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Unit 2 Honors Review
2015-2016
Objective: to review for unit 2 Quest
tomorrow
Warm-up:
Complete handout in the front of the classroom
**Note: if you have not completed your compound inequalities
project it needs to be completed by Friday (2 point classwork
grade)
Remember: QUEST TOMORROW!
Solving Multi-Step Equations
• Make sure to distribute everything first
Be careful: -(3+4) = -3 - 4
• Combine like terms
• Solve for your variable!
Common Mistake: 4x = 5 your solution should be
5
x=
4
Clearing with Fractions:
1. Find your LCD:
𝑞
15
1
5
− =
3
5
LCD = 15
2. Multiply each term by your LCD:
15 𝑞
15 1
15 3
∙
−
∙
=
∙
1 15
1 5
1 5
3. Cross Simplify:
q-3=9
4. Solve:
q= 12
Clearing Decimals:
1. 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!
Solving Equations with Variables on Both Sides
Steps:
1. Distribute everything on both
sides of your equal sign
• Example:
2. Combine like terms on each
side of the equal sign
3x + 15-9 = 2(x+2)
3. Move your SMALLEST
variable to the opposite side
using inverse operations
3x + 6 = 2x + 4
4. Move your constant to the
opposite side
**Variables should be on one
side, constants should be on the
other**
5. SOLVE for your variable
3x + 15 – 9 = 2x + 4
-2x
-2x
x+6=4
-6 -6
x = -2
Solving Equations with Variables on Both
Sides Special Cases
• Identity/ All Real Solutions
x + 4 -6x = 6-5x-2
-5x + 4 = -5x + 4
+ 5x
+5x
4=4
**This is a true statement
therefore, it is ALL REAL
SOLUTIONS
• No Solution
-8x + 6 + 9x = -17 + x
x + 6 = x -17
-x
-x
6 = -17
**This will never be a true
statement, therefore, it is NO
SOLUTION
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:
1. Isolate your variable (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}
Solving Absolute Value Equations (Special Cases)
One Case = 1 solution
No Solution:
|x+3|+4 = 4
2 - |2x-5|=7
|x+3| = 0
-|2x-5| = 5
**you can’t have a negative 0 so
there’s only one case**
Case 1:
x+ 3 = 0
x = -3
Solution: {-3}
|2x-5| = -5
**STOP!! Can you ever have an
absolute value = negative
number?!!
NO, therefore your solution is:
NO SOLUTION
INEQUALITIES!!
• 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
Solving Inequalities with Variables on Both Sides
• Just as with equations:
1. distribute everything
2. combine like terms
3. then move the “SMALLEST” variable to the other side of the inequality and the
constants to the other side
4. 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
Solving Inequalities with Variables on both
sides Special Cases
• All Real Numbers:
• No Solution
**True Statement, with no
variables**
• **False statement with no
variables**
Example:
Example:
x+5> x+3
-x
-x
5 > 3 (always true)
ALL REAL SOLUTIONS
**you can conclude this from
the original inequality, how?
2 (x+3) < 5 + 2x
2x + 6 < 2x + 5
-2x
-2x
6< 5
This will never be true therefore,
NO SOLUTION!
What is the difference between and
and or?
AND means
intersection
A
B
A
B
-what do the two items
have in common?
OR means union
-if it is in one item, it is in
the solution
Compound Inequalities
“AND”
“OR”
Solutions that they BOTH have
collaboratively
Solutions they the Each have + what
they have collaboratively
**Should be written as one
inequality..
**CAN NOT BE WRITTEN AS ONE
INEQUALITY**
Smallest value < variable < large value
Example:
Example:
4 < x+2 < 8
-2
-2 -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”
2x < 6 or
x < 3 OR
3x > 12
x> 4
7) 2x < -6 and 3x ≥ 12
1. Solve each inequality for x
2. Graph each inequality
3. Combine the graphs
4. Where do they
intersect?
5. They do not! x cannot be
greater than or equal to 4
and less than -3
Solution!!
No
2 x 6

2
2
x  3
o
-3
3x 12

3
3
x4
-6
1
o
●
4
0
7
Graph x ≥ -1 or x ≤ 3
-5
0
5
The whole line is shaded!!
Graph:
x > 4 or x > -2
Solving Absolute Value Inequalities
1. Isolate the absolute value
2. Create two inequalities
1.
2.
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
x > 14
Case 2:
x-8 < -6
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
Solving Absolute Value Inequalities (Special Cases)
All Real Numbers
No Solution
**True Statement**
**False Statement**
Example:
Example:
|x-6| +7 > 2
|x+12| -5 < -6
|x-6| > -5
|x+12| < -1
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
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