above the line y = 3xand below (or on) the
liney = –5. The system of linear inequalities
for which the shaded region is the solution set
is given byy 3 x,y –5.
- b.The slope-intercept form of the lines 9(y– 4)
= 4xand –9y= 2(x+ 9) are, respectively,y =
^49 x+ 4 andy = –^29 x– 2. The graphs of both
lines are dashed, so the inequality signs used in
both inequalities are eitheror. Next, note
that points in the shaded region lie below the
liney = ^49 x+ 4 and above the liney = –^29 x– 2.
This tells us that the system of linear inequali-
ties for which the shaded region is the solution
set is given byy^49 x+ 4,y–^29 x– 2. This sys-
tem is equivalent to 9(y– 4) 4 x,–9y2(x
+ 9), which can be seen by reversing the sim-
plification process used to obtain the slope-
intercept forms of the lines in the first step. In
doing so, remember that multiplying both
sides of an inequality results in a switching of
the inequality sign.
- c.The slope-intercept forms of the linesy – x
= 6 and 11y= –2(x+ 11) arey = x+ 6 andy =
- 121 x– 2, respectively. The graphs of both lines
are solid, so the inequality signs used in both
inequalities are either or. Next, points in
the shaded region lie above (or on) the liney =
x+ 6 and above (or on) the liney = – 121 x– 2.
The system of linear inequalities for which the
shaded region is the solution set is given by
y x+ 6,y – 121 x– 2. This system is equiv-
alent toy – x 6, 11y –2(x+ 11), which
can be seen by reversing the simplification
process used to obtain the slope-intercept
forms of the lines in the first step.
- c.The slope-intercept forms of the lines 5x–
2(y+ 10) = 0 and 2x+ y= 3 are, respectively,y
= ^52 x– 10 andy = –2x3. The graphs of both
lines are solid, which means that both inequal-
ity signs are either or. Points in the
shaded region lie above (or on) the line y=
^52 x– 10 and below (or on) the line y= –2x– 3,
so the system of linear inequalities for which
the shaded region is the solution set is given by
y ^52 x– 10,y –2x– 3 This system is equiv-
alent to 5x– 2(y+ 10) 0, 2x+y –3, which
can be seen by reversing the simplification
process used to obtain the slope-intercept
forms of the lines in the first step. In doing so,
remember that multiplying both sides of an
inequality results in a reversing of the inequal-
ity sign.
- b.The slope-intercept forms of the lines
7(y– 5) = –5xand –3 = ^14 (2x– 3y) are, respec-
tively,y= – x+ 5 and y= ^23 x+ 4. The graphs
of both lines are dashed, so the inequality
signs used in both of the inequalities compris-
ing the system are eitheror. Points in the
shaded region lie below the line y= –^57 x+ 5
and below the liney = ^23 x+ 4. This tells us that
the system of linear inequalities for which the
shaded region is the solution set is given byy
–^57 x+ 5,y^23 x+ 4. This system is equiva-
lent to 7(y– 5)–5x,–3^14 (2x– 3y), which
can be seen by reversing the simplification
process used to obtain the slope-intercept
forms of the lines in the first step. In doing so,
remember that multiplying both sides of an
inequality results in a reversing of the inequal-
ity sign.
- d.The solution set for the system in choice ais
the empty set. The solution set for the system in
choice bconsists of only the points that lie on
the line y= 3x+ 2, and the solution set of the
system in choicecconsists of only the points
that lie on the line y= x. So, the solution sets of
^5
7
ANSWERS & EXPLANATIONS–
