none of these systems span the entire Cartesian
plane. In fact, it is impossible for such a system
of linear inequalities to have a solution set that
spans the entire Cartesian plane.
- b.Note that the graphs of the lines y= x+ 3
and y= x– 1 are parallel, where the graph of
y= x+ 3 lies strictly above the graph ofy= x– 1.
Using the first inequality specified in the system,
any point that it is in the solution set of the
system to yx + 3,yx– 1 would necessarily
be above the line y= x+ 3, and therefore, by
the previous observation, also above the line
y= x– 1. However, the second inequality in
the system requires that the point be belowthe
line y= x– 1, which is not possible. Hence, the
solution set of this system is the empty set.
- d.The boundaries of Quadrant III are the
x-axis and y-axis; the equations of these axes
are y= 0 and x= 0, respectively. Since points
in the solution set are not to be on either axis,
both inequalities comprising the system we
seek must involve one of the signsor.
Next, note that the sign of both the x- and
y-coordinate of a point in Quadrant III is neg-
ative. We conclude that the system with this
solution set is given by x0,y0.
- b.A system of linear inequalities whose solu-
tion set consists of the points on a single line
must be of the form y mx + b,y mx+ b,
assuming that the lines are not vertical. Observe
that the first inequality in the system 2y– 6x
4, or y 2 + 3x, is equivalent to y 2 + 3x,so
this system is of the form specified. The solu-
tion set consists of those points on the line
y= 3x+ 2.
Section 3—Polynomial
Expressions
Set 23(Page 66)
- d.
(x^2 – 3x+ 2) + (x^3 – 2x^2 + 11) =
x^3 + x^2 – 2x^2 – 3x+ 2 + 11 =
x^3 – x^2 + 3x+ 13
- a.
(3x^2 – 5x+ 4) – (–^23 x+ 5)
= 3x^2 – 5x+ 4 + ^23 x– 5
= 3x^2 – 5x + ^23 x+ 4 – 5
= 3x^2 – ^135 x+ ^23 x– 1
= 3x^2 – ^133 x– 1
- b.
(^13 x^2 – ^15 x– ^23 ) – (^23 x^2 – 170 x+ ^12 )
= ^13 x^2 – ^15 x– ^23 – ^23 x^2 + 170 x– ^12
=^13 x^2 – ^23 x^2 – ^15 x+ 170 x– ^23 – ^12
= –^13 x^2 – 120 x+ 170 x–^46 –^36
= –^13 x^2 + 150 x– ^76
= –^13 x^2 + ^12 x– ^76
- c.
(9a^2 b+ 2ab– 5a^2 ) – (–2ab– 3a^2 + 4a^2 b)
= 9a^2 b+ 2ab– 5a^2 + 2ab+ 3a^2 – 4a^2 b
= 9a^2 b– 4a^2 b+ 2ab+ 2ab– 5a^2 + 3a^2
= 5a^2 b+ 4ab– 2a^2
- a.
(^16 x^2 + ^23 x+ 1) + (2x– ^23 x^2 + 4) – ^72 + 3x+ ^12 x^2 )
= ^16 x^2 + ^23 x+ 1 + 2x– ^23 x^2 + 4 – ^72 – 3x– ^12 x^2
= ^16 x^2 – ^23 x^2 – ^12 x^2 + ^23 x+ 2x– 3x+ 1 + 4 – ^72
= ^16 x^2 – ^46 x^2 – ^36 x^2 + ^23 x– x+ 5 – ^72
= –x^2 – ^13 x+ ^32
ANSWERS & EXPLANATIONS–
