SECTION 1-3
Linear & Absolute Equations
Linear & Compound Inequalities
Functions
SECTION 4-6
Linear Functions and Slopes
Equations of Line
Systems of equations graphically
SECTION 7-9
Systems of Equations Algebraically
Exponents
GCF and Factoring
SECTION 10-12
Factoring Trinomials
Factoring Special Forms
Polynomials by factoring
SECTION 13-15
Rational Functions
Rational Expressions
Rational Equations
100

Distributive Property

3(4π‘₯ + 3)

12π‘₯ + 9

100

Calculate the slope of the line passing through the following points: (βˆ’πŸ‘, βˆ’πŸ’) and (βˆ’πŸ, πŸ”)

m = 5 

100

Solve the following system of equations algebraically. Check your solution. 

4π‘₯ + 5𝑦 = 24 (𝐄πͺ. 𝟏) 

6π‘₯ + 7𝑦 = 34 (𝐄πͺ. 𝟐)

𝑦 = 4

π‘₯ = 1  

100

Factor. 

5π‘₯2 + 13π‘₯ + 6

(π‘₯ + 2)(5π‘₯ + 3)

100

Find the domain. Express using interval notation.

𝑓(π‘₯) =     3 

          ______

          π‘₯2 βˆ’ 1

Hint: exclude values that make denominator 0.

Domain of 𝑓 = (βˆ’βˆž, βˆ’1) βˆͺ (βˆ’1, 1) βˆͺ (1, ∞)

200

Solve for x:

3(3π‘₯ + 7) = βˆ’5π‘₯ + 7

x= -1

200

Write an equation of the line that passes through the points (βˆ’2, 4) and (2, βˆ’8).

𝑦 = βˆ’3π‘₯ βˆ’ 2

200

Solve the following system of equations algebraically.

4π‘₯ + 6𝑦 = 22 (𝐄πͺ. 𝟏) 

2π‘₯ + 3𝑦 = 5 (𝐄πͺ. 𝟐)

No Solution

Why?

⟹ 2π‘₯ + 3𝑦 = 5 (𝐄πͺ. 𝟐) 

⟹ (βˆ’πŸ)(2π‘₯) + (βˆ’πŸ)(3𝑦) = (βˆ’πŸ)(5)

⟹ βˆ’4π‘₯ βˆ’ 6𝑦 = βˆ’10 (𝐄πͺ. πŸ‘) 

This gives us a new version of Eq. 2 which we have labeled Eq. 3. We don’t have to multiply Eq. 1 by anything. We will now add Eq. 1. to Eq. 3 

       4π‘₯ + 6𝑦 = βˆ’22 (𝐄πͺ. 𝟏) 

+ βˆ’ 4π‘₯ βˆ’ 6𝑦 = βˆ’10 (𝐄πͺ. πŸ‘) 

       0!=12 

Both variables were eliminated! The bad news is we are left with a false statement since 0 does not equal 12. Therefore, there is no solution. 

200

Factor

8π‘₯2 βˆ’ 11π‘₯ + 3

(π‘₯ βˆ’ 1)(8π‘₯ βˆ’ 3)

200

Simplify. 

π‘₯2 βˆ’ 4π‘₯ + 3 

_________

    π‘₯ βˆ’ 1

= π‘₯ βˆ’ 3

300

Find the solution set of the equation.

|3π‘₯ βˆ’ 8| = |5π‘₯ βˆ’ 6|

{βˆ’1, 7/4 } 

OR

x= -1, x= 7/4

300

Write an equation of the line that is parallel to 3π‘₯ βˆ’ 2𝑦 = 5 and passes through the point (6, βˆ’2).

𝑦 = 3/2 π‘₯ βˆ’ 11

300

(2 pm-1q0)-4β‹… 2m-1p3

________________

           2pq2

    m3

_______

16p2q2

300

Predict next 3 difference of squares:

π‘₯2 βˆ’ 1 = (π‘₯ + 1)(π‘₯ βˆ’ 1) 

π‘₯2 βˆ’ 4 = (π‘₯ + 2)(π‘₯ βˆ’ 2) 

π‘₯2 βˆ’ 9 = (π‘₯ + 3)(π‘₯ βˆ’ 3)

 

π‘₯2 βˆ’ 16 = (π‘₯ + 4)(π‘₯ βˆ’ 4) 

π‘₯2 βˆ’ 25 = (π‘₯ + 5)(π‘₯ βˆ’ 5) 

π‘₯2 βˆ’ 36 = (π‘₯ + 6)(π‘₯ βˆ’ 6)

300

Equation on board 

  5π‘₯(π‘₯ + 2) 

__________

    π‘₯ + 6

400

Express in interval notation, and in set-builder notation. 5π‘₯ βˆ’ 2 < βˆ’17 𝑂𝑅 1 ≀ 4π‘₯ + 1

Solution in interval notation: (βˆ’βˆž, βˆ’3) βˆͺ [0, ∞) Solution in set-builder notation: {π‘₯|π‘₯ < βˆ’3 or π‘₯ β‰₯ 0}

400

Solve the following system of equations and find out if they have a solution 

2π‘₯ βˆ’ 3𝑦 = 6 

4π‘₯ βˆ’ 6𝑦 = βˆ’6

Both lines have the same slope, that is 2/3 , and different 𝑦-intercepts. This means that the lines are parallel and they will never intersect! This system of equations has no solution. We say that this system of equations is inconsistent.

400

(8π‘₯2𝑦 + 13π‘₯𝑦 + 7) βˆ’ (12π‘₯2𝑦 βˆ’ 4π‘₯𝑦 + 3)

βˆ’4π‘₯2𝑦 + 17π‘₯𝑦 + 4

400

Factor completely. 50π‘₯2 βˆ’ 18𝑦2

2(5π‘₯ + 3𝑦)(5π‘₯ βˆ’ 3𝑦)

400

Equation on board

π‘₯ = 38

500

Use the graph of the function to determine the following: 

Domain: 

Range: 

𝑓(βˆ’4) = 

𝑓(1) = 

For what values of π‘₯ is 𝑓(π‘₯) = 5? 

Determine the coordinates of the π‘₯- and 𝑦- intercepts. 

For what values of π‘₯ is 𝑦 negative? Express in interval notation.  

Domain: (-∞ , 5]

Range: [-3,8]

𝑓(βˆ’4) = 4

𝑓(1) = -2

For what values of π‘₯ is 𝑓(π‘₯) = 5?  4

Determine the coordinates of the π‘₯- and 𝑦- intercepts. x (-2,0) and (2,0) y (0,-3)

For what values of π‘₯ is 𝑦 negative? Express in interval notation.  (-2,2)

500

Solve the following system of equations and find out if they have a solution

2π‘₯ + 4𝑦 = 8 

π‘₯ + 2𝑦 = 4

Even though we set out to graph two separate lines, we ended up with just one line! This means that the two lines actually overlap. They touch at infinitely many points. This system of equations has Infinitely many solutions. We say that this system of equations is dependent.

500

(12π‘₯6𝑦3 βˆ’ 16π‘₯4𝑦2 + 4π‘₯𝑦) Γ· (4π‘₯𝑦)

3π‘₯5𝑦2 βˆ’ 4π‘₯3𝑦 + 1

500

Find all solutions to the equation. 

5π‘₯2 = 80

The solutions are π‘₯ = 4 or π‘₯ = βˆ’4

500

Equation on board

x= 6