1. Unit 2 Honors Review

2. Solving Multi-Step Equations
Make sure to distribute everything first
Be careful: -(3+4) =
Combine like terms
Solve for your variable!
Common Mistake: 4x = 5 your solution should be x= 5 4

3. Clearing with Fractions:
Find your LCD: 𝑞 15 − 1 5 = LCD = 15
Multiply each term by your LCD:
15 1 ∙ 𝑞 15 − ∙ 1 5 = ∙ 3 5
3. Cross Simplify:
q-3=9
4. Solve:
q= 12

4. Clearing Decimals:
Determine what the most amount of decimal places you have to move in each term to get each term to be non decimal.
**Note: when we move a decimal place we’re moving by 10, 100, 1000, …
2. Move that many decimal places for ALL TERMS (don’t forget about variables or constants)
3. Solve
Example: 1.5 = 1.2y-5.7 (1 decimal place) 15 = 12y – 57 (15+57) = 12y 72 =12y y = 6 **Note if a question has both fractions and decimals convert all of them to one or the other, if one is a repeating decimal convert all to fractions!

5. Solving Equations with Variables on Both Sides
Steps:
Distribute everything on both sides of your equal sign
Combine like terms on each side of the equal sign
Move your SMALLEST variable to the opposite side using inverse operations
Move your constant to the opposite side
**Variables should be on one side, constants should be on the other**
5. SOLVE for your variable
Example:
3x = 2(x+2)
3x + 15 – 9 = 2x + 4
3x + 6 = 2x + 4
-2x -2x
x + 6 = 4
-6 -6
x = -2

6. Solving Equations with Variables on Both Sides Special Cases
Identity/ All Real Solutions
x x = 6-5x-2
-5x + 4 = -5x + 4
+ 5x x
4 = 4
**This is a true statement therefore, it is ALL REAL SOLUTIONS
No Solution
-8x x = x
x + 6 = x -17
-x -x
6 = -17
**This will never be a true statement, therefore, it is NO SOLUTION

7. Solving Absolute Value Equations
They should have 2 solutions! Think back to |x| = 4, how many different values can x be? (4 or -4) 2 solutions…
Steps:
Isolate your absolute value (get it all by itself by using inverse operations)
**Note: If you have 7|x+2| DO NOT DISTRIBUTE!!
2. Remove your absolute value signs and create 2 equations
a) Original equation
b) Negate the solution (make it negative)
3. Solve each equation
4. Check to make sure the solutions work in the original equation
Example: 1-|10-2k|=-25 -|10-2k| = -26 |10-2k| = 26 Case1: 10-2k = 26 k = -8 Case 2: 10-2k= -26 k = 18 Check: plug values into original equation make sure it works Your solution: {-8,18}

8. Solving Absolute Value Equations (Special Cases)
One Case = 1 solution |x+3|+4 = 4 |x+3| = 0 **you can’t have a negative 0 so there’s only one case** Case 1: x+ 3 = 0 x = -3 Solution: {-3}
No Solution: 2 - |2x-5|=7 -|2x-5| = 5 |2x-5| = -5 **STOP!! Can you ever have an absolute value = negative number?!! NO, therefore your solution is: NO SOLUTION

9. INEQUALITIES!!

10. Reminders:
**When you divide or multiply by a negative switch the inequality sign
**If you have your variable on the left, that is the direction that you can shade on your graph
** If you’re not shading above your graph, make sure its really really dark so I can see it and don’t mark it wrong
**Open circles = < or >
**Closed Circle = < or >
**To rewrite an inequality, you flip the sign and the order
Example: x < 4 is equivalent to 4 > x
And graphed as

11. Solving Inequalities with Variables on Both Sides
Just as with equations:
distribute everything
combine like terms
then move the “SMALLEST” variable to the other side of the inequality and the constants to the other side
Then graph (moving the variable to the left of the inequality makes this a lot easier!)
Example: 5(4+x) < 3(2+x)
20 + 5x < 6 + 3x
20 + 2x < 6
2x < -14
x < -7

12. Solving Inequalities with Variables on both sides Special Cases
All Real Numbers:
**True Statement, with no variables**
Example:
x + 5 > x + 3
-x x
5 > 3 (always true)
ALL REAL SOLUTIONS
**you can conclude this from the original inequality, how?
No Solution
**False statement with no variables**
Example:
2 (x+3) < 5 + 2x
2x + 6 < 2x + 5
-2x x
6 < 5
This will never be true therefore,
NO SOLUTION!

13. What is the difference between and and or?
AND means intersection
-what do the two items have in common?
OR means union
-if it is in one item, it is in the solution A B

14. Compound Inequalities
“AND”
Solutions that they BOTH have collaboratively
**Should be written as one inequality..
Smallest value < variable < large value
Example:
4 < x+2 <
2 < x < 6
**If this confuses you solving simultaneously split them into two separate inequalities, and solve** Remember your solution should only be the values that are “SHARED”
“OR” Solutions they the Each have + what they have collaboratively **CAN NOT BE WRITTEN AS ONE INEQUALITY** Example: 2x < 6 or 3x > 12 x < 3 OR x > 4

15. ● ● 7) 2x < -6 and 3x ≥ 12 o o o o -3 -6 -3 -6 4 7 1 4 7 1
Solve each inequality for x
Graph each inequality
Combine the graphs
Where do they intersect?
They do not! x cannot be greater than or equal to 4 and less than -3 No Solution!! -3 -6 o -3 -6 o 4 7 1 o ● 4 7 1 o ●

16. The whole line is shaded!!
Graph x ≥ -1 or x ≤ 3
The whole line is shaded!!

17. Graph:
x > 4 or x > -2

18. Solving Absolute Value Inequalities
Isolate the absolute value
Create two inequalities
Remove the absolute value signs, keep absolute value expression, keep the inequality sign, and keep the value
Remove the absolute value sign, keep absolute value expression, change inequality sign and change value sign
Example:
|x-8| + 5 > 11
|x-8| > 6
Case 1: x-8 > 6 Case 2: x-8 < -6
x > x < 2
This was a greater than inequality therefore it will have an OR graph!
Follow the same steps for an AND graph
RECALL: GreatOR and LessthAND

19. Solving Absolute Value Inequalities (Special Cases)
All Real Numbers **True Statement** Example: |x-6| +7 > 2 |x-6| > -5 True, absolute values will always be a positive number, therefore any positive number will always be greater than -5, and therefore it is true for any value of x
No Solution **False Statement** Example: |x+12| -5 < -6 |x+12| < -1 False, this is saying that an absolute value is going to be less than a negative number, will this ever be true??? No because absolute values are always positive