Download Test 3 Quiz - Detailed Solutions

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Test 3 Quiz Explanation
1. −2𝑥 + 7 ≥ 9
First, we need to isolate the x. So we will subtract 7 from both sides and then divide by -2. Don’t forget when
you divide or multiply by a negative number, you change the direction of the inequality.
Subtract 7 from both sides:
-2x ≥ 2
Divide Both Sides by -2:
x ≤ -1
Graph:
2.
1−2𝑋
3
Interval:
+
3𝑋+7
7
(-∞, -1]
>1
Let’s recall that the Division Bar Groups the Numerator and the Denominator so it really looks like:
(1−2𝑋)
3
+
(3𝑋+7)
7
>1
The LCM of 3 and 7 is 21 (Note: Both 3 and 7 divide into 21 with no remainder), So we want to multiply every term by 21
21 ∗ (1 − 2𝑋) 21 ∗ (3𝑋 + 7)
+
> 21 ∗ 1
3
7
We can then simplify by dividing. Note all the denominators should go away
7(1 − 2𝑥) + 3(3𝑥 + 7) > 21
Next we distribute the 7 and 3 to all terms inside the parenthesis.
7 – 14x + 9x + 21 > 21
Simplify on each side of the inequality by combining like terms
28 – 5x > 21
Subtract 28 from both sides
-5x > -7
Divide Both Sides by -5 Note: Don’t forget to change the direction of the Inequality
x<
7
5
𝟕
Interval: (-∞, )
Graph:
𝟓
7
5
3. |5𝑋 + 3| = 7
First we always have to Isolate the Absolute Value Bars. We can see here that it isolated.
Next we have to make our 2 equations.
Equation 1: Drop Absolute Value Bars and Set Equal to Right Hand Side
5x + 3 = 7
Equation 2: Drop Absolute Value Bars and Set Equal to Opposite of Right Hand Side
5x + 3 = -7
Solve Each Equation for x:
Equation 1:
5x + 3 = 7
Subtract 3 from Both Sides:
5x = 4
Divide Both Sides by 5
x=
4
5
Equation 2:
5x + 3 = -7
Subtract 3 from Both Sides:
5x = -10
Divide Both Sides by 5
x = -2
So solution is
x = {−𝟐,
𝟒
𝟓
}
4. |9Y + 1| = |6Y + 4|
When we have Absolute Value = Absolute Value, we still need 2 equations:
Equation 1: Drop Absolute Value Bars on both sides and set equal to each other.
9y + 1 = 6y + 4
Equations 2: Drop Absolute Value Bars on both sides and set Left-Hand Side = to Opposite Sign as the Right Hand Side.
Note: The Absolute Value Bars also act Like Grouping Symbols, so you can treat them as if there are in Parenthesis: If you
don’t it is likely you will make a sign mistake on the second term on the Right Hand Side
(9y + 1) = -(6y + 4)
9y + 1 = -6y – 4
Solve Each Equation for y to get your two solutions
Equation 1:
9y + 1 = 6y + 4
Subtract 6y from Both Sides:
3y + 1 = 4
Subtract 1 from Both Sides
3y = 3
Divide Both Sides by 3
y=1
Equation 2:
9y + 1 = -6y - 4
Add 6y to Both Sides:
15y + 1 = -4
Subtract 1 from Both Sides
15y = -5
Divide Both Sides by 15
y=
−5
15
Simplify
y=
−1
3
So Solutions are:
𝟏
y = {− 𝟑, 1}
Note: Problems 5 through 10 are Special Cases
RECALL: THE DEFINITION OF ABSOLUTE VALUE IS DISTANCE FROM ZERO WHICH IS ALWAYS GREATER OR EQUAL TO 0
1
5. - |X + 1| > 1
2
As always, first we have to Isolate the Absolute Value Bars FIRST!
First Multiply Both Sides by -2 (Note could Divide by -1/2)
Also, since we Multiplied or Divided by a Negative Number We Have to Switch the Direction of the Inequality
|x + 1| < -2
Since by definition of Absolute Value |x
+ 1|≥ 0 , It can never be negative.
Therefore it can never be less than a
negative number. In this case -2. So there is no work to be done here. The answer is:
NO SOLUTION or Ø
6. |2X + ¼| > -2
Since by definition of Absolute Value |2X
+ ¼| ≥ 0, It is always > then a Negative Number.
Therefore here the solutions is:
ALL REAL NUMBERS or (-∞, ∞)
7. |X -5| +10 < 5
First we have to Isolate the Absolute Value Bars!
Subtract 10 from both sides:
|X -5| < -5
Since by definition of Absolute Value |X
- 5| ≥ 0, It is NEVER < then a Negative Number
So there is no work to be done here. The answer is:
NO SOLUTION or Ø
8. |X-1| ≤ 0
By Definition |X - 1| ≥ 0, so it can never be < 0, but it can equal 0 so that’s what we have to solve for here.
X–1=0
Add 1 to both sides
X = 1, So there is only one Solution
{1}
1
9. |X-1| + 2 > 2
First Isolate the Absolute Value Bars
Subtract 2 from both sides
|X-1| > 0
By Definition, |X-1| ≥ 0, so the only solution not included in this problem is when x-1 = 0. So we have to remove that from
the solution set. We can do this by solving x - 1 = 0
x-1=0
Add 1 to both sides
x= 1
So this is all Real Numbers Except 1
(-∞, 1) U (1, ∞)
)(
1
10. -5*|X-1| ≤ 0
First we need to isolate the Absolute Value Bars.
Divide Both Sides by -5 (Don’t Forget to Change the Inequality Direction)
|X-1| ≥ 0
By Definition, |X-1| ≥ 0, so the Solution is:
ALL REAL NUMBERS or (-∞, ∞)
11. |X-3| < 2
Here we do not have a Special Case. We see that it is “<” so we know it is an “AND” Statement
Equation 1: Drop the Absolute Value Bars and Keep the Inequality and Right Hand Side
X–3<2
Equation 2: Drop the Absolute Value Bars, Switch the Inequality and take the Opposite Sign of the Right Hand Side
X – 3 > -2
Solve Both Equations for X
Equation 1: X - 3 < 2
Add 3 to Both Sides
X<5
Equation 2: X – 3 > -2
Add 3 to Both Sides
X>1
So we have our Compound Inequality
X < 5 AND X > 1
Using the Three Line Method:
X<5
X>1
AND
)
(
(
1
)
5
(1, 5)
12.|Y| ≥ 4
Again, Absolute Value Bars are Already Isolated
Since “≥” We will use the “OR” Operator
Equation 1: Drop Absolute Value Bars and Keep Inequality and Right Hand Side
Y≥4
Equation 2: Drop Absolute Value Bars and Switch Direction of Inequality and Use Opposite Sign as Right Hand Side
Y ≤ -4
So we have our Compound Inequality
Y ≥ 4 OR Y ≤ -4
Using the Three Line Method:
Y≥4
Y ≤ -4
OR
[
]
]
[
-4
4
(-∞, -4) U (4,∞)