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Solving a system of linear equations in two variables. We're going to begin by determining whether an ordered pair is an actual
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solution to a specific system of linear equations. And before we do that, let's talk about what a solution to a system of equations actually is. So if you
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recall in solving a linear equation, your solution is the ordered pair that makes the equation true. When we incorporate this into a system of linear equations,
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meaning more than one linear equation, then the solution or the ordered pair of this system of linear equations must satisfy both equations in the system.
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We're going to begin by taking a look at this example here. Our job is to determine whether the ordered pair (five, four) is a solution to the system:
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two x minus five y equals negative ten and two x plus four y equals six. In order for this ordered pair (five, four) to be a solution, it will have to satisfy
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both equations in the system. We'll begin with the first equation: two x minus five y equals negative ten. Now my ordered pair is
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(five, four), so by substituting five for x and four for y, we're trying to figure out if what we have on the left side of
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the equation is equal to what we have on the right, which is negative ten. Well two times five is ten, minus twenty,
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and that is negative ten. So because I get a true statement, negative ten equals negative ten,
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then that tells us that (five, four) is a solution to this equation. But in order for it to be a solution to the system of equations,
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it must also satisfy the second equation in the system. So similar to what we did with the first equation,
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we're going to use our second equation: two x plus four y equals six,
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and we're also going to substitute five for x and four for y. Two times five is ten, plus sixteen. As you notice here, ten plus sixteen is twenty-six, and twenty-six
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definitely does not equal six. Therefore the ordered pair (five, four) does not satisfy the second equation in our system, and because (five, four)
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does not satisfy both equations, then what we say is that the ordered pair (five, four) is not a solution to our system of equations.
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Now we'll solve a system of linear equations by graphing. To solve a system of equations by graphing,
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it's necessary to graph each of the lines represented in the system of equations. Once the lines are graphed,
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we're looking for any point of intersection between the lines. That point of intersection, the x-value and the y-value,
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are the values that satisfy or make true each of the equations in the system. Let's take a look at this example. We want to solve this system by graphing.
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Negative two x plus y equals negative five, and two x minus four y is equal to eight. Notice both of these equations are in standard form:
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A x plus B y equals C. To find the slope of an equation in standard form,
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we're going to take negative A and divide it by B. To find the y-intercept of an equation in standard form, we're going to take C
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and divide it by B. So let's look at equation number one here and figure out the slope and the y-intercept.
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My A value is negative two, so the negative of negative two is positive two, and B is one. For the y-intercept our value of C is negative five,
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and again B is one, but it's not necessary to write it here. The reason I left the one here is because with slope
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we're looking at rise over run, or a change in y over a change in x. Now let's take this information and actually graph the first equation.
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My y-intercept is negative five, so along the y-axis I'm going to negative five and I'm putting my point.
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With the slope of positive two over positive one, from my y-intercept I'm going up two and over one to the right.
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Now I'm going to use slope again to get another point on this graph. So up two, over one. And now I can draw the line.
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So let's look at the second graph in our system: two x minus four y is equal to eight.
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To graph the second line, we're going to use x- and y-intercepts. So recall to find an x-intercept, we're going to substitute zero for y.
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So substituting zero for y here, leaves me with two x is equal to eight, which makes my x-intercept four. To find the y-intercept,
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we will substitute zero for x, and in substituting zero for x, I'm left with negative four y is equal to eight. Therefore, y is equal to negative two.
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Now let's go to the graph and plot these intercepts. So an x-intercept of four means
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I'm going on the x-axis to four, and I'm plotting the point. A y-intercept of negative two means on the y-axis I'm going
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down to negative two and plotting the point there. Now we can draw the line.
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What we recognize now is that this point right here (two, negative one) is the point of intersection. That means two for x
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and negative one for y makes each of these equations in the system true. So we say that (two, negative one) is the solution to this system of equations.
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We're going to continue graphing systems of linear equations. In graphing systems of linear equations, we can have either one solution,
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no solution, or infinitely many solutions. So this first system we're going to look at is two x minus y equals six
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and six x minus three y equals six. So to begin graphing the first equation, we're going to find the intercepts. To find the x-intercept,
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we're going to make the y-value zero leaving me with two x is equal to six. Therefore, I know that x is equal to three. My x-intercept is three.
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To find the y-intercept, we'll make x zero leaving us with negative y is equal to six
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which makes y equal to negative six. In our second equation, we can also find intercepts to graph this line easily. The x-intercept: six x is equal to six,
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which makes the x-intercept one. And for the y-intercept, negative three y is equal to six, making our y-intercept negative two.
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So let's begin by graphing the first equation. With an x-intercept of three, I'm going along my x-axis to three, and I'm plotting the point.
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With the y-intercept being negative six, down on the y-axis to negative six, and there's the y-intercept.
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I can now connect the two points. And there I have the line. Now doing the same for the second equation, an x-intercept of one
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and a y-intercept of negative two. Connecting the points gives me this line.
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Notice here that it looks like we have two parallel lines. We could only be sure of that
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by actually finding the slopes of each of these lines and comparing them, as well as comparing the y-intercepts.
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If you were to find the slope of line number one and line number two, you would realize that it's the same.
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And by looking at the graphs here, they obviously have different y-intercepts. Therefore making the two lines parallel.
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With parallel lines there is no point of intersection. Therefore, the solution to this system of equations is that there is no solution.
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Now if we move over to our second system of equations, in my first equation I notice that my x-intercept is three and my y-intercept is negative six.
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For equation number two, finding the x-intercept gives me three, and finding the y-intercept, negative six. So let's go ahead and get these lines on the graph.
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Starting with equation number one, I have an x-intercept of three and a y-intercept of negative six. Connecting these points and drawing the line,
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gives me that. Now for our second equation, I also have an x-intercept of three, and I also have a y-intercept of negative six.
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When I pick up the ruler to draw this line, what I notice is that it's the exact same line as equation number one.
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It's very hard for me to pick out only one point of intersection for both of these lines,
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and that's because they intersect at every point on the lines. Therefore, there is not only one solution to this system. There are infinitely many solutions.
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So here is an example of a system that has no solution, or no point of intersection. This is an example of a system that has infinitely many solutions,
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or infinitely many points of intersection.