There are several ways to solve systems of linear equations. The most common methods are by graphing, elimination, and substitution. Let's start off with one of the most basic methods, graphing.
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Graphing Method
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2x + y = 3
3x + 2y = 6
In order to solve this system using the graphing method, we first have to change the two equations into slope-intercept form.
2x + y = 3 --> y = -2x + 3
3x + y = 7 --> y = -3x + 7
Then, we graph these two lines. (Attached Below)
The solution set of a system of linear equations when graphing is actually the point at which the two lines intersect. So by graphing the two lines, we can obviously see that the solution set of this problem is (4, -5).
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Elimination Method
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The concept of elimination revolves around the concept of adding two equations. Using an example, let's see what happens when you add equations together.
2x + y = 3
3x + 2y = 6
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5x + 3y = 9
Do you see how this works? Now, let's say that the orientation of these two equations were different. What would you do then?
2x + y = 3
6 - 3x = 2y
If this situation occurs, you have to rearrange it in a way that the form of the equations match. For example, if you have one in standard form, you have to algebraically return the other equation to the same form.
2x + y = 3
6 - 3x = 2y --> 6 = 3x + 2y --> 3x + 2y = 6
Now that the equations are in the same form, we can begin to solve. However, let's start with a simpler system to demonstrate the concept.
2x - y = 5
3x + y = 5
The process of elimination involves adding equations in a way that one of the unknown variables disappears. In this first example, let's see what happens when we simply add them right away.
2x - y = 5
3x + y = 5
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5x + 0 = 10
Using the Identity Property of Addition, we can simply get rid of the 0.
5x = 10
Now it's all algebra.
5x = 10
x = 10/5
x = 2
Now we have one of the variables. You may be thinking "what about the other variable?". Well, in order to solve for the variable, all you have to do is plug what you have for this variable back into the equation like so:
3x + y = 5
3(2) + y = 5
6 + y = 5
y = -1
It doesn't matter which equation we plug it into because you will get the same result. Therefore, our solution set is (2, -1).
Now, this seems very simple, but what would you do if you had something like this:
2x + y = 3
3x + 2y = 7
Let's see what happens when you simply add the two equations.
2x + y = 3
3x + 2y = 7
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5x + 3y = 10
It doesn't look pretty does it? In fact, we can't even solve for the variables this way. So what do we do? Well, we do something called manipulating the equation. We multiply one of the equations in the system by a number so that we can cancel one of the variables when we add them.
2x + y = 3
3x + 2y = 7
-2(2x + y = 3)
3x + 2y = 7
-4x - 2y = -6
3x + 2y = 7
Now, if we were to add these two equations, we would be able to cancel one of the variables.
-4x - 2y = -6
3x + 2y = 7
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-x + 0 = 1
At this point, using the information above, we can once again find the solution set.
-x = 1
x = -1
2x + y = 3
2(-1) + y = 3
-2 + y = 3
y = 5
Solution Set: (-1, 5)
If you're wondering why this works, allow me to tell you. By multiplying both sides of the equation by a number, you are actually balancing an equation. If we were to only multiply one side, then the values of x and y would change. This is a basic rule of algebra: "What you do to one side of the equation, you must also do to the other"
Finally, to demonstrate a mastery of the elimination method, let's try solving this system of linear of equations.
4x – 3y = 25
–3x + 8y = 10
Now right off the bat, you will notice that no matter what you multiply one of the equations by, you can't get to a place where one of the terms is canceled. So, we have to multiply both equations in order to solve this.
4x – 3y = 25
–3x + 8y = 10
8(4x – 3y = 25)
3(–3x + 8y = 10)
32x - 24y = 200
-9x + 24y = 30
Now we can solve this system.
32x - 24y = 200
-9x + 24y = 30
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23x + 0 = 230
23x = 230
x = 10
4x – 3y = 25
4(10) – 3y = 25
40 - 3y = 25
-3y = -15
y = 5
Solution Set: (x, y) = (10, 5)
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Substitution Method
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The substitution method is a method that is based off a rule of algebra. Basically, it's similar to using a formula. You substitute values into the formula. The same thing applies here.
Let's say we have the following system:
y = 2x + 3
5x + 2y = 24
Before we turn over to elimination, let's try using our new friend substitution. It clearly says that y is equal to 2x + 3. This means that we can substitute 2x + 3 in for y.
5x + 2y = 24
5x + 2(2x + 3) = 24
5x + 4x + 6 = 24
Now that we have eliminated a variable, we can solve the system.
9x + 6 = 24
9x = 18
x = 2
Now let's plug this back into one of the equations. I prefer to plug it into the equation that has a variable isolated.
y = 2x + 3
y = 2(2) + 3
y = 4 + 3
y = 7
Therefore, the solution set is (x, y) = (2, 7).
Now, if we don't have a variable already isolated, we have to algebraically solve for a variable. It is very similar to converting standard form to slope-intercept form when graphing.
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Special Cases
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There are two special cases when dealing with systems of equations. Let's say we have the given system.
-2x + y = 5
-4x + 2y = 10
Let's try solving this.
-2(-2x + y = 5)
-4x + 2y = 10
4x - 2y = -10
-4x + 2y = 10
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0 + 0 = 0
0 = 0
Uh-oh. Both of the variables disappeared! This is what's called an identity case. An identity case occurs when both variables disappear and both sides have the same value. This means that 6 = 6 is also an identity case. Now, what does an identity case mean? It means that it's the same line. Since it's the same line, there are infinitely many solutions because the lines intersect in every way possible. This is also called an infinitely many solutions case.
Now, let's take a look at the second special case.
2x + y = 9
4x + 2y = 4
-2(2x + y = 9)
4x + 2y = 4
-4x – 2y = -18
4x + 2y = 4
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0 + 0 = -14
0 = -14
Once again, both variables disappeared. This is called a case of no solution. This occurs when the two lines are parallel. Since parallel lines never intersect, there is no solution.