Algebra

Standard form for linear equations | Algebra I | Khan Academy

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We have already considered several ways to write linear equations. You can write them according to given angular coefficient and intersection point, where they will be in the form y = mx + b, where m and b are constants. m is the coefficient of this term here, mx, and m will represent the angular coefficient (slope). And then from b you can find the intersection point with the y-axis. You can find the intersection point with the y-axis of b. Literally the graphic made up of the pairs hu, which satisfy this equation, intersects the y-axis at the point (0; b). And its angular coefficient (slope) is m. We have already seen this many times. We have also seen that we can to write the equation by given angular coefficient and point of the line. Let me clarify it. This is given the angular factor and the intersection point. Equation for given angular coefficient and intersection point. These are just different ways writing the same equations. You can algebraically process the equation from one type to another. Another way to write is given an angular factor and a point of the line. Equation for given angular coefficient and point of the line. As for the type of given angular coefficient and intersection point … For an equation for which the rights which presents its decisions, has an angular coefficient (slope) m … The slope is equal to m. But if you know that x = a and y = b satisfy this equation, then by a given angular factor and point you can express the equation as y – b = m * x – a. This is the kind of equation by given angular coefficient and point of the line. But in this video I want to get another form that you may have already seen it. This is the normal type of equation. Normal appearance. The normal look has the shape Ah + Bu = C, where A, B and C are integers. What I want to do in the video, as we did in the equations for given angular coefficient and point of the line and angular factor and intersection point, is to find out when the normal look is useful, and what is it less suitable for? Let’s give a real example here. We have the linear equation, which is normal, 9x + 16y = 72. And we want to draw it. The thing that the normal look is really suitable for, is finding not only the point of intersection with the y-axis, for the intersection with the y-axis it is better to we use the equation for given angular coefficient and intersection point, but we can find it quite easily from the normal look the intersection points with the x and y axes. The intersection point with the x-axis is not so easy to find of these other forms here. So how do we do that? To find the intersections with the x and y axes, let’s just do it small table here, for x and y. The point of intersection with the x-axis is at y = 0. And the intersection point with the y-axis is at x = 0. When y = 0, what is x? When y = 0, 16 * 0 = 0, this article disappears and you are left with 9x = 72. So if 9x = 72, 72/9 = 8. Therefore x = 8. Once again, it was pretty easy to find. This article disappears and you only need to calculate: 9x = 72, x = 8. When y = 0, x = 8. So the point, let’s see y is 0; x is 1, 2, 3, 4, 5, 6, 7, 8. This is the point here. This point here is the intersection point with the x-axis. When we talk about the point of intersection with the x-axis, we mean the point in which the line intersects the x-axis. Now, what about the intersection with the y-axis? Okay, x = 0, that disappears, we are left with 16y = 72. And that’s how we can solve this. Okay, we have 16y = 72. We can divide the two sides into 16. We get y = 72/16. Let’s see what this equals? This is equal to … both are divided by 8, so that’s 9/2. Or we can say that it is 4.5. When x = 0, y = 4.5. So, we can make that point as well. x = 0, y is 1, 2, 3, 4,5. And only with these two points, two points are enough to draw a line, now we can draw it. So, let’s do it. Let … oops, I thought I was using the tool, which will draw a straight line. Let’s see if I can … So the rights will look like this. Here we got it. I just drew, the line representing all pairs x and y, which satisfy the equation 9x + 16y = 72. I mentioned that the normal look is suitable for certain things, and where the normal appearance is appropriate, where perhaps in some unique way it connects with the other forms we have considered, it is very easy to find the intersection with the x-axis. It was very easy to find the intersection with the x-axis of the normal form of the equation. And it was not so difficult to find the intersection with the y-axis. If we consider the equation by given angular coefficient and intersection point, the point of intersection with the y-axis jumps straight at you. In the form of a given point of the line and angular factor neither the point of intersection with the x-axis nor the one with the y-axis is visible. The thing in which forms at a given angular coefficient and intersection point and given the angular coefficient and the point of the line are better, is finding the angular coefficient (slope) in them, while in the normal form you will need to make some calculations. You can use these two points, you can use the intersection points with x and y as two points, and find the angular coefficient (slope) of them. Therefore, you can literally say: “Okay, if I went from this point to this point, the change in x going from 8 to 0 is -8. And to go from 0 to 4.5 … I wrote this little delta there unnecessarily. If you go from 8 to 0, the change in x is equal to -8. And to go from 0 to 4.5, the change in y will be 4.5. So the angular coefficient (slope) is the change in y, which is 4.5, on the change in x, i.e. on –8. And since I don’t want to have a decimal point here, let me multiply the numerator and denominator by 2. You get minus 9/16. Now once again, we need to rework a bit here. Or we use these two points, we don’t see it right away from that, although you may see some pattern for what is happening here. But we still have to think about whether is this negative? Is it positive? You need to do a little algebraic processing. What we usually do when looking for the angular coefficient (slope), is to turn it into one of the other forms. Especially given the angular coefficient and intersection point. But the normal look in itself is great for finding the intersections of x and y at the same time and honestly, it’s not that hard to turn it into equation for given angular coefficient and intersection point. Let’s do it, just to clarify it. If we start with 9x, let me write it in yellow. If we start with 9x + 16y = 72 and we want to turn it into an equation with given angular coefficient and intersection point. we can subtract 9x from both sides. We get 16y = –9x + 72. And then we divide the two sides into 16. So we divide everything by 16. And we will stay with y = – 9 / 16x, this is the angular coefficient (slope), you see it right there, +72 / 16, we have already found that it is 9/2 or 4.5. So I can just write 4.5. Of this kind, it is much easier to find the slope here and the intersection with y is also visible. But the intersection with the x-axis is not so obvious.

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