will use equations 1 and 3:

x – 4y – z = 6	Equation 1
3(x – 4y – z) = 3(6) 3x – 12y – 3z = 18	Multiply both sides of Equation 1 by 3
3x – 12y – 3z = 18 -3x + 2y - z = -4 -10y – 4z = 14 Equation 5	Add the result from the previous step to Equation 3; we will call the result Equation 5

We now have two equations, 4 and 5, which contain only the variables y and z. We will now consider these equations as a system of two equations in two variables:

7y + 5z = -12 4(7y + 5z) = 4(-12) 28y + 20z = -48	Multiply both sides of Equation 4 by 4
-10y – 4z = 14 5(-10y – 4z) = 5(14) -50y - 20z = 70	Multiply both sides of Equation 5 by 5
28y + 20z = -48 -50y – 20z = 70 -22y = 22 y = -1	Add the two equations obtained above; solve for y

Back-substitute y = -1 into either Equation 4 or 5 to find z:

7y + 5z = -12

7(-1) + 5z = -12

-7 + 5z = -12

5z = -5

z = -1

Back-substitute y = -1 and z = -1 into one of the three original equations to find x:

x – 4y – z = 6

x – 4(-1) – (-1) = 6

x + 4 + 1 = 6

x + 5 = 6

x = 1

The solution is (1, -1, -1).