Ok, im a freshman at a chicago high school, and in my latest report card i got an F in algebra. can somebody teach me how to graph linear inequalities please. any help would be great.

Originally posted by molotov5
Ok, im a freshman at a chicago high school, and in my latest report card i got an F in algebra. can somebody teach me how to graph linear inequalities please. any help would be great.

hmm.... that s tough....

can you be more specific about what part of this is giving you trouble?

i will outline the general procedure: if you have an inequality like:

y > mx + b, then first you should graph the line y=mx+b, which is a line that passes through the y axis at y=b, and goes up with slope m.

when you graph the line, make it a dotted line. if this were y>= mx+b, then you would draw a solid line.

and then because y can take on values greater than this line, you should shade everything above this line.

You are given an inequality, say, (3x+2)/(x-1) >= 2

Usually with inequalities you will be asked to solve for x.

First thing to do is have a good look at your inequality and work out if there are some values for x that will not work. Take x = 1 for instance. The inequality is not defined for x = 1 because the denominator is zero.

Next step: get you inequality in a nice position to do work on it. In the above example it is a quotient. We can make everything look nice by multiplying both sides of the inequality by (x-1)^2. Remember multiplying does not change the sign of the inequality.

Now we have (x-1)(3x+2) >= 2(x-1)^2

Expand as much as you can...

3x^2 + 2x - 3x - 2 >= 2x^2 - 4x + 2

Now you must let the inequality equal zero...

x^2 + 3x - 4 >= 0

Here we now have a nice looking inequality with no brackets and one side equals zero. This is what you want to get your inequality looking like before you graph it. If your initial inequality is already like this then you don't have to worry about solving it.

As you can see this is a quadratic and by various methods you will need to calculate the roots of the quadratic. You can use my method and say to yourself: "What two numbers multiply to give -4 and what two numbers add to give +3" The answer is simply 4 and -1. So (x+4)(x-1) >= 0

Therefore x <= -4 or x > 1.

Note how I did not write x >= 1 since x cannot equal 1. So we exclude 1 from the final inequality. If you did see why x should be LESS than or equal to -4 then I will show you...

Graphing (x+4)(x-1) >= 0 is very easy. Draw your x and y axes. Then draw a parabola on it. The reason why we draw a parabola is because a parabola IS a graphical picture of a quadratic! Anyway, your parabola will cut the x-axis at 1 and -4. NOT -1 and +4, remember to reverse you signs when you take the number out of the inequality...

(x+4) >= 0
x +4 >= 0
x <= -4 See how the sign changes. As a rule of thumb you should remember this!

Anyway, so you have your concave UP parabola sitting on your xy axes cutting the x-axis at 1 and -4. Your inequality states that the quadratic is GREATER than or equal to zero. This means that on your parabola we want all values of x that are ABOVE the x axis. If the inequality stated LESS than your equal to zero then we would want all values of x BELOW the x-axis. In our case the values of x that are ABOVE the x-axis are those points on the parabola that are less than or equal to -4 or greater than +1.

EASY STEPS TO GRAPHING INEQUALITIES...
1. Simplify your inequality so that it has no brackets and one side equals zero.
2. Find roots of the inequality.
3. Draw your parabola such that it cuts the x-axis at the roots (remember to reverse the signs!)
4. Look back at your inequality after step one is completed and determine whether you want values of x GREATER than or LESS than zero.
5. Your final product should be a nice parabola sitting on the x-axis cutting the axis at the roots of the quadratic.

REMEMBER!!!
1. Always exclude the values of x that make the denominator zero.
1/(x+1) >=0 (x cannot equal -1)
1/(x+1)(x-2) <=0 (x cannot equal -1 or 2)

2. Never divide by a function of x
If you divide by a function of x you will eliminate one of the quadratic roots.

I hope this helps.

Cheers. Ben.

ok, what i dont understand about this is everything. And i appreciate the help.

Hi Molotov5, can you give us some more information about yourself and what you are doing. Then post an example of a question that you need help with and I'm sure someone will be able to run through it with you.

Cheers pal.

Suppose you have y < 5x - 3.

First, graph the line y = 5x - 3. To do that, find the intercepts. When x=0, y=-3, and when y=0, x=3/5. Plot those two points and draw a line through them (in this case a dashed line, since we have a greater-than, as opposed to greater-than-or-equal).

Next, detemine whether the required region is above or below the line. The easiest way is to test any point not on the line. Often, the point (0,0) is an easy one. In this case, put x=y=0, and we have:

0 < -3

which is true. Therefore, the point (0,0) is in the required region. It is conventional to shade the region of the graph which is NOT required, so in this case, shade everything on the side of the line which doesn't include the point (0,0).

Does that help?

Originally posted by James R
0 < -3

which is true.

And 0 = 1

oxymoron, I have to say that I think your explanation, while correct, is too complicated- also it only applies to quadratic problems while you headed it "graphing inequalities".

In the first place, Molotov5 never said whether he was talking about inequlities on the line (such as (x-4)(x+ 3)< 0) or in the plane (such as y> x^2 + 3x- 2).

Assuming, as James R did, that we are talking about graphing inequalities in two variables, the simplest way to graph an inequality is to first graph the equation:

If the inequality is y> x^2+ 3x- 2, first graph y= x^2+ 3x- 2. That's a parabola. All points that satisfy y> x^2+ 3x- 2 are on one side of that parabola, all points that satisfy y< x^2+ 3x- 2 are on the other side.

If the inequality is in one variable, as oxymoron assumed, using (3x+2)/(x-1) >= 2. "Graph" (3x+2)/(x-1)= 2: that is, mark the points on which the equation is true. This is the same as
3x+ 2= 2(x-1)= 2x-2 or x= -4. There is also a "break" where the denominator is 0: x= 1. Those two points separate "> 2" from "< 2".
If we check x= -5 (less than -4) we get (3(-5)+2)/(-5-1)= -17/-6, just slightly less than 3. Yes, this is greater than 2. All values of x less than or equal to -4 satisfy this inequality.
If we check x= 0 (between -4 and 1) we get (3(0)+2)/(0-1)= 2/-1= -2. This is NOT greater than 2. NO points between -4 and 0 satisfy it.
If we check x= 1 (larger than 0) we get (3(1)+2)/(2-1)= 5/1= 5. Yes, this is larger than 2. All values of x greater than 0 satisfy the inequality.

The graph consists of all points on the number line below and including -4 and all points on the number line larger than 0.

4DHyperCubix:

Oops. That's what happens when you're jet-lagged.