Systems of Inequalities

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Systems of Inequalities Created By Richard Gill
and Jeannie Taylor For Mth 163 Precalculus
1 Funded by a Grant from the VCCS LearningWare
Program
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This presentation starts with a few review slides
on graphing linear inequalities and a few review
slides on graphing non-linear inequalities. To
skip either or both review sessions, click the
appropriate button below and you will bypass the
review. You can certainly come back to the review
later if you need to. At the end of the
presentation there will be a set of exercises
with answers and occasional complete solutions.
Good luck.
Click here to see the review on linear
inequalities.
Click here to see the review on non-linear
inequalities.
Click here to skip both review sessions.
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A short review on graphing inequalities.
In order to graph the inequality y gt 3 x first
graph the equation y 3 x. This line will be
the borderline between the points that make y gt 3
x and the points that make y lt 3 x.
In y mx b form we have y -x 3. In this
case we have a line whose slope is 1 and whose
y-intercept is 3.
Now we have to decide which side of the line
satisfies y gt 3 x.
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A short review on graphing inequalities.
All we have to do is to choose one point that is
off the line and test it in the original
inequality. If the point satisfies the inequality
then we are on the correct side of the line and
we shade that side. If the point does not satisfy
the line, we shade the other side.
The most popular point to use in the shading test
is (0, 0).
THE TEST substitute (0, 0) into y gt 3 x and
see if you get a true statement.
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Since (0, 0) did not satisfy the inequality y gt 3
x we conclude that (0, 0) is on the wrong side
of the tracks and we shade the other side. Our
conclusion is that every point in the shaded area
is part of the solution set for y gt 3 x.
Each point that we pick in the shaded area
generates a true statement.
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A short review on graphing non-linear
inequalities. This same shading technique can be
used for shading nonlinear inequalities such as
We can start by solving for y and graphing the
borderline as a dotted line.
Do you know what kind of graph this equation will
generate?
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Take a bow if you thought the borderline was
going to be a parabola.
As we did with the linear inequality, we can
determine which side to shade with a test point.
But (0, 0) will not work as a test point this
time. Why?
(0, 0) does not work as a test point because it
is not on either side of the curve.
We need a test point that is on one side or the
other.
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Take a minute to test (2, 1) in the original
inequality
(2, 1) does not solve the original inequality so
we will shade on the other side of the curve.
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So here is the region that solves the original
inequality.
You can use the label at the top or the one at
the bottom. Why is the curve dotted?
You are so shrewd. The curve is dotted to
indicate that points on curve are not part of our
solution set. When we dont have the equal to
option, the borderline is dotted.
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Graphing Systems of Inequalities Finally, we will
take up the mission of this presentation.
Consider the following system on inequalities
We will solve each inequality individually and
then look for points that satisfy both
inequalities at the same time. As before, we
graph the border and then decide which side of
the curve to shade.
The first function has a restriction on its
domain. The contents of the root have to be
greater than or equal to zero. What is the domain
of the first function?
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Since the contents of the root have to be
non-negative, the domain will include values of x
no larger than 3. What kind of shape will the
graph of this equation take on?
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(0, 0) passes the test, so (0, 0) is on the
correct side of the curve and we shade in that
direction.
The test for (0, 0)
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(0, 2) passes the test so we shade in that
direction.
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We now have the shading for each inequality in
the system
We now put the two graphs together and look for
the points that are shaded in both inequalities.
In other words, we look for the points that solve
the inequalities simultaneously.
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One more example and you can try your luck with a
few exercises.
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First a few tips. You will frequently see systems
of inequalities with some of the restrictions
below. Try to visualize each one before you click.
x gt 0
x gt 0 and y gt 0
y gt 0
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So spend a few minutes with the system below
before you proceed. Graph each borderline on the
same axis, use a test point to shade each
inequality and highlight the region that is
shaded in each inequality.
It will stick with you better if you try the work
yourself before you proceed.
If you have not already done so figure the
shading before you click.
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We can use (0, 0) as a test point for each
inequality.
This is not true so we shade on the other side of
the curve.
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We can use (0, 0) as a test point for each
inequality.
This is not true so we shade on the other side of
the curve.
Similarly
This is true so we shade on the side of the curve
that contains (0, 0).
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The area that gets shaded twice is our solution
set. It contains all points that solve both
inequalities simultaneously.
More precisely, our solution set looks like this
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Each point in the shaded area solves each
inequality simultaneously.
One more example and then you can spend some time
on the exercises.
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• See what you can do with the following system of
  inequalities. 1. Graph each borderline.
• 2. Shade each inequality.
• 3. Highlight the area that is common to
  each shading.

Congratulations if you knew this was a circle
centered at the origin with radius of 5.
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In order to get the whole circle to show on
Winplot or on a graphing calculator you have to
solve for y and graph each root.
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In order to shade, we go back to the first
inequality.
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In order to shade, we go back to the first
inequality.
Now we can focus on the absolute value
inequalities. Rewrite as
Note that (0, 0) satisfies each of the linear
inequalities.
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With all four linear boundaries in place, can you
see which regions will be shaded?
x 5
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With all four linear boundaries in place, can you
see which regions will be shaded?
Note that points in the shaded region like (4, 4)
and (-5, -3) satisfy each inequality in the
system.
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Now check out the following exercises. All of the
answers are listed at the end of the presentation
along with a few of the complete solutions.