# 5-1 Graphing systems of linear inequalities ```5-1 Graphing systems of linear inequalities
GRAPHING A SINGLE INEQUALITY
Given:
Note:
Objective:
inequality 3x + y &lt; 9
(1, 2) satisfies it, (3, 1) does not satisfy it
show (on a coordinate system)
all points (x, y) that satisfy it
Steps:
1. GRAPH: graph the equation 3x + y
=
9
(a) all the points on one side or the other side of the line
satisfy the equation. As for the line itself:
(b) if the inequality is strict (&lt; or &gt;)
the line itself will not be included in the solution set.
Indicate this by drawing the line as a dashed line.
(c) if the inequality is non-strict ( or )
the line is included. Draw as a solid line.
2. TEST POINT: pick any point not on the line. (0, 0) is best
(if it is not on the line). Otherwise pick (0,1) (or
whatever). Substitute the test point in the inequality.
(a) if the test point satisfies the inequality:
shade the same side of the line as the test point
(b) if the test point does not satisfy the inequality:
shade the other side of the line
Tip: you can do it without writing anything down except the
solution!
5-1
p. 1
Example:
5-1
3x + y &lt; 9
p. 2
GRAPHING SYSTEMS OF INEQUALITIES
E. g.
y  x-4
(AND)
y &lt; -2x + 4
Objective (as always): describe all points (x, y) that satisfy
both inequalities (by the implicit AND above).
METHOD IS SIMPLE:
1. Graph both inequalities on the same coordinate system
2. Result is where the two individual shadings overlap
TECHNIQUE IS SIMPLE (don't write anything, just the answer!):
1. Graph both lines -- but don't shade yet; just indicate where
you would shade by using little arrows:
5-1
p. 3
Corner points
The corner points of the solution of a system of equations are
those places where shaded area has a “corner”, and are where
the lines you have graphed intersect. The above solution has
one corner point.
The corner points are easy to find. Remember,
 the point where two lines intersect . . .
 is the point . . .
 that satisfies both equations
So all we need to do is solve (using algebra) the system of
equations represented by the two lines!
y = x - 4
y = -2x + 4
Using elimination, we multiply the second equation by -1 and
to get
or
in turn
y =x - 4
-y = 2x - 4
----------------0
= 3x - 8
x
= 8/3
y
= -4/3
So the corner point is: (8/3, -4/3)
Frequently, a system will have the two additional inequalities
x  0, y  0. This, of course, means that all solutions lie in
Quadrant I, and more corner points will be introduced
(including the origin (0, 0)).
5-1
p. 4
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