- c.The graph of the line is solid, so it is
included in the solution set and the inequality
describing the shaded region must include
equality (it must be either or ). Next,
since the graph of the line rises from left to
right at the rate of one vertical unit up per one
horizontal unit right, its slope is 1. Since it
crosses the y-axis at (0,2), the equation of the
line is y= x+ 2. Finally, since the shaded
region is above the liney= x+ 2, the inequal-
ity illustrated by this graph is y x+ 2, which
is equivalent tox– y –2. We can verify this
by choosing any point in the shaded region,
such as (0,3), and observing that substituting
it into the inequality results in the true state-
ment –3 –2.
- c The fact that the graph of the line is solid
means that it is included in the solution set, so
the inequality describing the shaded region
must include equality (it must be either or
≤). Next, since the graph of the line rises from
left to right at the rate of one vertical unit up
per one horizontal unit right, its slope is 1.
Since it crosses the y-axis at (0,0), the equation
of the line is y= x. The shaded region is above
the liney= x, so we conclude that the inequal-
ity illustrated by this graph is y x, which is
equivalent to y– x 0. Substituting a point
from the shaded region, such as (0,3) into the
inequality results in the true statement 3 0.
- a.The graph of the line is dashed, so it is not
included in the solution set, and the inequality
describing the shaded region must not include
equality (it must be eitheror). Next, since
the graph of the line falls from left to right at
the rate of one vertical unit down per six hori-
zontal units right, its slope is –^16 . It crosses the
y-axis at (0,– 21 ), so the equation of the line is y
= –^16 x– ^12 . Finally, since the shaded region is
above the liney= –^16 x– ^12 , the inequality illus-
trated by this graph isy–^16 x– ^12 . Multiplying
both sides of this inequality by 2 and moving
the x-term to the left results in the equivalent
inequality ^13 x+ 2y–1.We can verify this by
substituting any point from the shaded region,
such as (0,1), into the inequality, which results
in the true statement 2 > –1.
- a.Because the graph of the line is solid, we
know that it is included in the solution set, so
the inequality describing the shaded region
must include equality ( it must be either or
≤). Next, the graph of the line falls from left to
right at the rate of three vertical unit down per
one horizontal unit right, so its slope is –3.
And, since it crosses the y-axis at (0, 4), we
conclude that the equation of the line is y=
–3x+ 4. The shaded region is below the line y
= –3x+ 4, so the inequality illustrated by this
graph is y –3x+ 4 Multiplying both sides of
the inequality by 2 and moving the x-term to
the left results in the equivalent inequality 2y+
6 x 8. This can be further verified by choosing
a point from the shaded region, such as (0,0),
and observing that substituting it into the
inequality results in the true statement 0 0.
- c.The graph of the line is dashed, so it is not
included in the solution set and the inequality
describing the shaded region must not include
equality (it must be eitheror). Next,
since the graph of the line rises from left to
right at the rate of three vertical units up per
one horizontal unit right, its slope is 3. And,
since it crosses the y-axis at (0,–2), the equa-
tion of the line is y= 3x– 2. Finally, since the
shaded region is above the line y= 3x– 2, we
conclude that the inequality illustrated by this
graph isy 3 x– 2. Moving the y-term to the
right results in the equivalent inequality 3x– y
ANSWERS & EXPLANATIONS–
