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Suppose you graphed every single point of the form \((2t + 3, 3-3t).\) For example, when t=2, we have \(2t + 3 = 7 \) and \(3-3t = -3\), so (7,-3) is on the graph. Explain why the graph is a line, and find an equation whose graph is this line.

 

In order for this graph to be a line, we need to verify two things: first, that all the points on the graph are on the proposed line and second, that all points on the proposed line are on the graph. Be sure to deal with both.

 Apr 2, 2019
 #1
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Also i'm not the guest who has been spamming random stuff

 Apr 2, 2019
 #2
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As users we understand only some do and some dont

Nickolas  Apr 2, 2019
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can u help me with this problem?

 Apr 2, 2019
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I can look at it but I am probally not able to do it what grade is this for 

Nickolas  Apr 2, 2019
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Well to start then do you now about the X and Y poinet 

 Apr 2, 2019
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\(\text{Let }x= 2t+3\\ t = \dfrac{x-3}{2}\\ y = 3-3t = 3 - 3\left(\dfrac{x-3}{2}\right)\\ y = 3 -\dfrac 3 2 x+\dfrac 9 2 \\ y = -\dfrac 3 2 x + \dfrac{15}{2}\)

 

I'll let you work through what you need to verify.  It should be trivial given the formula for the line.

For the first part, pick any t and show that the resulting x and y obey the last formula above.

For the second part pick any x and calculate the associated y using that formula.

Then show that the a single value of t produces them.

 Apr 2, 2019

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