I have an exam on problems like these and my teacher gave me a worksheet with these problems but I have no idea how to solve them. If you can please help me I'd really appreciate it. Thanks for any help. :)

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- May 9th 2007, 06:53 PMalwaysalillostGraphing Inequalities
I have an exam on problems like these and my teacher gave me a worksheet with these problems but I have no idea how to solve them. If you can please help me I'd really appreciate it. Thanks for any help. :)

- May 9th 2007, 07:00 PMJhevon
Here is the first:

let's simplify it a bit to see if the functions looks like something we can recognize

y - 3 >= 5

=> y >= 8

this is just a horizontal line bounding a region. we drae the line y = 8 and shade the region above it. note that the line is a part of the solution since we can be equal to 8. see the diagram below - May 9th 2007, 07:04 PMJhevon
Here's the second, again, let's see if we can make it simpler:

x - 5 < 3

=> x < 8

now this is a vertical line, everything in the region to the left of it will be shaded since we must be less than 8. note that the line is NOT a part of the solution, so we must draw it as a broken line. see the diagram below. - May 9th 2007, 07:09 PMJhevon
Here's the third. same routine, simplify

2x + 7 > 5

=> 2x > -2

=> x > -1

so we draw the vertical line x = -1 and shade everything to the right of it, since x must be greater than this line. again, the line is NOT included since we can't be equal to it, and so we must draw the line broken - May 9th 2007, 07:12 PMJhevon
And the fourth has arrived. let's simplify him first.

5x <= 20

=> x <= 4

so we draw the line x = 4 and shade the region to the left of it, since we must be less than. the line IS included this time since we can be equal to it. so do not draw it broken, draw a solid line. - May 9th 2007, 07:15 PMJhevon
- May 9th 2007, 07:27 PMalwaysalillost
Thans Jhevon! Once again you have explained to me very clearly and I was able to understand my work a lot better. Thanks again!