Lesson BASICS - Graphing Inequalities

This Lesson (BASICS - Graphing Inequalities) was created by by longjonsilver(2297)

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About longjonsilver: I have a new job in September, teaching

Introduction
This discussion is focussed on straight line graphs. However, the theory can be applied to any type of graph.

I shall show you the very simple process of graphing inequalities and find the "region" bounded by several straight line inequalities.

Method
When asked to graph an inequality, first you plot (or sketch) the equality version, to get the straight line. Then we need to figure out which half of the graph is the required area.

So, lets look at an example...

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Example: Graph y < 3x + 1.
Solution
First we will plot the equation y = 3x + 1, to get the straight line. I am not going to teach you how to plot a graph, that is another Lesson. Hopefully, you can either plot it from coordinates, or if your maths skills are sufficient, you can sketch it by looking at the equation.

The graph is graph%28300%2C300%2C-5%2C5%2C-10%2C10%2C3x%2B1%29

graph%28300%2C300%2C-5%2C5%2C-10%2C10%2C3x%2B1%29

Explanation: Now, if you walk along this line, every point is where y EQUALS 3x+1: that is what the equation said, and we have plotted that as a graph.

Now, how about the equation y < 3x+1? This is saying "where is y LESS THAN 3x+1. The answer will be either EVERYWHERE ABOVE THE LINE or EVERYWHERE BELOW THE LINE. The issue is how to figure out this.

THe way i started doing it was to pick a point either below or above the line... ANY point will do, but best if you pick an "easy" point. For that reason i pick one on the y-axis. Here, lets pick (0,0) the origin. The only thing to remember is pick a point that doesn't lie on the line.

So,
put (0,0) into y and 3x + 1
--> 0 and 3(0) + 1
--> 0 and 0 + 1
--> 0 and 1
so this says 0 is less than 1... just by looking at the numbers: zero IS less than one.

--> 0 < 1
so (0,0) is in the region y < 3x + 1, which is the region we were asked for...so we want the whole of the space BELOW the straight line.

To show this, shade the region above the line and write a key saying "unshaded region is the required region"

Question: why shade the region we don't want?
Answer: i shall show you with the next question.

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Example: Show the region bounded by y%3C3x%2B1

and

.
Solution: We have already done the first equation.

Second equation - plot y = -2x - 1 on the same graph, and you get graph%28300%2C300%2C-5%2C5%2C-10%2C10%2C3x%2B1%2C+-2x-1%29

graph%28300%2C300%2C-5%2C5%2C-10%2C10%2C3x%2B1%2C+-2x-1%29

So we already know that the region we want is below y=3x+1. We now need to figure out which part of that is satisfied by the second equation.

So again, pick a point not on the line, again i pick (0,0):
--> y and -2x-1
--> 0 and -2(0)-1
--> 0 and 0-1
--> 0 and -1
--> zero is greater than -1
--> y > -2x-1 which is the region we were asked for. This is the part above and to the right of y=-2x-1 (since this is where point (0,0) is, so shade the OTHER part, the part we don't want --> below and to the left... what remains is a region, unshaded, that is the region asked for in the question.

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Summary
Hopefully you can see from this, that the process is one that builds up. You have to do a certain amount of work, but that work is systematic and quite easy.

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Additional Example

The following example answers just the second part of the students' question.

Question 28678: 1. Solve the following system of linear inequalities by graphing.
x + 2y <= 3
2x � 3y <= 6

2. Solve the following system of linear inequalities by graphing.
3x + 4y <=12
x + 3y <= 6
x => 0
y => 0

Please help!

See answer to question 28678

Answer #15636 by longjonsilver(2297)

You can put this solution on YOUR website!
First, plot or sketch the first 2 straight lines:

graph%28300%2C300%2C-3%2C10%2C-2%2C6%2C%28-3%2F4%29x%2B3%2C+%28-1%2F3%29x%2B2%29

graph%28300%2C300%2C-3%2C10%2C-2%2C6%2C%28-3%2F4%29x%2B3%2C+%28-1%2F3%29x%2B2%29

is the brown line, also quoted as y+%3C=+%28-3%2F4%29x+%2B+3

is the green line, also quoted as y+%3C=+%28-1%2F3%29x+%2B+2

So, for the first equation... pick a point that is NOT on the line. And pick a nice easy point ie one with small numbers...i choose the origin (0,0).

y and (-3/4)x + 3
--> 0 and (-3/4)(0) + 3
--> 0 and 0 + 3
--> 0 and 3
--> zero is less than 3
so y+%3C=+%28-3%2F4%29x+%2B+3

is the region below the line... SHADE EVERYTHING ABOVE THE LINE, as this is NOT wanted.

Repeat for the second line, and pick a point, again i pick (0,0):
y and (-1/3)x + 2
--> 0 and (-1/3)(0) + 2
--> 0 and 0+2
--> 0 and 2
--> zero is LESS than 2
so y+%3C=+%28-1%2F3%29x+%2B+2

is the region below that line... SHADE EVERYTHING ABOVE THE LINE, as this is NOT wanted.

And now, the third line, x%3E=0

. Well x=0 is the y-axis. And we want every point/place on the graph that is greater or equal to that line...this is everything to the right of the y-axis. SHADE EVERYTHING TO THE LEFT OF THE y-axis, as this is NOT wanted.

And now, the last line, y+%3E=+0

. Well, y=0 is the x-axis. And we want every point/place on the graph that is greater or equal to that line... this is everything above the x-axis. SHADE EVERYTHING BELOW THE x-axis, as this is NOT wanted.

What region is left clear? is is the little 4-sided shape in the middle, bounded by the x- and y-axis and the 2 straight lines forming a ^ shape.

jon.

Question 28917: I am totally lost,
I'm suppose to solve the inequality and then graph
absolute value x+4 end of absolute value less than 6
please help
See answer to question 28917

Answer #15843 by longjonsilver(2297)

You can put this solution on YOUR website!
the graph of y=|x+4| looks like: +graph%28300%2C300%2C+-12%2C4%2C-4%2C10%2Cabs%28x%2B4%29%29+

it is just the graph of y = x+4 except where we have negative y-values...here the line is reflected back...the line is essentially y=-(x+4)

So, look at the V-shaped line. We need to know where this is less than 6... draw a horizontal line at y=6 and you get the answers:

+graph%28300%2C300%2C+-12%2C4%2C-4%2C10%2Cabs%28x%2B4%29%2C+6%29+

+graph%28300%2C300%2C+-12%2C4%2C-4%2C10%2Cabs%28x%2B4%29%2C+6%29+

OK, algebraically, we have in essence 2 equations to compare with y=6.

1. x+4<6
--> x < 2 is one answer (which matches with the graph)

2. -(x+4) < 6
--> -x - 4 < 6
--> -x < 10
--> x > -10 is the other answer (which matches with the graph too).

jon.

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