+610 Suppose you graphed every single point of the form (2t+3, 3-t). 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.
Hint: If the point (2t+3, 3-t) is on the graph, then x=2t+3 and y=3-3t. But the equations we're used to graphing don't have any variables besides x and y.
+118608 NOTE there is an error in this question, The y value is given as two different values. I used the first one.
Suppose you graphed every single point of the form (2t+3, 3-t). 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.
Hint: If the point (2t+3, 3-t) is on the graph, then x=2t+3 and y=3-3t.
But the equations we're used to graphing don't have any variables besides x and y.
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I really do not know what is wanted in this question but I shall discuss it a little.
I have used the value of y given in the original co-ordinates
(2t+3, 3-t)
This a a set of points where x and y are defined by the variable t such that
x=2t+3 and y=3-t
if I make x the subjects of these then I get
\(t=\frac{x-3}{2}\qquad and \qquad t=3-y\\ hence\\ \frac{x-3}{2}=3-y\\ y=3-\frac{x-3}{2}\\ y=\frac{6}{2}-(\frac{x}{2}-\frac{3}{2})\\ y=\frac{6}{2}-\frac{x}{2}+\frac{3}{2}\\ y=-\frac{x}{2}+\frac{9}{2}\\\)
This is clearly a linear graph with a gradient of -0.5 and a y intercept of 4.5
All the points on this line can be represented as (2t+3,3-t)
Any points not on this line cannot be represented by (2t+3,3-t) Where t is in the set of real numbers.
I do not know if something else is needed or wanted 
+118608 NOTE there is an error in this question, The y value is given as two different values. I used the first one.
Suppose you graphed every single point of the form (2t+3, 3-t). 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.
Hint: If the point (2t+3, 3-t) is on the graph, then x=2t+3 and y=3-3t.
But the equations we're used to graphing don't have any variables besides x and y.
-------------
I really do not know what is wanted in this question but I shall discuss it a little.
I have used the value of y given in the original co-ordinates
(2t+3, 3-t)
This a a set of points where x and y are defined by the variable t such that
x=2t+3 and y=3-t
if I make x the subjects of these then I get
\(t=\frac{x-3}{2}\qquad and \qquad t=3-y\\ hence\\ \frac{x-3}{2}=3-y\\ y=3-\frac{x-3}{2}\\ y=\frac{6}{2}-(\frac{x}{2}-\frac{3}{2})\\ y=\frac{6}{2}-\frac{x}{2}+\frac{3}{2}\\ y=-\frac{x}{2}+\frac{9}{2}\\\)
This is clearly a linear graph with a gradient of -0.5 and a y intercept of 4.5
All the points on this line can be represented as (2t+3,3-t)
Any points not on this line cannot be represented by (2t+3,3-t) Where t is in the set of real numbers.
I do not know if something else is needed or wanted
