# Thread: I'm having some trouble with graphing inequalities

1. ## I'm having some trouble with graphing inequalities

I'm having a very hard time with graphing and I can't seem to figure out how to graph inequalities. I appreciate any help that you have to offer me.

2. Originally Posted by alwaysalillost
I'm having a very hard time with graphing and I can't seem to figure out how to graph inequalities. I appreciate any help that you have to offer me.
the same principles apply with these as in your other thread about graphing inequalities, in that, we will attempt to solve the inequalities in such a way that they look like formulas we recognize.

1)
x + y < 1

let's solve for y
=> y < -x + 1
hmm, if not for the inequality sign,this would look like a line, with slope -1 and y intercept 1, and that's what we will draw. however, since it is an inequality, we have to shade the region where y is less than the values given by the line. the line is not included in the solution since we have a strictly less than sign, so we must draw a broken line. see below

3. Originally Posted by alwaysalillost
I'm having a very hard time with graphing and I can't seem to figure out how to graph inequalities. I appreciate any help that you have to offer me.
the second one, same story

y <= 2x + 1

well, this isn't exactly the same story, since we don't have to simplify anything, they gave us in the useable form.

what does this remind you of?...i hope it reminds you of a line with slope 2 and y-intercept + 1, we will graph that line and shade the region under it, since we want y less than or equal to the value of the line. since we can be equal to the values of the line, the line we draw is solid and is included in the solution

4. Originally Posted by alwaysalillost
I'm having a very hard time with graphing and I can't seem to figure out how to graph inequalities. I appreciate any help that you have to offer me.
the third times a charm, so maybe i should leave this one to you

just kidding, i'm kind of bored at the moment anyway, besides, this is a bit different from the other problems.

we have 2 inequalities, and we will graph both at the same time

y > x + 1
y < (3/2)x + 3

so, these are in the nice form again. let's graph these lines.

now here i drew the lines bright and bold, howerever, both should be broken, i had to do it like that so you could see them. the grey region satisfies the first inequality, the blue region satisfies the second, the pink region satisfies both simultaneously, and the unshaded region satisfies neighter. i hope you understand the graph and why the regions are shaded the way they are. if you are unclear about it, say so

5. Thanks Jhevon for your help but I was wondering if you could explain to me how you got the last answer a little bit further. I'm a bit confused and don't quite understand. I'm greatful for all your help.

6. Originally Posted by alwaysalillost
Thanks Jhevon for your help but I was wondering if you could explain to me how you got the last answer a little bit further. I'm a bit confused and don't quite understand. I'm greatful for all your help.
i simply did each line one at a time. so i graphed the first line, and shaded the required region, just as i did in previous questions, then i graphed the second line and shaded the required region, just as i did in previous questions again. and that's it. i used different colors to differentiate the solutions. the grey area is the region that is a solution for only one of the lines, the blue region is the solution for the other and the pink region works for both problems. so for this question you pretty much do two solutions on a single graph, nothing too fancy. understand?

7. Yes. Thank you very much!