1. Absolute Value Equations & Inequalities
INSTRUCTIONS
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2. What is absolute value?
You ALWAYS have to consider the POSITIVE and the NEGATIVE solutions when working with absolute values.
Example 1:
|7| = +7 AND -7
Example 2:
|x+7| = 2
X+7 = +/- 2 means we have to solve 2 equations when you remove the absolute value sign.
X+7 = AND x+7 = -2
x = AND x = -9
Solution: { -5 , -9 }
-5 AND -9 both work in the equation.
Let's first return to the original definition of absolute value: “
|x | is the distance of x is from zero on a number line.“ The |2| , on a graph, would be
We have to include both –2 and 2 because they are two units from zero ( | 2 | = 2 = -2 )

3. Solving absolute value equations
To find the x values that will work for
|x+5| -2 = 13 , we must:
First, isolate the absolute value expression.
|x+5| -2 = add 2 to both sides
|x+5| = 15
When we remove the absolute value signs, we have to consider the positive and negative answer, we get this.( it’s like the |2| = 2 AND -2 example earlier. )
x+5 = +/-15
Set up two equations to solve ( the positive and negative )
x+5=15 AND x+5= -15 , now solve for x in each to get
x = { 10 , -20} , both of these answers work.

4. Solving absolute value equations
3|x+5| -2 = 13 Isolate absolute value
3|x+5| = added 2 to both sides
|x+5| = 5 Divide both sides by 3
x+5 = +/-5 Take off absolute value sign
Consider positive and negative answers.
x+5 = 5 AND x+5 = -5 Solve for x’s
x = { 0 , -10 }
Try the next two before you get the solution.
Try this one before you click.
Isolate absolute value
Set up +/- answer when you take off absolute value sign
Solve TWO equations to find answers.

5. 6|5x + 2| = 312 6|5x + 2| = 312 |5x + 2| = 52 5x + 2 = 52 5x + 2 = -52
Isolate the absolute value expression by dividing by 6.
6|5x + 2| = 312
|5x + 2| = 52
Set up two equations to solve.
Try this one before you click.
Isolate absolute value
Set up +/- answer when you take off absolute value sign
Solve TWO equations to find answers.
5x + 2 = 52 5x + 2 = -52
5x = x = -54
x = 10 or x = -10.8
Check: 6|5x + 2| = |5x + 2| = 312
6|5(10)+2| = |5(-10.8)+ 2| = 312
6|52| = |-52| = 312
312 = = 312

6. 3|x + 2| -7 = 14 3|x + 2| -7 = 14 3|x + 2| = 21 |x + 2| = 7
Isolate the absolute value expression by adding 7 and dividing by 3.
3|x + 2| -7 = 14
3|x + 2| = 21
|x + 2| = 7
Set up two equations to solve.
Try this one before you click.
Isolate absolute value
Set up +/- answer when you take off absolute value sign
Solve TWO equations to find answers.
x + 2 = x + 2 = -7
x = or x = -9
Check: 3|x + 2| - 7 = |x + 2| -7 = 14
3|5 + 2| - 7 = |-9+ 2| -7 = |7| - 7 = |-7| -7 = 14
= = 14
14 = = 14

7. Graphing Absolute Value.
Suppose you're asked to graph the solution to | x | < 3.
The solution is going to be all the points that are less than three units away from zero. Look at the number line:
Problems involving < (less than) sign are AND inequalities( conjunctions), and involve solutions where lines overlap or intersect each other.

8. Review of the Steps to Solve a Compound Inequality:
2x+3>2 and 5x<10 looks like this when solved.
x>-1/2 and x<2
Example:
This is a conjunction because the two inequality statements are joined by the word “and”.
You must solve each part of the inequality.
The graph of the solution of the conjunction is the intersection of the two inequalities. Both conditions of the inequalities must be met.
In other words, the solution is wherever the two inequalities overlap.
If the solution does not overlap, there is no solution.

9. “and’’ Statements can be Written in Two Different Ways
< m+6 and m+6 < 14
These inequalities can be solved using two methods.

10. Method One Example : 8 < m + 6 < 14
Rewrite the compound inequality using the word “and”, then solve each inequality.
8 < m and m + 6 < 14
2 < m m < 8
m > and m < 8
2 < m < 8
Graph the solution: 8 2

11. Method Two Example: 8 < m + 6 < 14
To solve the inequality, isolate the variable by subtracting 6 from all 3 parts.
8 < m + 6 < 14
2 < m < 8
Graph the solution. 8 2

12. Solving a < ( Less Than) Absolute Value Inequality
For example
| x + 7 | < 2 would become this when you take off absolute value sign.
X+7 <2 AND x+7 > -2
- Because it’s absolute value, we need POSITIVE and NEGATIVE solution
In the second equation you negate and reverse the right hand side.
Then solve and graph as you’ve learned earlier in year.
Remember, NO OVERLAP means no solution.
Step 1: If you have a ( Less thand or less thand or equal to) you are working with a conjunction or an ‘and’ statement.
Remember: “Less thand”
Rewrite the inequality as a conjunction .
Step 2: Because we need the POSITIVE and NEGATIVE solution, in the second equation you must negate the right hand side and reverse the direction of the inequality sign.
Solve as a compound inequality.
Let’s try one.

13. Solving an Absolute Value Inequality
Try this one before you click.
Isolate absolute value
Set up +/- inequalities when you take off absolute value sign
Remember to negate and reverse 2nd inequality
Solve TWO inequalities to find answers.
Graph if needed.
Solve 3|x-3| < 18 Step 1: Isolate the Absolute value sign by dividing by 3 to both sides..
|x-3| < 6
Step 2: Rewrite the inequality without absolute value sign as a conjunction with your positive and negative inequality. Remember: “Less thand” Also, remember to negate and reverse 2nd inequality.
x-3 < AND x-3 > -6
Step 3: Solve to find positive and negative solution
x < AND x > -3 also written -3 < x < 9
Step 4: Graph if asked
Now you try one.

14. Solving an Absolute Value Inequality
Remember, if you see a < (less than) inequality like this
3|y-3|-2 < -6
Less than a negative number, there is NO SOLUTION!!
Absolute Values are ALWAYS positive, so it can’t be less than 0.
Solving an Absolute Value Inequality
Try this one before you click.
Isolate absolute value
Set up +/- inequalities when you take off absolute value sign
Remember to negate and reverse 2nd inequality
Solve TWO inequalities to find answers.
Graph if needed.
Solve 2|x+1|+4 < 12
|x+1| < 4 Step 1: Isolate the Absolute value sign. Subtract 4, divide by 2
Step 2: Rewrite the inequality without absolute value sign as a conjunction with your positive and negative inequality. Remember: “Less thand” Also, remember to negate and reverse 2nd inequality.
x+1 < AND x+1 > -4
Step 3: Solve to find positive and negative solution
x < AND x > -5 also written -5 < x < 3
Step 4: Graph if asked
Now ONTO > ( Great”OR” than) inequalities.

15. ‘or’ Statements Example: x - 1 > 2 or x + 3 < -1
x > 3 x < -4
x < -4 or x >3
Graph the solution. 3 -4

16. Review of the Steps to Solve a Compound Inequality:
Example:
This is a disjunction because the two inequality statements are joined by the word “or”.
You must solve each part of the inequality.
The graph of the solution of the disjunction is the union of the two inequalities. Only one condition of the inequality must be met.
In other words, the solution will include each of the graphed lines. The graphs can go in opposite directions or towards each other, thus overlapping.
If the inequalities do overlap, the solution is all real numbers.

17. If you have a > ( greatOR than) you are working with a disjunction or an ‘or’ statement. Remember: “Greator”
Example 1:
This is an ‘or’ statement. (Greator). Rewrite.
In the 2nd inequality, reverse the inequality sign and negate the right side value.
Solve each inequality.
Graph the solution.
|2x + 1| > 7
2x + 1 > 7 or 2x + 1 >7
2x + 1 >7 or 2x + 1 <-7
x > 3 or x < -4 3 -4

18. You Try this one: 5|3x -2| > 20 |3x -2| > 4
Try this one before you click.
Isolate absolute value
Set up +/- inequalities when you take off absolute value sign
Remember to negate and reverse 2nd inequality
Solve TWO inequalities to find answers.
Graph if needed.
This is an ‘or’ statement. (Greator). Rewrite.
In the 2nd inequality, reverse the inequality sign and negate the right side value.
Solve each inequality.
Graph the solution.
5|3x -2| > 20
|3x -2| > 4
3x -2 > 4 or 3x -2 < -4
x > 3 or x < -2/3 3
-2/3

19. Try to solve and graph these
before you
look at
solutions.
1) |y – 3| > 1
2) |p + 2| < 6
3) |4m – 5| > 7
4) 3|x – 2| < 9

20. Solve and Graph 1) |y – 3| > 1 2) |p + 2| < 6 3) |4m – 5| > 7
4) 3|x – 2| < 9
Y-3>1 OR y-3 <-1
y>4 OR y<2
P+2<6 AND p+2>-6
P<4 AND p>-8
4m-5>7 OR 4m-5 < -7
m>3 OR m < -1/2
3|x-2|<9 becomes |x-2|<3
Then
X-2<3 AND x-2> -3
x<5 AND x> -1

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