or 1 – 4x –7. We solve these two inequalities
separately:1 – 4x 7 1 – 4x –7
–4x 6–4x –8
x – ^64 = –^32 x 2So, the solution set is (– ,–^32 ][2, ).- d.First, we rewrite the given inequality in an
equivalent form:
|1 – (–2^2 + x) – 2x| | 3x– 5
1 – (–4 + x) – 2x| | 3x– 5
|1 + 4 – x– 2x| | 3x– 5
|5 – 3x| |3x– 5|Now, note that the left side of the last inequal-
ity is equivalent to|5 – 3x| = |–1(3x– 5)| = |–1||3x– 5)| = |3x– 5|Thus, the original inequality is actually equiva-
lent to |3x– 5| |3x–5|. Since the left and
right sides of the inequality are identical, every
real number xsatisfies the inequality. So, the
solution set is the set of all real numbers.Set 15 (Page 26)
- c.The coordinates of points in the third
quadrant are both negative. - e.The x-coordinate ofJis –3 and the
y-coordinate is 4. So,Jis identified as the
point (–3, 4). - b.Since ABCDis a square, the x-coordinate
ofBwill be the same as the x-coordinate ofA,
namely –1, and the y-coordinate ofBwill be
the same as the y-coordinate ofC, namely 4.
So, the coordinates ofBare (–1,4). - e.Since ABCDis a square, the x-coordinate of
D is the same as the x-coordinate ofC, which
is 6, and the y-coordinate ofDis the same as
the y-coordinate ofA, which is –3. So, the
coordinates ofDare (6,–3).- d.Points in Quadrant IV have positive x-
coordinates and negative y-coordinates.
Therefore, (2,–5) lies in Quadrant IV. - a.For all nonzero real numbers, both x^2 and
(–y)^2 are positive, so points of the form (x^2 ,(–y)^2 )
must lie in Quadrant I. - d.First, note that |–x–2| = 0 only when x= –2
and |–x– 1| = 0 when x= –1. For all other val-
ues ofx, these expressions are positive. For all
real numbers x–2, we conclude that |–x– 2|
0 and –|–x–1|0. Therefore, points whose
coordinates are given by (|–x–2|, –|–x–1|)
must lie in Quadrant IV. - b.The fact that xis a positive real number
requires that the point (x,y) lie to the right of
the y-axis, so it cannot lie in Quadrants II or
III, or be on the y-axis. It can, however, lie in
Quadrants I or IV, or be on the x-axis. The fact
that ycan be any real number does not further
restrict the location of (x,y). Hence, the correct
choice is b. - c.Because yis a non-negative real number, the
point (x,y) must lie on or above the x-axis, so
it cannot lie in Quadrants III or IV. The point
can also be on the y-axis if it is the origin. The
fact that xcan be any real number does not
further restrict the location of (x,y), so the
correct choice is c. - a.We need to choose the selection that has a
positive x-coordinate and negative y-coordinate.
Since a0, it follows that –a0. Thus, the
choice that lies in Quadrant IV is (–a,a). - b.The correct selection will have a negative x-
coordinate and negative y-coordinate. Note
that for any nonzero real number, that –a^2 0
and (–a)^2 0. The choice that lies in Quadrant
III is (a,–a^2 ).
ANSWERS & EXPLANATIONS–