Do all sections in NB: All work must be shown for sections I-IV!!!
Section I: Solving Equations
1) 20 - (5/8)x = 40
2) 6(7x - 2) = 8(4x + 1)
3) 2(5x - 4) - 3(4x + 3) = -43
4) x2 + 44 = 15x
5) 3x2 + 18x = 81
6) 3x2 = 5x + 5
Section II: Factoring
7) x2 – 676
8) 4x2 – 12x – 112
10) 18x2 – 9x – 5
11) 18x2 + 15x – 12
9) 6x3 + 72x2 + 216x
Section III: Absolute Value
12) 3|9x – 6| - 11 = 88
13) ¾ |4x + 8| + 13 = 37
14) 3|7x + 14| - 4 _ = 16
5
15) 4|12x – 18|
16) 5|6x - 3| + 7 < 82
17) -½ | 5x + 10 | - 41 > -56
18)
19) 2 |7x + 14 | - 14 > -7
10
2 | 3x - 12 | + 4 > 2
11
- 11 = 29
3
Section IV: Systems
(must be checked in both original equations)
20) x = 4- 3y
21) 5x + 3y = -7
7x + 10y = -5
7x + 2y = -34
22) y = x2 – 13x + 22
y = x - 18
Simplifying and Combining Like Terms
Exponent
4x
Coefficient
2
Variable (or Base)
* Write the coefficients, variables, and exponents of:
a) 8c2
b) 9x
c) y8
d) 12a2b3
Like Terms: Terms that have identical variable parts {same variable(s) and same
exponent(s)}
When simplifying using addition and subtraction, combine “like terms” by keeping the
"like term" and adding or subtracting the numerical coefficients.
Examples:
3x + 4x = 7x
13xy – 9xy = 4xy
12x3y2 - 5x3y2 = 7x3y2
Why can’t you simplify?
4x3 + 4y3
11x2 – 7x
6x3y + 5xy3
Simplify:
1) 7x + 5 – 3x
7x)
2) 6w2 + 11w + 8w2 – 15w
3) (6x + 4) + (15 –
4) (12x – 5) – (7x – 11)
5) (2x2 - 3x + 7) – (-3x2 + 4x – 7)
6) 11a2b – 12ab2
WORKING WITH THE DISTRIBUTIVE PROPERTY
Example:
3(2x – 5) + 5(3x +6) =
Since in the order of operations, multiplication comes before addition and subtraction, we
must get rid of the multiplication before you can combine like terms. We do this by
using the distributive property:
3(2x – 5) + 5(3x +6) =
3(2x) – 3(5) + 5(3x) + 5(6) =
6x - 15 + 15x + 30 =
Now you can combine the like terms:
6x + 15x = 21x
-15 + 30 = 15
Final answer: 21x + 15
Solving Equations
Golden Rule of Algebra:
“Do unto one side of the equal sign as you will do to the other…”
Whatever you do on one side of the equal sign, you MUST do the same exact
thing on the other side. If you multiply by -2 on the left side, you have to multiply by -2
on the other. If you subtract 15 from one side, you must subtract 15 from the other. You
can do whatever you want (to get the x by itself) as long as you do it on both sides of the
equal sign.
Solving Single Step Equations:
To solve single step equations, you do the opposite of whatever the operation is.
The opposite of addition is subtraction and the opposite of multiplication is
division.
Solve for x:
1) x + 5 = 12
4) 5x = -30
2) x – 11 = 19
5) (x/-5) = 3
3) 22 – x = 17
6) ⅔ x = - 8
Solving Multi-Step Equations:
3x – 5 = 22
To get the x by itself, you will need to get rid of the 5 and the 3.
+5 +5
We do this by going in opposite order of PEMDAS. We get rid
of addition and subtraction first.
3x
3
= 27
3
Then, we get rid of multiplication and division.
x =9
We check the answer by putting it back in the original equation:
3x - 5 = 22, x = 9
3(9) - 5 = 22
27 - 5 = 22
22 = 22 (It checks)
Simple Equations:
1) 9x - 11 = -38
2) 160 = 7x + 6
3) 32 - 6x = 53
4) -4 = 42 - 4x
5) ¾x - 11 = 16
6) 37 = 25 - (2/3)x
7) 4x – 7 = -23
8) 12x + 9 = - 15
9) 21 – 4x = 45
10) (x/7) – 4 = 4
11) (-x/5) + 3 = 7
12) 26 = 60 – 2x
Equations with more than 1 x on the same side of the equal sign:
You need to simplify (combine like terms) and then use the same steps as a multi-step equation.
Example:
9x + 11 – 5x + 10 = -15
4x + 21 = -15
Now it looks like a multistep eq. that we did in the 1 st
-21 -21
Use subtraction to get rid of the addition.
4x
= -36
4
4
Now divide to get rid of the multiplication
9x – 5x = 4x and
11 + 10 = 21
x
13) 15x - 24 - 4x = -79
= -9
14) 102 = 69 - 7x + 3x
15) 3(2x - 5) - 4x = 33
16) 3(4x - 5) + 2(11 - 2x) = 43
17) 9(3x + 6) - 6(7x - 3) = 12
18) 7(4x - 5) - 4(6x + 5) = -91
19) 8(4x + 2) + 5(3x - 7) = 122
Equations with x's on BOTH sides of the equal sign:
You need to "Get the X's on one side and the numbers on the other." Then you can solve.
Example: 12x – 11 = 7x + 9
-7x
-7x
5x – 11 = 9
+11 +11
5x
= 20
5
5
x
Move the x’s to one side.
Now it looks like a multistep equation that we did in the 1 st section.
Add to get rid of the subtraction.
Now divide to get rid of the multiplication
=4
20) 11x - 3 = 7x + 25
21) 22 - 4x = 12x + 126
23) ¾x - 12 = ½x -6
24) 5(2x + 4) = 4(3x + 7)
25) 12(3x + 4) = 6(7x + 2) 26) 3x - 25 = 11x - 5 + 2x
Solving Quadratic Equations
Solving quadratic equations (equations with x2 can be done in different ways. We will
use two
different methods. What both methods have in common is that the equation has to be set
to = 0. For instance, if the equation was x2 – 22 = 9x, you would have to subtract 9x from
both sides of the equal sign so the equation would be x2 – 9x – 22 = 0.
Solve by factoring: After the equation is set equal to 0, you factor the trinomial.
x2 – 9x – 22 = 0
(x-11) (x+2) = 0
Now you would set each factor equal to zero and solve. Think about it, if the product of
the two binomials equals zero, well then one of the factors has to be zero.
x2 – 9x – 22 = 0
(x-11) (x+2) = 0
x – 11 = 0
+11
x = 11
x+2=0
+11
-2
or
-2
x = -2
* Check in the ORIGINAL equation!
Solving Quadratics by Factoring:
20) x2 - 5x - 14 = 0
21) x2 + 11x = -30
22) x2 - 45 = 4x
23) x2 = 15x - 56
24) 3x2 + 9x = 54
25) x3 = x2 + 12x
26) 25x2 = 5x3 + 30x
27) 108x = 12x2 + 216
29) 10x2 - 5x + 11 = 9x2 + x + 83
28) 3x2 - 2x - 8 = 2x2
30) 4x2 + 3x - 12 = 6x2 - 7x - 60
Solve using the quadratic formula:
When ax2 + bx + c = 0
x=
-b ± √b2 – 4ac .
2a
a is the coefficient of x2 b is the coefficient of x c is the number (third term)
Notice the ±
factoring)
is what will give your two answers (just like you had when solving by
x2 – 9x – 22 = 0
a=1
x=
-b ± √b2 – 4ac .
2a
b= - 9
c = -22
x=
-(-9) ± √ (-9)2 – 4(1)(-22)
2(1)
x=
9 ± √81 + 88
2
x=
9 ± √169 .
2
-4(1)(-22) = 88
Split and do the + side and - side
9 – 13
2
9 + 13
2
x = 11
or
x = -2
* Check in the ORIGINAL equation!
Solving Quadratics Using the Quadratic Formula:
31) 2x2 - 6x + 1 = 0
32) 3x2 + 2x = 3
33) 4x2 + 2 = -7x
34) 7x2 = 3x + 2
35) 3x2 + 6 = 5x
36) 9x - 3 = 4x2
HOW TO FACTOR TRINOMIALS
Remember your hints:
A. When the last sign is addition
subtraction
x2 - 5x + 6
1)Both signs the same
2) Both minus (1st sign)
(x - )(x - )
B. When the last sign is
x2 + 5x – 36
(x - )(x + )
3) Factors of 6 w/ a sum
of 5. (3 and 2)
1) signs are different
2) Factors of 36 w/ a
difference of 5 (9
and 4)
3) Bigger # goes 1st
sign, +
(x - 3)(x - 2)
(x - 4)(x + 9)
FOIL Check!!!!!
Case II Factoring
Factoring a trinomial with a coefficient for x2 other than 1
6x2 + 5x – 4
1) Look for a GCF:
a. There is no GCF for this trinomial
b. The only way this method works is if you take out the GCF (if
there is one.)
2) Take the coefficient for x2 (6) and multiply it with the last term (4):
6x2 + 5x – 4
6 * 4 = 24
x2 + 5x – 24
3) Factor the new trinomial:
x2 + 5x – 24
(x + 8)(x – 3)
4) Take the coefficient that you multiplied in the beginning (6) and put it back
in the parenthesis (only with the x):
(x + 8)(x – 3)
(6x + 8)(6x – 3)
5) Find the GCF on each factor (on each set of parenthesis):
(6x + 8)
 2(3x+ 4)
(6x – 3)
 3(2x – 1)
6) Keep the factors left in the parenthesis:
(3x + 4)(2x – 1)
7) FOIL CHECK
Factor:
Solving Absolute Value Equations
Solving absolute value equations is almost the exact same as solving regular equations
with one major difference. In most cases you have 2 solutions.
Example:
|x|=5
We know that when x = 5, | 5 | will also equal 5, but it is also true that | -5 | will equal 5.
So, for |x | = 5, x = {-5, 5}. They both work.
How to solve absolute value equations
1) Isolate the absolute value.
2) Split into two separate equations, setting one to the negative and one to the positive.
Example:
| 2x + 6 | - 3 = 13
1) Isolate the absolute value:
** The steps are the same as if you were getting the x by itself. You move away all other
numbers by doing the opposite operation:**
| 2x + 6 | - 3 = 13
+3 +3
| 2x + 6 |
= 16
2) Now split into two
separate equations and
solve each.
2x + 6 = -16
-6 -6
2x
= -22
2
2
x = -11
3) Check by substituting in the original equation.
2x + 6 = 16
-6 -6
2x
= 10
2
2
x=5
Absolute Value Inequalities
Solving absolute value inequalities combine the strategies you used in:
1) Solving and Graphing Compound Inequalities
2) Solving Absolute Value Equations
Every absolute value inequality is a compound inequality. The 2 separate inequalities
come from when you split the inequality once the absolute value is isolated.
So first, you isolate the absolute value following all the same steps as you did when
isolating the absolute value when solving an absolute value equation. Then you split and
set to the negative and the positive.
The new stuff:
1) Once the absolute value is isolated, you choose whether it will be an AND problem or
an OR problem.
- Greater than (> , > ) will be an OR problem. (Graph both and keep)
- Less than (<, < ) will be an AND problem. (Graph both and keep the
intersection)
2) When you split the absolute value YOU MUST TURN THE INEQUALITY
AROUND WHEN SETTING TO THE NEGATIVE.
Example:
Isolate the abs. val.:
| 2x – 3 | - 10 > -5
+10 +10
| 2x – 3 |
Split and turn the ineq.
around (to less than)
when setting to -5 and
solve each ineq:
2x – 3 < -5
+3 +3
2x
< -2
2
2
> 5
2x – 3 > 5
+3 +3
2x
>8
2
2
Since its a Greater than,
x<-1
OR
it is an OR problem and
will be graphed accordingly.
x>4
Solving Systems of Equations
In order to solve for two variables, you need to have two equations. If you only have one
equation there are an infinite amount of ordered pairs (x,y) that will work. For example:
4x – 2y = 16 you can have x = 4 and y = 0 (4,0) and (2, -2) and (0, -4) and an infinite
amount of others. To be able to solve for a single ordered pair, you need a second
equation.
When we introduce the second equation, we will be able to solve for a single ordered
pair that will work in both equations. There are two ways to solve a system of equations
(algebraically and graphically). We will focus on solving algebraically. There are two
methods of solving algebraically (substitution and elimination). The key to both of them
is changing one (or both) equations so there is only one variable to solve for. Then you
follow all the rules of solving for the one variable. Then plug the value back into one of
the original equations to find the value of the second variable. Always state your answer
as an ordered pair.
SUBSTITUTION
Example: x = 3y + 8
5x + 2y = 6
5(3y + 8) + 2y = 6 Substitute 3y +8 for the x in the 2nd
equation
15y + 40 + 2y = 6 Distribute and solve.
17y + 40 = 6
17y = -34
y = -2
x = 3(-2) + 8
substitute the value for y back in to find x.
x= -6 + 8
x=2
(2, -2)
Check in BOTH ORIGINAL EQUATIONS!
Solve each system and check (in both equations):
a) x = 2y + 1
5x – 6y = 13
9x + 2y = -37
d) x + 3y = 11
e) 7x + 9y = -74
6x – 5y = 20
c) 5x – y = 7
b) y = 3x + 4
4x + 2y = 28
f) 10x – y = 1
4x + y = -5
8x + 3y = -8
Solving Systems with Linear Combinations (“Elimination”):
Sometimes solving a system of equations using substitution can be very difficult. For
these problems we solve using Linear Combinations (or Elimination). With elimination
you solve by eliminating one of the variables. This is accomplished by adding the 2
equations together. Before you can add the equations together, you need one of the two
variables to have two things:
1) Same Coefficient
2) Different Signs (one positive and one negative)
When you add terms with the same coefficient and different signs, the term drops out.
You then solve for the variable that is left. After you have solved for one variable, you
plug the value into one of the original equations and solve for the 2nd variable (just like
Substitution). Then, you check the solution in both original equations. The only
difference between Substitution and Elimination is how you solve for the 1st variable.
After that they are the same.
Examples:
A) Sometimes it works out that the 2 equations already have a variable with the same
coefficient and different signs. You can then just add the equations:
3x + 4y = 10 (The +4y and -4y cancel out
5x – 4y = -58 leaving you with just 8x.)
8x
= -48
8
8
x
= -6
Plug x = -6 in:
3(-6) + 4y = 10
-18 + 4y = 10
+18
+18
4y = 28
4
4
y=7
Final Solution: (-6, 7) CHECK IN BOTH!!!!
B) Sometimes (usually) the equations do not have same coefficient and different signs,
so we have a little bit of manipulating to do.
3x + 8y = 25
With this system, nothing will drop out if we just add the
5x + 4y = 23
equations. So we will multiply the bottom one by (-2).
-2(5x + 4y = 23)
Now the y’s have the same coefficient with different signs.
- 10x -8y = -46
3x + 8y = 25
Now plug x = 3 in:
- 10x -8y = -46
3(3) + 8y = 25
- 7x
= -21
9 + 8y = 25
-7
-7
-9
-9
8y = 16
x
=3
8
8
y=2
Final Solution: (3,2) CHECK IN BOTH!!!!
C. Sometimes we need to manipulate both equations. We can do this by “criss crossing
the coefficients.”
6x + 7y = 11
This is different than Example B, because no coeffcient
5x – 6y = -50
goes into another evenly.
-5(6x + 7y = 11)
You need the negative sign to change the 6x to negative
6(5x – 6y = -50)
so the signs will be different.
You can also use 5 and -6. You can also “criss cross”
the y coefficients.
-30x – 35y = -55
30x – 36y = -300
- 71y = -355
-71
-71
y=5
Plug in y = 5
5x – 6(5) = -50
5x – 30 = -50
+30
+30
5x
= -20
5
5
x = -4
Final Solution: (-4, 5) CHECK IN BOTH!!!!