Download 1. Solve the given inequality for y: y – 8 > 2y – 4 y – 8 +

1.
Solve the given inequality for y:
y – 8 > 2y – 4
y – 8 + 4 > 2y – 4 + 4
y – 4 > 2y
y – y – 4 > 2y – y
-4 > y
y < -4
Now compare each of the given values to -4. If they are less than -4 then they are
solutions to the original inequality:
5: is not less than -4  NO
-14: is less than -4  YES
-20: is less than -4  YES
-1: is not less than -4  NO
2.
The given interval includes all real numbers from 2 to infinity. Since 2 is
included, the interval uses a square bracket at the left end. However, since
infinity is not an actual number, but more like a concept, the right end of the
interval has a parenthesis. Infinity will never have a square bracket next to it
in an interval notation.
A. [2, ∞)
3.
Rewrite the given set as:
{x | -3 < x < -1}
Nothing is changed by doing this, it’s just the form that you are probably used to
seeing, with the lower value number on the left.
Clearly, when written this way, the interval includes values from -3 to -1, without
including the endpoints. Since the endpoints are not included, the interval will use
parentheses rather than square brackets.
B. (-3, -1)
4.
This graph shows that all values from -4 to 8 are included in the interval. Since
the graph shows parentheses at each end, the end points are not included in the
interval.
A. (-4, 8)
5.
Solve the inequality:
t + 19 ≥ 6
t + 19 – 19 ≥ 6 – 19
t ≥ -13
Since the endpoint is included (you can tell from the ≥ sign – if the endpoint was
not included then it would be > instead of ≥), the graph will show a square bracket
at -13. The interval includes values greater than -13, so the portion of the number
line to the right of -13 will be shaded to show that it is in the interval.
Solution: { t | t ≥ -13}, Graph C
6.
Solve this inequality by multiplying by the reciprocal of -9/4.
Remember that multiplying or dividing by a negative number requires that the
inequality sign be reversed.
9
1
− x≥−
4
8
# 4 &# 9 & # 1 &# 4 &
% − (% − ( x ≤ % − (% − (
$ 9 '$ 4 ' $ 8 '$ 9 '
x≤
4
72
x≤
1
18
Solution: { x | x ≤ 1/18 }
7.
Solve the inequality:
5x + 5 < 50
5x + 5 – 5 < 50 – 5
5x < 45
x < 45/5
x<9
Solution: { x | x < 9 }
8.
Solve the inequality:
0.4x + 5 ≤ 1.2x – 6
0.4x – 0.4x + 5 ≤ 1.2x – 0.4x – 6
5 ≤ 0.8x – 6
5 + 6 ≤ 0.8x – 6 + 6
0.8x ≥ 11
x ≥ 11/0.8
x ≥ 13.75
Solution: { x | x ≥ 13.75 }
9.
Solve the inequality:
2
(3x −1) > 8
5
" 5 %" 2 %
" 5%
$ '$ ' (3x −1) > 8 $ '
# 2 &# 5 &
#2&
3x −1 > 20
3x −1+1 > 20 +1
3x > 21
x>
21
3
x>7
Solution: A. The solution set is { x | x > 7 }
10.
Substitute the given values and compute I:
I=
700 (167)
( 71)
2
= 23.1898
Round this off to 23.
For the second part, set I = 25, and solve for W:
700W
( 71)
2
< 25
700W < 25 ( 71)
25 ( 71)
W<
700
2
W < 180.03
Round this off to 180.
Solution: 23, 180
2
11.
Under Plan A, she would pay: 180 + 0.40(M – 180), where M is the total amount
of her medical bills.
Under Plan B, she would pay: 190 + 0.35(M – 190).
If Plan B will save her money, then:
190 + 0.35(M – 190) < 180 + 0.40(M – 180)
Solve for M:
190 + 0.35M – 66.50 < 180 + 0.40M – 72
190 – 66.50 – 180 + 72 < 0.40M – 0.35M
0.05M > 15.50
M > 15.50 / 0.05
M > 310
Plan B will save her money if she has more than $310 in bills.
12.
The intersection of two sets is the collection of values that are present in both sets.
Since there are no values that are in both sets here, the intersection is the empty
set.
B. The intersection is the empty set.
13.
The solution set includes all values that are less than 6 and also greater than or
equal to -3.
In interval notation, this is written as [-3, 6).
The square bracket indicates that -3 is included in the interval.
The graph is C.
14.
Solve the two inequalities:
-28 ≤ 4x + 4
4x + 4 < 20
-28 – 4 ≤ 4x + 4 – 4
4x + 4 – 4 < 20 – 4
-32 ≤ 4x
4x < 16
-32 / 4 ≤ x
x < 16 / 4
-8 ≤ x
x<4
x ≥ -8
The solution set includes all values between -8 and 4. The value -8 is also
included.
The solution is { x | -8 ≤ x < 4 }.
The correct graph is B.
15.
Solve the compound inequality:
-8 < x + 4 < 9
This can be divided into two inequalities which can be solved separately, or it can
be done as a compound inequality.
Here is the compound inequality method:
-8 – 4 < x + 4 – 4 < 9 – 4
-12 < x < 5
Here is the divided inequality method:
-8 < x + 4
and
x+4<9
-8 – 4 < x + 4 – 4
x+4–4<9–4
-12 < x
x<5
Putting the two parts together gives:
-12 < x < 5
which is the same as the previous answer.
The solution is { x | -12 < x < 5 }
16.
Solve the compound inequality:
-15 ≤ 3x – 4 ≤ -3
-15 + 4 ≤ 3x – 4 + 4 ≤ -3 + 4
-11 ≤ 3x ≤ 1
-11/3 ≤ x ≤ 1/3
The solution is { x | -11/3 ≤ x ≤ 1/3 }
17.
This interval includes two parts. One part is all values less than -3. The other part
is all values greater than 2. Writing this in interval notation requires using the
union operator to join two intervals:
(-∞, -3) U (2, ∞)
The graph is B
18.
Solve the two inequalities:
x + 14 < -8
or
x + 14 > 3
x + 14 – 14 < -8 – 14
x + 14 – 14 > 3 – 14
x < -22
x > -11
The solution is { x | x < -22 or x > -11 }
The correct graph is D
19.
Solve the compound inequality:
4x − 9
≤ −8
10
4x − 9
≥6
10
4x − 9 ≤ −80
4x − 9 ≥ 60
4x ≤ −80 + 9
4x ≤ −71
x≤
−71
4
or
4x ≥ 60 + 9
4x ≥ 69
x≥
69
4
The solution is { x | x ≤ -71/4 or x ≥ 69/4 }
20.
Since there are 6 10-second periods in 1 minute, if you count your pulse for 10
seconds you have to multiply by 6 to get the rate per minute.
Applying this in reverse, if her pulse should be between 145 and 165, then she
should be counting 1/6 of those numbers in 10 seconds:
145
= 24.1667 ≈ 24
6
165
= 27.50 ≈ 28
6
The number of beats should be between 24 and 28.
21.
Substitute the given values for x and y into the inequality:
5(-3) + 4 < -3
-15 + 4 < -3
-11 < -3
This is a valid statement, so (-3, 4) is a solution.
Yes.
22.
Plot the points (0, 0) and (2, 8) and connect with a dashed line.
Shade the region above the line.
The graph looks like this:
23.
Plot the intercepts (5, 0) and (0, 4) and connect with a solid line.
Shade the region below the line.
The graph looks like this:
24.
Solve the inequality for y:
6x – 5 ≤ 11x + y
6x – 11x – 5 ≤ 11x – 11x + y
-5x – 5 ≤ y
y ≥ -5x – 5
Plot the intercepts at (0, -5) and (-1, 0) and connect with a solid line.
Shade the region above the line.
The graph looks like this:
25.
Graph 1: B. y + x ≤ 3
Graph 2: A. 3x ≥ 4y – 12
Graph 3: D. 5x – 3y < 15
26.
Graph 1: A. y + x ≤ 3
Graph 2: D. 3x ≥ 4y – 12
Graph 3: A. 2x – 3y < 6
27.
Plot the points (10, -5) and (-10, -5) and connect with a solid line.
Plot the poins (6, 10) and (6, -10) and connect with a solid line.
Shade the region to the upper right of the point where the two lines intersect.
The graph looks like this:
28.
Plot the points (9, 0) and (0, 3) and connect with a solid line.
Plot the points (3, 0) and (0, 9) and connect with a solid line.
Plot the points (0, 0) and (0, 3) and connect with a solid line.
Plot the points (0, 0) and (3, 0) and connect with a solid line.
Shade the region between all four lines.
The graph looks like this: