Reply to Student
shwbtjy10x9
respond_mat222_wk2.docxRESPONDING to (2) CLASSMATES
|
Your “and” compound inequality is : |
Your “or” compound inequality is : |
|
10 ≤ 2 + 4x < 20 |
4x + 3 < 16 or –2x < 3 |
Assigned Number 14
Compound Inequalities are:
10 2 4x 20 This is my Compound Inequality
10 2 2 4x 2 20 2 Add 2 to all three parts of the inequality
12 4x 22
12 4 4x 4 22 4 Divide all three parts by 4
3 x 5.5 Written algebraically
So a value of x that is less than or equal to 3 and less than or equal to 5.5 can make the inequality true. This 3 x 5.5 is how this compound inequality is written. As an intersection of sets it would come to look like [ 3, ) (-, 5.5] which equals [3, 5.5] in interval notation.
-----------|------[-------]----
0 3 5.5
Here is the number line graph of the solution set. The brackets as well as the end points are included with the solution set and the green lighting is highlighted within.
4x + 3 < 16 or –2x < 3
4x + 3 -3 < 16 – 3 Subtract 3 from both sides
4x 4 < 13 4 or -2x 2 < 3 2 Divide from both sides to get the variable
The -x (flipped symbol)
x < 3.25 or x > 3/2 Complete solution set written algebraically
The solution set written in interval notation is the union of two intervals (, 3.25] ( 3/2)
A number line graph of the solution set:
-------|-----]-----(---------
0 3.25 3/2
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Your “and” compound inequality is : |
Your “or” compound inequality is : |
|
0 ≤ –3 + 8x ≤ 22 |
2x – 9 ≤ 3 or 3x – 8 > 28 |
Assigned Number 27
My Week 2 Discussion compound inequalities are as follows;
1. 0 ≤ -3 + 8x ≤ 22 First step is to add 3 to all parts of this inequality
1. 0 +3 ≤ -3 + 8x +3 ≤ 22 +3
1. 3 ≤ 8x ≤ 25 Now divide by 8
1. 3 ÷ 8 ≤ 8x ÷ 8 ≤ 25 ÷ 8
1. 3/8 ≤ x ≤ 25/8 or .375 ≤ x ≤ 3.125 Written algebraically
This states that x is greater than or equal to .375 and less than or equal to 3.125.
-----------│---[------------------]----
0 .375 3.125
The intersection of the sets would be written as [.375, ∞)ᴖ(-∞, 3.125]
Now the “or” compound inequality;
1. 2x – 9 ≤ 3 or 3x -8 > 28
1. 2x – 9 +9 ≤ 3 + 9 or 3x -8 + 8 > 28 + 8 I must add in both compound inequalities
1. 2x ≤ 12 or 3x > 36 Now, divide in order to solve for the variable
1. 2x/2 ≤ 12/2 or 3x/3 > 36/3
1. x ≤ 6 or x > 12 Written algebraically
This compound inequality written in set notation as the union of the two intervals is
(-∞, 6]ᴗ(12, ∞).
----------│--------]------------(----------
0 6 12
RE
SPONDING
to (2) CLASSMATES
MAT222: Intermediate Algebra
Your “and”
compound
inequality is
:
Your “or” compound inequality is
:
10
=
2 + 4x < 20
4x + 3 < 16
or
–
2x < 3
Assigned Number 14
Compound Inequalities are:
10
=
2
+
4x
<
20
This is my
Compound Inequality
10
+
2
=
2
+
4x
+
2
<
20
+
2 Add 2 to all three parts of the
inequality
12
=
4x
<
22
12
÷
4
=
4x
÷
4
<
22
÷
4 Divide all three parts by 4
3
=
x
<
5.5 Written algebraically
So a value of x that is less than
or
equal to 3 and less than
or
equal to 5.5 can mak
e the
inequality true. This 3
=
x
<
5.5 is how this
compound inequality
is written. As an
intersection
of sets it would come to look like [ 3,
8
)
n
(
-
8
, 5.5] which equals [3, 5.5] in interval notation.
ß
-----------
|
------
[
-------
]
----
à
0 3 5.5
Here is the number line graph of the solution set. The brackets as well as the end points are
included with the solution set
and
the green lighting is highlighted within.
4x + 3 < 16 or
–
2x < 3
4x + 3
-
3 < 16
–
3
Subtract 3 from both sides
4x
÷
4 < 13
÷
4 or
-
2x
÷
2 < 3
÷
2
Divide from both sides to get the variable
The
-
x (flipped symbol)
x < 3.2
5 or
x > 3/2
Complete solution set written algebraically
The solution set written in interval notation is the
union
of two intervals (
-
8
, 3.25]
?
(
-
8
,
3/2)
RESPONDING to (2) CLASSMATES
MAT222: Intermediate Algebra
Your “and”
compound inequality is:
Your “or” compound inequality is:
10 = 2 + 4x < 20 4x + 3 < 16 or –2x < 3
Assigned Number 14
Compound Inequalities are:
10 = 2 + 4x < 20 This is my Compound Inequality
10 + 2 = 2 + 4x + 2 < 20 + 2 Add 2 to all three parts of the inequality
12 = 4x < 22
12 ÷ 4 = 4x ÷ 4 < 22 ÷ 4 Divide all three parts by 4
3 = x < 5.5 Written algebraically
So a value of x that is less than or equal to 3 and less than or equal to 5.5 can make the
inequality true. This 3 = x < 5.5 is how this compound inequality is written. As an intersection
of sets it would come to look like [ 3, 8) n (-8, 5.5] which equals [3, 5.5] in interval notation.
-----------|------[-------]----
0 3 5.5
Here is the number line graph of the solution set. The brackets as well as the end points are
included with the solution set and the green lighting is highlighted within.
4x + 3 < 16 or –2x < 3
4x + 3 -3 < 16 – 3 Subtract 3 from both sides
4x ÷ 4 < 13 ÷ 4 or -2x ÷ 2 < 3 ÷ 2 Divide from both sides to get the variable
The -x (flipped symbol)
x < 3.25 or x > 3/2 Complete solution set written algebraically
The solution set written in interval notation is the union of two intervals (-8, 3.25] ? (-8, 3/2)
