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A table of solutions for a linear equation is shown below. When constructing the graph of the equation, how would you scale the $x$ -axis and the $y$ -axis?$$\begin{array}{|c|c|c|}\hline x & {y} & {(x, y)} \\ \hline-20 & {600} & {(-20,600)} \\ {5} & {100} & {(3,100)} \\ {35} & {-500} & {(35,-500)} \\ \hline\end{array}$$
a) along $x-$ axis $1 \mathrm{cm}=10$ units.b)along $y$ - axis lem
Algebra
Chapter 3
Graphing Linear Equations and Inequalities in Two Variables; Functions
Section 2
Graphing Linear Equations
Graphs and Statistics
Equations and Inequalities
Linear Functions
Systems of Equations and Inequalities
Campbell University
Oregon State University
Idaho State University
Lectures
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Let's say we have a graph that doesn't have a scale on it for the X or Y axis, and we are given three different points. Negative. 5 135 native. 500. How might we scale are X and Y access to fit these three points on there and see a linear relationship? Let's start first with the X axis and the values for X. Those include negative 20 five and 35. So at this point, we want to ask ourselves what would be a common increment or a common factor between these three different values? Each of these is divisible by five. So if we were to scale our X axis by five, we'd be looking at negative five negative 10 native, 15 negative, 20 equal increments and then, of course, on the positive side, going by five, we can fit all of these values on here with uneven increment. Then, considering R Y axis, our range is from negative 500 to 600 with 100 in between. So it would be ideal to have an equal increment if we were to use a common factor of maybe 100. Let's think about where that would put us on our graph. Here's 100 200 300 400 and so on. And this fits pretty nicely, going in the negative direction. Negative 100 negative. 200. 300 data. 400 negative. 500 So we can see that we have these increments that would still show a linear result. For instance, let's go ahead and plot the point. Negative. 2600. They get a 20 on X 600 on why five and 100 and then 35 negative 500. So if we were to plot the graph associated with these three solutions, they would all lie on the same line. But I didn't see if I can do a nice sketch here. There we go.
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