WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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In today's lesson, we will be covering linear equations.
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Again, this is some review from Algebra I, which we will discuss here.
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And if you need further detail, go back to the Algebra I lectures and check those out for this subject.
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So, a linear equation is an equation of the form ax + by = c, where, a, b, and c are constants.
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For example, 2x + 5y = 3, and here a = 2, b = 5, and c = 3.
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Another example could be x - 7y = 4; here, even though it is not written, there actually is a 1 in front of the x.
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Just by convention, we don't write it out when the coefficient is 1.
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So, a = 1, b = -7, and c = 4.
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We will talk a little bit more about what this form is.
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A linear equation represents a function, which is known as a linear function; and that can be written in the form f(x) = mx + b.
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And this is a very useful form of the equation that will be discussed in another segment.
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And an example of this would be something such as f(x) = 4x + 2; you may also see it written as y = 4x + 2.
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And here, m = 4; b = 2; and these two numbers represent certain elements that we will talk about in a few minutes.
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OK, so the first form of the linear equation that we talked about just a minute ago is also called standard form.
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So, a linear equation is in standard form if it is written in the form ax + by = c,
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where a, b, and c are integer constants (and this is important) with no common factor.
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If a common factor remains, it is not in standard form.
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For example, if I have an equation 9x + 4y = 10, this is in standard form.
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I have ax (9 is a), plus by (b is 4), equals c (c is 10), and there is no common factor.
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Consider another possible equation: 6x + 8y = 12; this is not in standard form, because 6, 8, and 12 have a common factor.
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What I need to do is divide both sides of this equation by 2 to get 3x + 4y = 6; now, it is in standard form.
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Another example would be if I had 7y + 5 = -x; again, it is not in standard form.
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I can add x to both sides to get x + 7y + 5 = 0, and then subtract 5 from both sides to get x + 7y = -5; this is standard form.
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We can graph linear equations using the intercept method.
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Recall that the point where a line crosses the x-axis is its x-intercept.
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And the point where the graph crosses the y-axis is the y-intercept.
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For example, if I had a line such as this, first the x-intercept is right here; the x-intercept is -2.
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And at the x-intercept, y is always going to be 0; so, the coordinate pair would be (-2,0), because it crosses the x-axis, and so y is 0.
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The y-intercept is right here: y equals 4, and at this point, x is going to equal 0, so (0,4) is the y-intercept.
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And we can use this knowledge in order to graph a linear equation.
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For example, if I have a linear equation 2x + 3y = 6, all I need in order to plot this are two points on the line.
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Now, I can easily get two points by finding the x- and y-intercepts.
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I know that, with the y-intercept, x will be 0; so if I let x equal 0, that is going to give me 0 times 2 is 0,
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so that will be 3y = 6; 6 divided by 3--y equals 2.
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OK, to find my second point, I am going to find the y-intercept (this is actually the x-intercept right here)...0.
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Excuse me, this is the y-intercept right here; now I am looking for the x-intercept.
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The x-intercept is the point at which y equals 0, and the line of the graph crosses the x-axis.
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So, this time, I am going to let y equal 0; so I am going to have 2x, plus 3 times 0, equals 6, or 2x + 0 = 6.
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So then, I end up with x = 3; this is the x-intercept.
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Now, I have two points; I can plot the line.
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At the y-intercept, x is 0; y is 2; the y-intercept is right here, at (0,2); the x-intercept is right here at (3,0).
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Now, I have two points; and when I have two points, that means I can connect them to form a line.
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And I did that by finding my x-intercept at (3,0) and my y-intercept at (0,2).
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The technique is to find the x- and y- intercepts by letting x equal 0, and putting that into the equation,
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and solving for y to get the y-intercept, and letting y equal 0
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to find the x-intercept, and then using those two points to plot the line.
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In the first example, is the function linear? Well, recall that a linear function can be written in standard form.
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And standard form is ax + by = c: I look at what I have here, and I have ax (so it seems like I am going along OK),
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minus by (or that would be the same as plus -by, so that is fine), equals c; but then I have this over here.
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And this is not part of standard form.
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And you could attempt to get rid of that; maybe you say, "OK, I will multiply both sides by x to get rid of it."
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But then, see what happens: all right, let's see what happens if I try to multiply both sides by x to get rid of this x in the denominator.
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OK, I end up with 2x² - 3xy = 4x + 1; so I am no better off.
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I still don't have it in standard form; there is no way for me to get this in standard form; therefore, this is not a linear function.
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It is not in standard form; I can't get it into standard form.
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OK, is this function linear? Well, recall that we talked earlier about the form g(x) = mx + b.
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And remember, g(x) is the same idea as f(x): you can use different letters, so it is still just that we are looking for this form for a linear function.
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Well, I am looking at this, and I have what looks like an mx and a b, but the problem is that I also have this.
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And this is not a part of the form for a linear function--this x² term--therefore, this is not a linear function.
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Once again, I can't get it into the right form for a linear function.
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OK, in Example 3, we are asked to write the equation in standard form.
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And recall that standard form is ax + by = c.
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And I am dealing with some fractions, and I need to get rid of those in order to get this in standard form.
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I can do that by multiplying first both sides of the equation by 3; this is going to give me 3y/6 =...here, the 3's will cancel out; -7 times 3--that is -21.
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I can simplify this to y/2 = x - 21; so, I am still left with a fraction.
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I am going to multiply both sides of the equation by 2 to eliminate that fraction.
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The 2's cancel out, so I have y = 2x...-21 times 2; that is -42.
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Next, I am going to subtract 2x from both sides to get the x on the left side of the equation.
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And just switching around the order of the x and y to match this: -2x + y = -42.
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And you could stop here, or you could multiply both sides by -1; and that is more conventional, to have the a term be positive.
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So, that is going to leave me with 2x - y = 42.
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I started out with this equation, and I wanted to get it into this form.
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And I could achieve that by first multiplying by 3 to get rid of this fraction, then (I ended up with this)
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multiplying by 2 to get rid of the y fraction, then simply subtracting 2x from both sides to get the 2x (the ax)
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on the left side of the equation, and finally just a little more work to clean this up, multiplying both sides by -1.
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This is the equation in standard form, where a equals 2, b equals -1, and c equals 42.
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Example 4: Find the intercepts and graph the equation: 4x - 5y = 20.
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We already talked about how, with the x- and y-intercepts, you can find those, and then you will have two points, and you can graph an equation.
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So, first, in order to find the y-intercept, I am going to let x equal 0; this will give me the y-intercept, the point at which the line crosses the y-axis.
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So, to find the y-intercept, I am going to substitute 0 for x.
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This is going to give me 0 - 5y = 20, or -5y = 20; divide both sides by -5 to get y = -4.
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OK, next I want to find the x-intercept: to find that, I am going to let y equal 0.
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I am going to go back and do this again, this time letting y equal 0.
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This is for the y-intercept, and this is for the x-intercept.
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This is going to give me 4x - 0 = 20, or 4x = 20, which is x = 5.
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Now that I have my x- and y-intercepts, I actually have two points on the line.
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My first point is (0,-4), the y-intercept; my second point is at (5,0), and that is the x-intercept.
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OK, connecting these two points gives me the graph of this equation.
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We use the intercept method in order to graph, finding the x- and y-intercepts,
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plotting those out on the coordinate plane, and then using those to form a line.
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That concludes this lesson for Educator.com; and I will see you here again soon.