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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!

HelpPLZ  Oct 1, 2018
 #1
avatar+93675 
+2

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

Melody  Oct 2, 2018
 #2
avatar+144 
0

Why do you multiply by 6? 

HelpPLZ  Oct 3, 2018
 #3
avatar+144 
0

Oh nvrm I get it

HelpPLZ  Oct 3, 2018
 #4
avatar+93675 
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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

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