2006-11-19 14:27:43 · 7 answers · asked by wanda_brown62 1 in Science & Mathematics ➔ Mathematics
I hope this could help..
2006-11-19 14:58:04 · answer #1 · answered by M. Abuhelwa 5 · 0⤊ 0⤋
2x+y=3
y=3-2x
this is your equation...3 is where you need to start, so graph (0,3), then since your slope is -2/1, from (0, 3) go down 2 and over to the right 1. Repeat that process again from (1,1). That will give you three points. Your points should be (0,3), (1,1), (1,-1)...hope this helps any.
2006-11-19 16:00:51 · answer #2 · answered by Kaylin 4 · 0⤊ 0⤋
Change to:
y = -2x + 3
The y-intercept is solved by making x = 0. So...
y = -2(0) + 3
y = 3
From there, -2x is your slope.
-2x is really -2/1x so you'd move rise over run.
Down two, over one is a point, down to, over one is a point. Go back to your y-intercept, go up two, left one, up two, left one, etc.
Connect a few dots, and draw arrows on both sides of the line.
2006-11-19 14:30:40 · answer #3 · answered by Anonymous · 1⤊ 0⤋
The graph should be a straight line. You plug in numbers for x and y values. Here are some sample points: (-2,7), (-1,5), (0,3), (1,1), (2,-1).
2006-11-19 14:36:31 · answer #4 · answered by Anonymous · 0⤊ 0⤋
first make the equation equivalent, then get x via utilising itself, and graph it. if (0,0) isnt on the line, placed that via the time of the approach the equation and if it works, shade that part of the line want that helped in any have fun with
2016-12-17 12:54:21 · answer #5 · answered by Anonymous · 0⤊ 0⤋
ok you are never gonna pass the class by getting your answers off of yahoo answers. how will you even pass the tests?
2006-11-19 14:30:16 · answer #6 · answered by keep it real 4 · 0⤊ 1⤋
a straight line thru (0,3) and (1.5,0)
2006-11-19 14:30:01 · answer #7 · answered by Anonymous · 1⤊ 0⤋