Once you've learned how
to graph linear
inequalities, you
can move on to solving systems of linear inequalities.
A "system" of
linear inequalities is a set of linear inequalities that you deal with
all at once. Usually you start off with two or three linear inequalities.
The technique for solving these systems is fairly simple. Here's an example.
Solve the following
system:
2x
3y<12 x
+ 5y<20 x
> 0
Just as with solving
single linear inequalities, it is usually best to solve as many of the
inequalities as possible for "y"
on one side. Solving the first two inequalities, I get the rearranged
system:
y> ( 2/3 )x 4
y< ( 1/5 )x + 4
x
> 0
Copyright Elizabeth Stapel 2000-2011 All Rights Reserved
"Solving" systems
of linear inequalities means "graphing each individual inequality,
and then finding the overlaps of the various solutions". So I graph
each inequality, and then find the overlapping portions of the solution
regions.
The line for
the first inequality in the above system, y> (
2/3 )x
4,
looks like this:
This inequality
is a "greater than" inequality, so I want to shade
above the line. However. since there will be more than one inequality
on this graph, I don't know (yet) how much of that upper side
I will actually need. Until I know, I can keep track of the
fact that I want the top region by drawing a little "fringe"
along the top side of the line, like this:
Now I graph the
line for the second inequality above, y< ( 1/5 )x + 4:
...and, since
this is a "less than" inequality, I'll draw the fringe
along the bottom of the line:
The last inequality
is a common "real life" constraint: only allowing
x
to be positive. The line "x
= 0"
is just the y-axis,
and I want the right-hand side. I need to remember to dash the
line in, because this isn't an "or equal to" inequality,
so the boundary (the line) isn't included in the solution:
The "solution"
of the system is the region where all the inequalities are happy;
that is, the solution is where all the inequalities work, the
region where all three individual solution regions overlap.
In this case, the solution is the shaded part in the middle:
