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How am I suppose to graph 2x - y + 3 greater than or equal to 0? Do I need to solve for x and y? If so how in the hell do I do it? Please help! I really don't understand. Show your work and please also explain along the way.

Thanks!

2006-12-19 01:38:41 · 3 answers · asked by 2good4hem 3 in Science & Mathematics Mathematics

3 answers

Solve for y,
2x - y + 3 >= 0
2x + 3 >= y
y <= 2x + 3

Now draw the line y = 2x + 3 and shade everything below it.
The line is in slope-intercept form. Two points are (0, 3) and (-3/2, 0). Draw the line that passes through both of them.

2006-12-19 01:45:27 · answer #1 · answered by computerguy103 6 · 0 0

As a convention let's represent "greater than or equal to" with ">=", okay?

You can put 2x - y + 3 >= 0 in slope-intercept form:
Add 'y' to both sides:
2x + 3 >= y,
Rewrite the inequality so that 'y' is on the left side:
y <= 2x + 3 (note the change in the inequality symbol),

You can draw the line y = 2x + 3; the slope is 2 and the line passes through (0, 3).
Because we have that y is also less than all this, you can shade the entire region lying below the line.

If the question was originally phrased with "greater than" instead of "greater than or equal to," you would draw a dotted line instead of a solid line (and still shade the underlying region).

2006-12-19 01:54:31 · answer #2 · answered by Bugmän 4 · 0 0

first treat the equation as if it was an equals sign. Re arrange the equation to get y=2x+3. Now you know that all of the points on this line fit the equation 0=2x-y+3 as it is a rearrangement.

For the inequality you want x to be big and y to be small so that the equation is positive so this is the region below the line. (you can test this by plugging in a point in this area eg(0,0).)

If it was just a greater than sign you would make th line a dotted one as it it would be all values below the line but not the line itself, in this case as the line contains solutions to the inequality you leave it as a solid line.

2006-12-19 01:57:00 · answer #3 · answered by tom s 1 · 0 0

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