A system of inequalities is a set of two or more inequalities that must be true
at the same time. In two variables, the solution is not usually a single number.
Instead, it is a set of ordered pairs \((x,y)\). When we graph those ordered pairs,
we get a feasible region.
1. Boundary curves
Each inequality has a boundary. For a linear inequality such as
\[
y>x+2,
\]
the boundary is the line
\[
y=x+2.
\]
For a non-linear inequality such as
\[
x^2+y^2\le25,
\]
the boundary is the circle
\[
x^2+y^2=25.
\]
The inequality tells us which side of the boundary is allowed. With strict inequalities
\((<,>)\), the boundary is not included. With non-strict inequalities \((\le,\ge)\),
the boundary is included.
2. Solid and dashed boundaries
3. Feasible region
For one inequality, the solution is one side of a boundary line or curve. For a system,
we must satisfy every inequality at the same time. Therefore, the feasible region is the
intersection of all allowed regions.
\[
\text{Feasible region}
=
R_1\cap R_2\cap R_3\cap\cdots
\]
If the shaded regions overlap, the overlap is the solution. If they do not overlap,
the system has no feasible solution in the chosen graph window.
4. Linear systems and vertices
A linear inequality in two variables can be written as
\[
ax+by+c\le0.
\]
Its boundary is the line
\[
ax+by+c=0.
\]
A system of linear inequalities creates a region made from half-planes. The corners of
that region are called vertices. Vertices occur where two boundary lines intersect.
For two lines
\[
a_1x+b_1y+c_1=0,
\qquad
a_2x+b_2y+c_2=0,
\]
the intersection point is found by solving the two equations simultaneously. The point
is a true vertex only if it satisfies all inequalities in the system.
5. Non-linear systems
Non-linear systems can have curved boundaries, such as parabolas, circles, absolute-value
graphs, or trigonometric curves. For example:
\[
y\ge x^2-2,
\qquad
y\le4.
\]
The feasible region is above the parabola and below the horizontal line. A system like
\[
x^2+y^2\le25,
\qquad
y\ge x
\]
is the part of the disk of radius \(5\) that lies above the line \(y=x\).
6. Test-point method
A useful way to decide which side of a boundary to shade is to choose a test point.
Substitute the point into the inequality. If it makes the inequality true, shade the
side containing the test point. If it makes the inequality false, shade the opposite side.
For a system, the same point must pass every test. If even one inequality fails, that
point is outside the feasible region.
7. Worked example
Solve and graph:
\[
y>x+2,
\qquad
y<-x+6.
\]
First draw the boundary lines:
\[
y=x+2,
\qquad
y=-x+6.
\]
Since both inequalities are strict, both boundary lines are dashed. The first inequality
asks for points above \(y=x+2\). The second asks for points below \(y=-x+6\).
The feasible region is the overlap: above the first line and below the second line.
The two boundaries intersect where
\[
x+2=-x+6.
\]
Then
\[
2x=4
\quad\Longrightarrow\quad
x=2.
\]
Substituting into \(y=x+2\), we get
\[
y=4.
\]
So the boundary lines meet at \((2,4)\), but because the inequalities are strict,
the point itself is not included in the feasible region.
8. Summary
