1. Algebra II – Pre-requisite Skills Mr. Rosilez

2. Solving Equations – Key Concepts
Addition Property – If a = b, then a + c = b + c
Subtraction Property - If a = b, then a – c = b – c
Multiplication Property – If a = b, then ac = bc
Division Property – If a = b, and 𝑐≠0, then 𝑎 𝑐 = 𝑏 𝑐
Distributive Property – a(b ± c) = ab ± ac
Substitution – If a = b, then b may replace a
Proportion – A statement that two ratios are equal. To solve, cross multiply. Ex. 4 𝑥 = →12𝑥=120→𝑥=10
Inequalities – Solving inequalities work the same way as solving equations except for one difference. Whenever you multiply or divide by a negative number, you must change the direction of the sign.

3. Solving Equations - Examples
3. 2(𝑥−3) 5 = 𝑥 4 8 𝑥−3 =5𝑥 8𝑥−24=5𝑥 −24=−3𝑥 8=𝑥 4. −2 𝑥−3 ≤12 −2𝑥+6≤12 −2𝑥≤6 𝑥≥−3
1. 2 𝑥−6 =5𝑥+9 2𝑥−12=5𝑥+9 −12=3𝑥+9 −21=3𝑥 −7=𝑥 𝑥+ 3 4 =2𝑥− ( 1 3 𝑥+ 3 4 =2𝑥− 1 2 ) 4𝑥+9=24𝑥−6 15=20𝑥 𝑥= = 3 4

4. Variables and Exponents
Variables may only be combined if like terms. Only add or subtract the coefficients, not the variables!
3 𝑎 2 +5 𝑎 2 =8 𝑎 2
An exponent tells you how many times to multiply a base. The expression is called a power with base 6 and exponent 3.
6 3 =6×6×6=216
Rules for exponents:
Product of Powers
𝑎 𝑚 ∙ 𝑎 𝑛 = 𝑎 𝑚+𝑛
Power of a Product
𝑎∙𝑏 𝑚 = 𝑎 𝑚 ∙ 𝑏 𝑚
Power of a Power
𝑎 𝑚 𝑛 = 𝑎 𝑚𝑛
Quotient of Powers
𝑎 𝑚 𝑎 𝑛 = 𝑎 𝑚−𝑛 , 𝑎≠0
Power of a Quotient
𝑎 𝑏 𝑚 = 𝑎 𝑚 𝑏 𝑚 , 𝑏≠0
Negative Exponent
𝑎 −𝑛 = 1 𝑎 𝑛 , 𝑎≠0
Zero Exponent
𝑎 0 =1, 𝑎≠0

5. Examples – Solving Variables & Exponents
1. 3𝑥 8 ∙2 𝑥 4
(3∙2) (𝑥 8+4 )=6 𝑥 12
2. 𝑚 −2 −3
𝑚 (−2∙−3) =𝑚 6
𝑛 ∙ 3𝑛 2
2 (1∙4) 𝑛 (2∙4) ∙ 3 (1∙2) 𝑛 (1∙2)
16 𝑛 8 ∙9 𝑛 2 =144 𝑛 10
𝑥 2 𝑦 2𝑥𝑦
16 2 𝑥 (2−1) 𝑦 (1−1) =8𝑥
𝑟 −3 𝑠 𝑠 = 3 (1∙2) 𝑟 (−3∙2) 𝑠 (1∙2) 10𝑠
3 2 𝑟 −6 𝑠 2 10𝑠 = 𝑟 −6 𝑠 (2−1) = 9𝑠 10 𝑟 6
6. 3 𝑎 2 𝑏 0 𝑐 21 𝑎 −3 𝑏 4 𝑐 2 = 𝑎 (2−−3) 𝑏 (0−4) 𝑐 (1−2)
= 1 𝑎 5 𝑏 −4 𝑐 −1 7 = 𝑎 5 7 𝑏 4 𝑐
7. −6𝑟+3𝑠−5𝑟+8
−6r−5r +3s+8=−11𝑟+3𝑠+8
8. 3𝑟 5𝑟+2 −4(2 𝑟 2 −𝑟+3)
15 𝑟 2 +6𝑟−8 𝑟 2 +4𝑟−12
15 𝑟 2 −8 𝑟 𝑟+4𝑟 +12
7 𝑟 2 +10𝑟−12

6. Working with Square Roots
A square root of a number “n” is a number “m” such that 𝑚 2 =𝑛. Every positive number has two roots, a “+” and a “-”. The primary root is considered the positive root. Negative numbers have no square roots. The square root of “0” is “0”. Adding, subtracting, multiplying and dividing roots is similar to working with variables. However, you can’t have a root in the denominator of your final answer.
Properties:
𝑎𝑏 = 𝑎 ∙ 𝑏
𝑎 𝑏 = 𝑎 𝑏 = 𝑎𝑏 𝑏
Examples - (Primary answers for problems 3-5):
=±11
= 16 ∙ 3 =±4 3
− 28
( 25 ∙ 7 )−( 4 ∙ 7 )
5 7 −2 7 =3 7
∙2 2
10∙ 36 =10∙6=60
= ∙ = =2 6

7. Linear Equations
The following are the three methods of writing linear equations:
1. Standard Form
𝐴𝑥+𝐵𝑦=𝐶
𝑠𝑙𝑜𝑝𝑒= −𝐴 𝐵 , 𝑌 𝑖𝑛𝑡.= 𝐶 𝐵
2. Slope-intercept Form
𝑦=𝑚𝑥+𝑏
𝑚=𝑠𝑙𝑜𝑝𝑒, 𝑏=𝑦 𝑖𝑛𝑡.
3. Point-slope Form
𝑦− 𝑦 1 =𝑚(𝑥− 𝑥 1 )
𝑚=𝑠𝑙𝑜𝑝𝑒, 𝑥 1 , 𝑦 1 =𝑝𝑜𝑖𝑛𝑡
Slope (m) – The steepness of a line.
𝑚= 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 = ∆𝑦 ∆𝑥 = 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1
where 𝑥 1, 𝑦 1 and 𝑥 2, 𝑦 2 are
points on the line. Horizontal
lines have zero slope, vertical
lines have undefined slope.
X-intercept – Where the graph crosses the x-axis. To solve, substitute zero for “y” in the equation & solve.
Y-intercept – Where the graph crosses the y-axis. To solve, substitute zero for “x” in the equation & solve.

8. Examples for Slope & Linear Equations
Find the slope for each:
1. 4, 2 (6, −3) , 6 (3, 4)
𝑚= −3−2 6−4 = − 𝑚= 4−6 3−3 = −2 0 =𝑢𝑛𝑑𝑒𝑓.
Find the intercepts for:
3. 2𝑥−6𝑦= 𝑦=3𝑥+6
X-int: 2𝑥−6 0 =12 X-int: 0=3𝑥+6
2𝑥=12→𝑥= −6=3𝑥 →−2=𝑥
Y-int: 2 0 −6𝑦=12 Y-int: 𝑦=3 0 +6
−6𝑦=12→ 𝑦=− 𝑦=6

9. Examples for Slope & Linear Equations (Continued)
Writing Equations based on given info:
1. 𝑚=3, 𝑦−𝑖𝑛𝑡. =6
𝑦=3𝑥+6
2. 𝑚= 1 2 , 𝑝𝑜𝑖𝑛𝑡 (4, 3)
𝑦−3= 1 2 𝑥−4
𝑦−3= 1 2 𝑥−2
𝑦= 1 2 𝑥+1
3. −2, 4 (2, 16)
Find slope: 16−4 2−−2 = 12 4 =3
Method 1: Find y-intercept using
𝑦=𝑚𝑥+𝑏 ,
choosing either point for x and y.
4=3 −2 +𝑏→4=−6+𝑏→10=𝑏
Write the equation: 𝑦=3𝑥+10
Method 2: Use either point and the slope with the point slope equation.
𝑦−16=3 𝑥−2
𝑦−16=3𝑥−6
𝑦=3𝑥+10

10. Graphing Linear Equations
The basic method for graphing equations is to make a table of values where you choose the x values to substitute into the equation to solve for the y values. You then plot the points and connect the dots.
Example: 𝑦=3𝑥−1
Hence, you would plot the
points (-1, -4), (0, -1), and
(1, 2) to form the graph.
Other hints on graphing lines:
If written in slope –intercept form, graph by using the y- int. and using the slope from this point.
If written in standard form, graph by finding the x and y intercepts.
If no intercept is given (y=3x), then the intercept is zero.
If the equation is x = _# , it is a vertical line through the given number.
If the equation is y = _#_, it is a horizontal line through the given number.

11. Graphing Linear Equations
Graph the following:
𝑦=2𝑥−4
slope = 2, y-int. = -4
2𝑥−3𝑦=6
x-int. = 3, y-int. = -2
𝑦=−4
horizontal line through -4
𝑥=−3
vertical line through -3
𝑦−4=− 1 2 (𝑥+4)
Convert to slope int. form, 𝑦=− 1 2 𝑥+2, & graph.

12. Systems of Equations
The goal of systems is to find a point, (x, y), that is the solution for both equations. There are three basic methods for solving systems of equations: graphing, substitution, and elimination. We will focus on substitution and elimination.
Substitution is best used when one of the equations is already solved for one variable (ex. 𝑦=−2𝑥+6). You will substitute this equation into the other equation to solve. You will then use the answer and substitute it back into the first equation to solve and complete the process.
Example:
2𝑥+3𝑦=22
𝑦=2𝑥+2
Substitute 2𝑥+2 for y in the first equation and solve.
2𝑥+3 2𝑥+2 =22
2𝑥+6𝑥+6=22
8𝑥+6=22
8𝑥=16
𝑥=2
Then, substitute this value into 𝑦=2𝑥+2 to solve.
𝑦=2 2 +2
𝑦=6
Thus, the solution is (2, 6).

13. Systems of Equations - Continued
Elimination – When neither of the equations has a variable that is already “solved” for a variable, then elimination is the preferred method of solving systems. The goal of elimination is two get rid of one of the variables in the two equations by multiplying one or both equations. Then, you either add or subtract the equations to eliminate a variable in order to solve for the other. You then substitute this value into one of the original equations to solve for the eliminated variable.
Example
2𝑥+3𝑦=4
3𝑥+5𝑦=8
3 2𝑥+3𝑦=4
2 3𝑥+5𝑦=8
6𝑥+9𝑦 =12
− 6𝑥+10𝑦=16
−𝑦=−4
𝑦=4
Substitute y = 4 to solve for x.
2𝑥+3 4 =4
2𝑥+12=4
2𝑥=−8
𝑥=−4
Hence, the solution is (-4, 4).

14. Factoring and Solving Equations
Rules for Factoring:
1. Pull out the GCF.
4 𝑥 2 −8𝑥+12=4( 𝑥 2 −2x+3)
2. For equations in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐 𝑤ℎ𝑒𝑟𝑒 𝑎=1, find 2 numbers such that when you multiply = ac, when you add = b.
𝑥 2 +4𝑥+3
𝑎𝑐=3, 𝑏=4, 𝑠𝑜 3, 1
(𝑥+3)(𝑥+1)
3. If 𝑎≠1, try grouping
6 𝑥 2 +13𝑥−5
𝑎𝑐=−30, 𝑏=13, 𝑠𝑜 15, −2
6 𝑥 2 +15𝑥−2𝑥−5
3𝑥 2𝑥 −1 2𝑥+5
(2𝑥+5)(3𝑥−1)
4. For difference of squares
4 𝑥 2 −16 𝑦 2
4 𝑥 2 =2𝑥 𝑦 2 =4𝑦
(2𝑥+4𝑦)(2𝑥−4𝑦)

15. Factoring and Solving Quadratic Equations
In order to solve quadratic equations, try factoring them first. Once factored, set each part equal to zero and solve.
𝑥 2 −12𝑥=0
3𝑥 𝑥−4 =0
3𝑥= 𝑥−4=0
𝑥=0,4
2. 𝑥 2 −6𝑥+8=0
𝑥−4 𝑥−2 =0
𝑥−4=0 𝑥−2=0
𝑥=4, 2
If you can’t factor, then use the quadratic formula to solve 𝑎 𝑥 2 +𝑏𝑥+𝑐.
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
Ex: 𝑥 2 −5𝑥−9
𝑎=3, 𝑏=−5, 𝑐=−9
𝑥= −(−5)± (−5) 2 −4(3)(−9) 2(3)
𝑥= 5±
𝑥= 5±

16. Transformations
A transformation is a function that moves or changes a figure in some manner to produce a new figure. The four major transformations are a translation (movement left to right and or up and down), a reflection (a flip over a specific line of reflection), a rotation (usually counterclockwise around a point such as the origin) and a dilation (reduction or an enlargement of a figure from a specified point).
Rules for Transformations
Translation
(𝑥, 𝑦)→(𝑥±#, 𝑦±#)
Reflection over:
X-axis 𝑥, 𝑦 → 𝑥, −𝑦
Y-axis 𝑥, 𝑦 → −𝑥, 𝑦
Y=x 𝑥, 𝑦 → 𝑦, 𝑥
Y=-x (𝑥, 𝑦)→(−𝑦, −𝑥)
Rotation-CC about the origin:
90° (𝑥, 𝑦)→(−𝑦, 𝑥)
180° 𝑥, 𝑦 → −𝑥, −𝑦
270° (𝑥, 𝑦)→(𝑦, −𝑥)
Dilation, Scale factor K:
(𝑥, 𝑦)→(𝑘𝑥, 𝑘𝑦)

17. Transformations - Continued
Using AB with vertices A(2, 3) B(3, 5) for each as the original (pre-image):
Translation 2 units right, 1 unit up
(𝑥, 𝑦)→(𝑥+2, 𝑦+1)
𝐴→ 𝐴 ′ 4, 4 , 𝐵→𝐵′(5, 6)
Reflection in y = -x.
(𝑥, 𝑦)→(−𝑦, −𝑥)
𝐴→ 𝐴 ′ −3, −2 , 𝐵→𝐵′(−5, −3)
Rotation 90° 𝐶𝐶
(𝑥,𝑦)→(−𝑦, 𝑥)
𝐴→𝐴′ −3,2 ,𝐵→𝐵′(−5, 3)
Dilation, SF = ½
(𝑥,𝑦)→( 1 2 𝑥, 1 2 𝑦)
𝐴→ 𝐴 ′ (1, 1.5), 𝐵→𝐵′(1.5, 2.5)