MAT 1302 Unit VII-Graphing Systems of Inequalities Part I Q13

5 days ago

English subtitle

Graphing systems of Inequalities, Part One.
Graph the solution set of the system of inequalities.
Here I have two linear inequalities.
In order to graph the system, I'm going to
graph each linear inequality separately on
the same coordinate plane and then shade the
overlap of the inequalities.
So first, we're going to take the top inequality,
which is 3x minus 4y is less than or equal
to 12, and I want to find the x and y-intercept
in order to graph the line.
The x-intercept is found by substituting a
0 in for y and I want to change the inequality
sign to an equal sign.
So, I have 3x minus 4 times 0 equals 12.
That's 3x minus 0 equals 12, which gives me
3x equals 12, and then, when we divide both
sides by 3, I get x equals 4.
So, my x-intercept is (4,0).
Next, we're going to find our y-intercept
and the y-intercept is found by substituting
in a 0 for x.
So, I have 3 times 0 minus 4y equals 12.
This gives me 0 minus 4y equals 12, which
is a negative 4y equals 12 and I'll divide
both sides by negative 4 and I get y equals
a negative 3.
So, my y-intercept is (0,-3).
Now we'll take these two points and plot them
on the graph: (4,0) is here and then we have
(0,-3), which is here.
If we look at my inequality sign, I have an
equal to line underneath the sign.
So, that means that my line is going to be
solid.
Now I want to decide where to shade on the
graph.
Do I shade above the line or below the line?
In order to determine this, we're going to
choose a test point and we can choose whatever
point we want as long as it's not on the line.
I'm going to choose a test point of (0,0)
and we'll plug this into the first inequality.
So, I have 3 times 0 minus 4 times 0 is less
than or equal to 12.
0 minus 0 is less than or equal to 12, so
0 is less than or equal to 12.
That is a true statement, so that means that
I'm going to shade above the line to include
that (0,0) point.
So, all this will be shaded.
That's the graph of my first inequality.
Now I'm going to graph the second inequality.
We'll look at our second inequality and first
you'll find the x-intercept.
So, I'll plug a 0 in for y and I get 4x minus
0 equals 8.
So, 4x equals 8.
Then, we'll divide both sides by 4 and we
get x equals 2.
So, my x-intercept is (2,0).
Next, we'll find the y-intercept.
We'll plug in a 0 for x and solve for y.
So, I have 4 times 0 minus 2y equals 8.
0 minus 2y equals 8 is negative 2y equals
8.
We'll divide both sides by negative 2 and
we get y equals negative 4.
So, my y-intercept is (0, -4).
Now I'm going to take both of these points
and plot them on the same graph that I plotted
the first inequality.
So, we'll find (2,0), which is here, and then
(0,-4), which is here.
We'll go to our inequality and look at the
sign.
We do not have an equal to line underneath
the sign, so I know my line is going to be
dashed or dotted.
Now I want to decide what side of the line
to shade.
We can choose whatever point we want as long
as it's not on the line.
I'm going to choose the point (1,2), which
is there.
So, I plugged in a 1 for x and a 2 for y into
the original inequality and we'll solve.
So, we get 4 minus 4 is greater than 8 and
then we have 0 is greater than 8.
That is not a true statement.
That means the point (1,2) is not going to
be included in my shading.
That means that I'll shade on the other side
of the line, which is over here.
Now, the solution to my equation is the graph
of the two lines plus the shading of the overlap.
So, I only want to include the shading of
the overlap of the two lines and as you can
see the overlap is going to be within these
two lines here.
So, my solution is what you see in the graph.
I have my first equation with a solid line,
that's a black line, and then I have my second
equation with a dotted line, that's the blue
line, and then the overlap of the shadings
is represented by the purple shading, which
is in the top middle of the two lines.