At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
solve for y y = (6-2x)/3
plug into graphing calc
thanks but i dont have one
try this: http://www.wolframalpha.com/
plug the equation i gave u in there and then just copy it down
Or you could just... you know... plug in two points and use a ruler?!?
thank you I give it a try I am just having a hard time understanding how to plot or knowing when to when i look at a problem
Ok, well pick an x that's nice (say 0). Then plug it in and find out what y is.
what do you get for y?
I did not know should I replace x with a 0?
y = 3
(x,y) = (0,3). Then do the same thing for y (plug in a 0) and find a second point
for (x,y) = (?,0)
Kool thank you so my answer is x=0,3 and y= 3,0?
When you plug in a 0 for x you get y = 3 that's correct. That gives you the point (x,y) = (0,3). If you plug in a 0 for y you have this equation: \[0 = (6-2x)/3\]
Which gives you \[x=3\] sorry, missread it. Ok, so now you have a second point (x,y) = (3,0)
So you draw a Cartesian grid and draw those two points, then connect them with a straight line. That line represents all the x,y pairs that are solutions to the equation.
You are so smart, that actually makes sense but what about y=5x-1 do i still use 0?
would I go up 5 and down 1??
If you are just being asked to graph it you can plug anything you want in for x or y (but not both!) then solve for the one you didn't plug in and you will have a point on the graph.