Question 77505
If you have y=-2 then you simply replace y with -2 and evaluate x. So for instance 

{{{y=5x-4}}} let y=-2

{{{(-2)=5x-4}}}  plug in y=-2

{{{2=5x}}} Add 4 to both sides

{{{x=2/5}}} So if y=-2, then {{{x=2/5}}} We do the same for every y that is in the table

We're basically going backwards. Instead of finding y with the given x values, we find x with the give y values. So to do this efficiently, we need to solve for x:

{{{y=5x-4}}}

{{{y+4=5x}}} Add 4 to both sides

{{{x=(y+4)/5}}} Divide both sides by 5 to solve for x.


Now if we evaluate each y we get:



{{{x=((-2)+4)/5}}} plug in y=-2

{{{x=(2)/5}}} So if y=-2, x=2/5

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{{{x=((-1)+4)/5}}} plug in y=-1

{{{x=(3)/5}}} So if y=-1, x=3/5

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{{{x=((0)+4)/5}}} plug in y=0

{{{x=4/5}}} So if y=0, x=4/5

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{{{x=(1+4)/5}}} plug in y=1

{{{x=5/5=1}}} So if y=1, x=1

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{{{x=(2+4)/5}}} plug in y=2

{{{x=6/5}}} So if y=2, x=6/5

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So here's the table filled out:
<pre>


x	y
2/5  |	-2
3/5  |	-1
4/5  |	0
1    |	1
6/5  |	2

</pre>