SOLUTION: Hi there. I have been stuck on this for hours! Here is the following question. The value of y for each is listed below "y", hence I need to find x. Could you show me how to do

Algebra ->  Linear-equations -> SOLUTION: Hi there. I have been stuck on this for hours! Here is the following question. The value of y for each is listed below "y", hence I need to find x. Could you show me how to do       Log On


   

Question 77505: Hi there. I have been stuck on this for hours! Here is the following question. The value of y for each is listed below "y", hence I need to find x. Could you show me how to do at least the first one so that I can figure out the rest? Thank you soooo much!

Fill in the chart given this linear equation:
y = 5x – 4
**Type all your work below the chart.
x y
-2
-1
0
1
2
Found 3 solutions by Earlsdon, stanbon, jim_thompson5910:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Find the value of y for each of the given values of x using the equation:
y+=+5x-4
All you need to do is to substitute each of the given values of x into the equation and solve for the corresponding value of y.
1) x = -2
y+=+5%28-2%29-4
y+=+-10-4
y+=+-14
2) x = -1
y+=+5%28-1%29-4
y+=+-5-4
y+=+-9
3) x = 0
y+=+5%280%29-4
y+=+0-4
y+=+-4
Do you see how it goes?

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Fill in the chart given this linear equation:
y = 5x – 4
Solve for "x":
5x=y+4
x=(y+4)/5
This is the rule for getting the x-value associated with a y-value:
**Type all your work below the chart.
For example:
If y= -2 then x=(-2+4)/5=2/5
If y=-1 then x=(-1+4)/5=3/5
etc.

x........y
2/5.....-2
3/5.....-1
4/5.... 0
5/5=1...1
6/5.....2
------------
Cheers,
Stan H.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
If you have y=-2 then you simply replace y with -2 and evaluate x. So for instance
y=5x-4 let y=-2
%28-2%29=5x-4 plug in y=-2
2=5x Add 4 to both sides
x=2%2F5 So if y=-2, then x=2%2F5 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%2B4=5x Add 4 to both sides
x=%28y%2B4%29%2F5 Divide both sides by 5 to solve for x.

Now if we evaluate each y we get:


x=%28%28-2%29%2B4%29%2F5 plug in y=-2
x=%282%29%2F5 So if y=-2, x=2/5
--------------------------------------------
x=%28%28-1%29%2B4%29%2F5 plug in y=-1
x=%283%29%2F5 So if y=-1, x=3/5
--------------------------------------------
x=%28%280%29%2B4%29%2F5 plug in y=0
x=4%2F5 So if y=0, x=4/5
--------------------------------------------

x=%281%2B4%29%2F5 plug in y=1
x=5%2F5=1 So if y=1, x=1
--------------------------------------------

x=%282%2B4%29%2F5 plug in y=2
x=6%2F5 So if y=2, x=6/5
--------------------------------------------
So here's the table filled out:


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