Transcript Systems of Inequalities

Systems of Inequalities
Created By
Richard Gill and Jeannie Taylor
For Mth 163: Precalculus 1
Funded by a Grant from the
VCCS LearningWare Program
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.
A short review on graphing inequalities.
In order to graph the inequality y > 3 – x first graph the equation
y = 3 – x. This line will be the borderline between the points that
make y > 3 – x and the points that make y < 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 yintercept is 3.
4.0
y
2.0
-4.0
-2.0
2.0
-2.0
Now we have to decide which side
of the line satisfies y > 3 – x.
-4.0
4.0
x
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.
4.0
The most popular
point to use in the
shading test is (0, 0).
THE TEST: substitute
(0, 0) into y > 3 – x
and see if you get a
true statement.
2.0
-4.0
-2.0
2.0
-2.0
-4.0
0>3-0
y
0 > 3, which is false.
4.0
x
Since (0, 0) did not satisfy the inequality y > 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 > 3 – x.
4.0
You can reinforce this idea
by testing several points in
the shaded area.
(2, 2)
2>3–2
2>1
(0, 3)
3>3–0
3>3
(4, 1)
1>3–4
y
2.0
-4.0
1 > -1
Each point that we pick in the shaded
area generates a true statement.
-2.0
2.0
-2.0
-4.0
4.0
x
A short review on graphing non-linear inequalities.
This same shading technique can be used for shading nonlinear
inequalities such as:
x2 - 4x  y  0
Once again the first step is to graph the equality which will serve as
the borderline between the points whose x-y coordinates make
x 2 - 4x  y  0 and the points whose x-y coordinates make
x 2 - 4x  y  0.
We can start by solving for y and graphing the borderline as a
dotted line.
x2 - 4x  y  0
y  4x - x2
Do you know what
kind of graph this
equation will
generate?
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.
y  4x - x2
We need a test point that is on one side or
the other.
2
x
- 4x  y  0.
Take a minute to test (2, 1) in the original inequality:
x - 4x  y  0 ?
2
2 2 - 42  1  0 ?
4 - 8 1  0?
-3  0
(2, 1) does not
solve the
original
inequality so we
will shade on the
other side of the
curve.
y  4x - x2
So here is the region that solves the original inequality.
y  4x - x2
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 don’t have the
“equal to” option,
the borderline is
dotted.
x2 - 4x  y  0
Graphing Systems of Inequalities
Finally, we will take up the mission of this presentation. Consider
the following system on inequalities:
 y  3 - x

 y  x 2
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?
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?
y  3- x
3- x  0
3  x or x  3
Any graph that can be written in the form y  mx b
will generate half of a horizontal parabola.
4.0
y
2.0
-4.0
-2.0
2.0
-2.0
-4.0
4.0
x
Now we must decide which
side of the curve to shade.
Why don’t you take a
minute to use (0, 0) as a test
point and see if (0, 0) is on
the side of the curve that
satisfies y  3 - x.
The test for (0, 0):
y  3- x
0  3-0 ?
(0, 0) passes the test, so
(0, 0) is on the correct
side of the curve and we
shade in that direction.
0  3!!
4.0
2.0
-4.0
-2.0
y
y  3- x
2.0
-2.0
-4.0
4.0
x
Next we turn our
attention to the second
inequality. Before you
click, see if you can
figure the border and
the shading for y  x 2 .
The border for the inequality is, of course, the world famous
parabola y  x 2 .
4.0
(0, 0) is not an
appropriate test point
but (0, 2) will suffice.
Does (0, 2) satisfy
y
2.0
y  x2 ?
-4.0
-2.0
2.0
-2.0
-4.0
4.0
x
yx
2
20 ?
2
2  0!!
(0, 2) passes the test so we shade in that
direction.
We now have the shading for each inequality in the system:
yx
4.0
2
4.0
y
y
2.0
y  3- x
2.0
-4.0
-2.0
2.0
4.0
x
-4.0
-2.0
2.0
-2.0
-2.0
-4.0
-4.0
4.0
x
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.
The intersection of
the two shadings is
the solution for the
system:
 y  3 - x

 y  x 2
One more example and you can try your luck with a few
exercises.
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>0
y>0
x > 0 and
y>0
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.
y
 y  x - 3

 y  4 - x
4.0
y  4- x
It will stick
with you better
if you try the
work yourself
before you
proceed.
y  x -3
2.0
-2.0
2.0
4.0
x
6.0
-2.0
If you have not already done so figure the shading before you click.
We can use (0, 0) as a test point for each inequality.
y  x-3
0  0-3 ?
y
0  3 ??
This is not true so we
shade on the other
side of the curve.
4.0
y  4- x
y  x -3
2.0
-2.0
2.0
-2.0
4.0
x
6.0
We can use (0, 0) as a test point for each inequality.
y  x-3
y
4.0
0  0-3 ?
0  3 ??
This is not true so we
shade on the other
side of the curve.
Similarly…
2.0
-2.0
2.0
4.0
-2.0
y  4- x
0 4-0?
0  2!!
This is true so we shade on the side of
the curve that contains (0, 0).
x
6.0
The area that gets shaded twice is our solution set. It contains all
points that solve both inequalities simultaneously.
y
 y  x - 3

 y  4 - x
4.0
2.0
More precisely, our
solution set looks like
this…
-2.0
2.0
-2.0
4.0
x
6.0
Each point in the
shaded area solves
each inequality
simultaneously.
y  x -3
y
 y  x - 3

 y  4 - x
4.0
y  4- x
2.0
-2.0
2.0
-2.0
One more example and then you can
spend some time on the exercises.
4.0
x
6.0
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.
 x 2  y 2  25

y 5

x 5
x 2  y 2  25 will generate the first border. What kind of graph is this?
Congratulations if you knew this was a circle centered at the
origin with radius of 5.
y  25 - x 2
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.
y
4.0
2.0
-4.0
-2.0
2.0
-2.0
-4.0
4.0
x
6.
y
4.0
In order to shade, we go back to
the first inequality.
x 2  y 2  25
y  25 - x 2
2.0
-4.0
-2.0
2.0
4.0
x
6.0
-2.0
Using test point (0, 0) you can see
that 0 + 0 > 25 so we will shade
outside the circle.
-4.0
y  - 25 - x 2
y  25 - x 2
y
In order to shade, we go back to
the first inequality.
4.0
2.0
x 2  y 2  25
-4.0
Using test point (0, 0) you can see
that 0 + 0 > 25 so we will shade
outside the circle.
-2.0
2.0
4.0
x
6.0
-2.0
-4.0
y  - 25 - x 2
Now we can focus on the absolute value inequalities. Rewrite as:
y  5  -5  y  5
x  5  -5  x  5
Note that (0, 0) satisfies each of
the linear inequalities.
With all four linear
boundaries in place,
can you see which
regions will be
shaded?
y
y=5
4.0
2.0
-4.0
-2.0
2.0
4.0
x
6.0
-2.0
-4.0
y = -5
x = -5
x=5
With all four linear
boundaries in place,
can you see which
regions will be
shaded?
So the shaded area
contains the points
that solve:
 x 2  y 2  25

y 5

x 5
x=5
Note that points in the shaded region like
(4, 4) and (-5, -3) satisfy each inequality
in the system.
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.