How would i graph a system of inequalities on a coordinate plane?

The system of inequalities, that i have to graph on a coordinate plane, has 5 inequalities: x%26gt;=9, x%26lt;=36, y%26gt;=55, y%26lt;=110 and x+y%26gt;=80.


I think that you have to put it in slope-intercept form...





%26lt;= means les than or equal to.|||Hi Chelsea--





Well, the first four form a square region in the coordinate plane. Every inequality is LTE (less than or equal to) or GTE (greater than or equal to) so all of your lines will be SOLID, not dashed.








So the top of the square is at y = 110, the bottom at y = 55, the left at y = 9, and the right at x = 36.





Once you do the cross-hatching it gets very messy, so it's good to do them lightly the first time through.





Now, to throw the line in there:





You can throw it in slope intercept form, but I prefer to just plot the x-intercept and the y-intercept. For your equation it works really well:





y-intercept is where x = 0, so 0 + y %26gt;= 80. Solving, y = 80. This is the point (0, 80).


x-intercept is where y = 0, so x + 0 %26gt;= 80. Solving, x = 80. This is the point (80, 0).





When you connect these points, you see that your line FALLS from the left to the right. (So, yes, in slope-intercept form it is y = (-1)x + 80 and your slope is -1.)





Now, you have to pick one side of the line or the other. Try an easy point to see if the inequality is true. (0,0) is your best choice:





0 + 0 %26gt;= 80 ***FALSE, you need to choose the other side of the line and shade that side.





Your answer will be the figure shaded TRUE for all of these.





Hope that helps. It would be much easier to explain with a few pictures.