### Linear Equations in Two Variables - Basics

This section deals with linear equations where you only have two variables. That we will deal with linear equations means that the variable will always be to the first power. The fact that there are two variables means that there will be two unknows. With this type of problem we will usually have two equations. We have a System of Equations with two being the simplest set. When we solve a set of equations involving two unknowns we find the point that satisfies both of the equations.
There may be
1. No point that satisfies both equations
2. Exactly one point that satisfies both equations
3. Many points that satisfy both equations

The images shown below depict the three cases. On the left you have two parallel lines. These lines never intersect so there is no point that simultaneously satisfies both equations. The center graph has two lines that cross at one point. The point where they cross is the solution point for this set of equations and the (x, y) pair show are the solution to the set. The last graph on the far right represents two lines that lie on top of one-another. They are slightly separated so you can see that there are two lines, but they really are on top of one-another. All points on the lines simultaneously satisfy both equations.

Here are three sample sets of equations :
• 2x + 3y = 11, 4x + 6y = 30, If you graph these two equations you will find that they do not intersect.
• x + y = 8, 2x - y = 1, If you graph these two equations you will find they have one common point (solution) at x = 3, y = 5.
• 4x + 8y = 7, 12x + 24y = 21, If you graph these equations you will see that they represent the same line and all of the points on the lines are common.

#### Solving Linear Equations in Two Variables

There are two commonly taught techniques for solving a two equation set of linear equations. These are the Substitution Method and the Elimination Method. Your text book covers both of these methods so we will focus here on working problems using those methods.

#### Solving Linear Equations in Two Variables Using the Substitution Method

Here is a system of two linear equations we want to solve: Eq 1: 2x + 4y = 6 and Eq 2: 3x - y = 2
• First solve one of the equations for the x or y variable. In this case it is easy to solve Eq 2 for y
• Subtracting 3x from both sides of the equation gives: 3x - y - 3x = 2 - 3x
• -y = 2 - 3x (but we want +y not -y)
• Multiplying both sides of the equation by -1 gives: y = -2 + 3x
• Now we will take that result and replace the y in Eq 1 with -2 + 3x since we just found that y = -2 + 3x
• Eq 1: 2x + 4(-2 + 3x) = 6
• Distributing the 4 gives: 2x + 4(-2) + 4(3x) = 6
• 2x - 8 + 12x = 6
• Adding 8 to both sides and adding the x terms: 14x = 6 + 8
• 14x = 14
• 14x/14 = 14/14
• x = 1
• To find y we subtitute our y into either equation. Let's use Eq 2 where we know y = -2 + 3x
• y = -2 + 3(1)
• y = -2 + 3
• x = 1
• So the final solution is (1, 1) To test it substitute x = 1 and y = 1 in both of the original equations.

Now it is your turn. The first equation is 5x + 3y = 7 and the second equation is x - 2y = 2 If you click the "Click to show problem" button this problem will appear. You can work the problem on the computer or on paper. The choice is yours. When you are done hit the button that says "Click to Show Answer" and the computer will display the answer in a step-by-step manner. After you finish the first problem and look at the answer you can click on the show problem button again and another problem will appear. At this time there are four problems available for you to work. It you want to try the problems again hit the "Reset" button.

#### Solving Linear Equations in Two Variables Using The Elimination Method

In this case we have two linear equations but we want to solve the system by adding or subtracting the equations.
• For this to work the coefficients of either the x terms or the y terms must be of the same magnitude in both equations.
• Thus is x has a coefficient of 2 in one equation and -2 in the other equation when we add the equaitons the x's dropout.
• Consider 2x + y = 7 and 4x - y = 5
• If we add the equations we get (2x + y) + (4x - y) = 7 + 5
• 2x + 4x + y - y = 12
• 6x = 12
• 6x/6 = 12/6
• x = 2 and substituting x = 2 into Eq 1 gives 2(2) + y = 7
• 4 + y = 7
• 4 + y - 4 = 7 - 4
• y = 3, so the solution is x = 2, y = 3
• If the coefficients don't match, you will need to multiply one or both equations by constants that make the x or y coefficients match.
• For example if you have 3x - 6y = -12 and x + 4y = 8 you might multiply Eq 2 by 3 to get 3x + 12y = 24
• Now we can subtract Eq 2 from Eq 1 to get: (3x - 6y ) - (3x + 12y) = -12 - 24
• 3x - 6y -3x - 12y = - 36
• -18y = -36
• -18y/-18 = -36/-18
• y = 2, to get x substitute y = 2 into Eq 2: x + 4(2) = 8
• x + 8 = 8
• x + 8 - 8 = 8 - 8
• x = 0, so the solution set is x = 0, y = 2