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Systems
of Linear Inequalities
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Solving and graphing systems of
linear inequalities
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Solving and graphing systems of
linear inequalities in two variables
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Rational
Inequalities |
Method of solving
rational inequalities
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The
graph of the translated equilateral (or rectangular) hyperbola
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Systems
of Linear Inequalities
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The solution of the system of
simultaneous inequalities is the intersection of sets
of the individual solutions.
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Example:
Solve and graph the
solution of the given system of
simultaneous inequalities.
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the solution of the system
is the open interval (3, +
oo
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Solving and graphing systems of
linear inequalities in two variables
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The set of points whose coordinates (x,
y)
satisfy the inequality ax+
by + c > 0 is a
half-plane of a Cartesian plane.
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Example:
Solve and graph the
solution of the given system of linear inequalities in two
variables.
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Solution:
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Plug the
coordinates of the origin |
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Thus, non shaded area of the Cartesian
plane where lies the origin, bounded by given lines, is the solution of the system of inequalities.
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Rational
Inequalities
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A rational inequality can be written
in one of the following standard forms:
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P/Q > 0 or P/Q <
0 (or P/Q > 0
or
P/Q < 0 ), where
Q is
not 0.
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The sign of the rational expression P/Q, where P and Q are polynomials, depends on the signs of P and Q.
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As the signs of the
polynomials change at the zeros, to solve a rational inequality we should find the zeros of
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both P and Q first and then we can determine the intervals of the independent variable that satisfy given
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rational inequality.
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Method of solving
rational inequalities
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The first step to solve a rational
inequality is to get a single rational inequality on the left side
of the inequality sign and have zero on the right side of the
inequality sign.
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The next step is to factor the numerator and denominator and find the values of
x
that make these factors equal to 0 to find critical points (boundaries, endpoints of intervals).
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Note that, by setting the numerator to
0 we get the zero points of the given rational expression, but by setting
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the denominator
to 0 we get points at which the rational expression or function is undefined
(i.e., when
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plugged into the expression give division by zero).
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The rational functions are not defined at the zeros of the denominator. Therefore, they have breaks or vertical
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asymptotes at these points,
and this is why these points cannot be included into the solution of the rational
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inequality.
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Example:
Solve the rational
inequality
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and
draw the graph of the rational function. |
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Solution:
The solution set of the given rational inequality includes all numbers
x
which make the inequality greater then or equal to 0, or which make the sign of the rational expression
to be positive or 0.
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A rational expression is positive if both the numerator and the denominator are positive
or if both are negative, and the rational expression equals 0 when
its numerator is equal to 0 that is
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therefore, we have to solve two simultaneous
inequalities:
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The solutions represented on the
number line are shown below.
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Thus, the solution set of the given
inequality written in the interval notation is (-
oo,
-1)
U
[2,
oo
).
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Example:
Find the solutions
of the inequality
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Solution:
The solution set of the given rational inequality includes all numbers
x
which make the inequality less then or equal to 0, or which make the sign of the rational expression
to be negative or 0.
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A rational expression is negative if the numerator and the denominator
have different signs, and the rational expression equals 0 when
its numerator is equal to 0 that is,
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therefore, we have to solve two simultaneous
inequalities.
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We graph the numerator and the
denominator in the same coordinate system to find all points of
the x-axis
that satisfy given inequality.
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The zero points of the numerator and
the denominator divide the x-axis
into four intervals at which given rational expression changes
sign.
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x2
+ 2x -
3 = 0, a = 1, b = 2 and
c = -3
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The solutions of the two pairs of the simultaneous
inequalities are intersections of sets of their partial solutions,
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as is shown below.
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Therefore, the solution set is (-
oo,
- 3]
U
(-
2,
1].
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Intermediate
algebra contents |
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