Notes, Lesson 5.1
Solving Systems of Equations Algebraically
Course Notes Course Home Page Sample Problems for this Lesson
  

There are two methods that will be used in this lesson to solve a system of linear equations algebraically. They are 1) substitution, and 2) elimination. They are both aimed at eliminating one variable so that normal algebraic means can be used to solve for the other variable. Once one variable is solved, then substitution will be used in both above methods to find the second variable.

Special Circumstances.
When solving inconsistent systems algebraically, you will find that all variables drop and the remaining statement is false. When you receive a false statement that has no variables in it, first check your algebra. If you have done it correctly, and there is a false statement remaining, then you know that you have no solution, and the system is inconsistent.

When solving consistent dependent systems algebraically, you will find that all variables drop and the remaining statement is true. When you receive a true statement that has no variables in it, first check your algebra. If you have done it correctly, and there is a true statement remaining, then you know that there are an infinite number of solutions, and the system is consistent dependent.

The Substitution Method.
To use the substitution method, you solve one of the equations for either variable, and then substitute that algebra expression in for the same variable in the other equation. This will allow you to solve for one variable. Then you would substitute that value into either original equation to find the value of the other variable.

Example.
 

Solve the system:
6x-2y=12; x+y=18
Given Problem.
Using the 2nd equation:
x+y=18
y=-x+18
Solve the 2nd equation for y.
6x-2(-x+18)=12
6x+2x-36=12
8x-36=12
8x=48
x=6
Substitute quantity for y that was just solved for in the last step into the other equation. Solve for x.
6+y=18
y=12
Substitute value for x into either original equation. Solve for y.
(6,12) is the solution
Put answer in the form of an ordered pair.
Check this by plotting both lines and visually checking the solution point. 

It does check.

The Elimination Method.
To use the elimination method, you use the ERAA to multiply both equations by the number necessary so that when adding or subtracting the equations, one of the variables is eliminated. Then you would substitute that value into either original equation to find the value of the other variable.

Example Problem.

Solve the system:
6x-2y=12; x+y=18
Given Problem.
2x+2y=36
Multiply the 2nd equation by 2.(I did this intentionally so that when the 2y would be added to the -2y from the other equation, the y variable would be eliminated
2x+2y=36
6x-2y=12
8x=48
Then add it to the 1st equation.
x=6
Then solve for the one variable remaining.
6+y=18
y=12
Substitute value for x into either original equation. Solve for y.
(6,12) is the solution
Put answer in the form of an ordered pair.
Check this by plotting both lines and visually checking the solution point. 

It does check.

Top of this PageCourse Home Page