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# I literally don't get this AT ALL

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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.

I'm incredibaly confused if someone could help that would be AMAZING!!
Thanks!

Oct 1, 2018

#1
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What you have here is a locus (set) of points such that

x=2t+3      and      y=3-3t   for t is any real value.

so what you need to do it make t the subject of both these equations

 x=2t+3 x-3=2t (x-3)/2  = t y=3-3t y-3=-3t 3t=3-y t=(3-y)/3

So now you know that

$$t=\frac{x-3}{2}=\frac{3-y}{3}\\ \frac{x-3}{2}=\frac{3-y}{3}\\ 6 \cdot \frac{x-3}{2}=6 \cdot \frac{3-y}{3}\\ 3(x-3)=2(3-y)\\ 3x-9=6-2y\\ 3x+2y-15=0 \qquad \text{which is a line}\\ or\\ 2y=-3x+15\\ y=\frac{-3x}{2}+\frac{15}{2}$$

which is still the same line.      gradient = -1.5    and    y intercept = -7.5

Oct 2, 2018
#2
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Why do you multiply by 6?

HelpPLZ  Oct 3, 2018
#3
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Oh nvrm I get it

HelpPLZ  Oct 3, 2018
#4
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Well I am glad that you got it.

Pity there was no 'thank you' thrown in there while you were at it though.

A 'thank you' is always appreciated.

Melody  Oct 3, 2018
edited by Melody  Oct 3, 2018