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Find the equations of the lines and plot the lines from Exercise $48$.

$7 x-5 y=59$

Algebra

Chapter 1

Functions and their Applications

Section 1

The Line

Functions

McMaster University

Harvey Mudd College

Baylor University

Idaho State University

Lectures

01:43

In mathematics, a function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. An example is the function that relates each real number x to its square x^2. The output of a function f corresponding to an input x is denoted by f(x).

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Find the equations of the …

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02:48

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02:08

Find an equation of the li…

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Finding an Equation of a L…

this problem is referencing the slope of the points that we looked at back and exercise number 48 as a reminder 48 had two points to negative nine and the 0.12 5. And in that exercise we found the slope of the line connecting them to be 7/5. Okay, Now let's take that information, and we're going to actually put that into thick way Asian of a line. Well, we often use slope intercept form. In this case, it's not as convenient because I don't have the intercept. I have the slope, but I don't have the intercept. So instead, we're gonna use point slope form because I have points. Fact, I have two points and I have a slope. So the point slope form is why minus why one equals m times X minus X one. So x one y one is a point that we've been given, and I can use either one of these points. They will both work. They'll both give me this exact same answer in the end and my slope, Well, it would be 7/5 here, So let's plug in some values. Um, I'm gonna take the first point just because it's first. Not for any other reason. Again. You could use the second point and it would work exactly the same way. So why, minus my wife value? So why minus negative? Nine becomes plus nine. My slope is 7/5 so X minus two. Now, I'm just going to simplify this. In fact, I'm gonna put this into slope intercept form because I will make it easier to graft my equation. So first I'm gonna get rid of my parentheses. That's 7/5 X minus 14 5th and then I'm going to subtract nine from both sides. When I do this, nine will be gone, and I'm going to subtract nine. If I want a common denominator, I could change that nine into 45 5th. So that gives me an equation. Why equals 7/5 X minus 59 fifth. So there's my equation. Like I said, I put this in the slope intercept form to make it easier to graph. You could put this into standard form. If you wanted Thio, that would be totally up to you. But here's our equation. When we go to graph it, we need to have a starting point which is going to be our Y intercept. Negative. 59 5th. That is almost negative. 12 negative. 65th to be negative. 12. So, on my sketch, this is almost going to be at the negative. 12 mark. 123456789 10, 11. Yep, it is right there at the very bottom of this grid, just a tiny bit above that line. And my slope is 7/5. That's a positive slope. So my lines should trend upward. As I go from left to right across my graph, I'm gonna go up 71234567 and over 512345 And I think I'm going to do one more point. I just find that makes a little bit easier when I go to connect them up. 7123456 7/5. 1234 five And we connect those points. Okay, So as you can see, we do go through the point to negative nine. That's right about here. And the 0.12 5 we haven't quite gotten to yet. Just as I'm leaving. Fact, I think it's 123 is probably about right up in here that we're hitting. So the points that were given with our slope this is the graph for these, um, for these conditions.

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