A powerpoint compilation

MIDTERM LESSONS IN

ADVANCE ALGEBRA

WORD PROBLEMSThe following steps must be followed in

order to solve word problems systematically.

Read and understand the problem.

1.Draw a diagram or table if possible.

2. Represent the unknown with a variable.

3. Write the other quantities in terms of that variable.

4. Write the equation that models the situation in the problem.

5. Solve the equation.

6. Answer the question in the problem.

7. Check the answer by using it in solving the original problem.

WORD PROBLEMS EXAMPLES

1NUMBER PROBLEMS

2MIXTURE

PROBLEMS

3RATE, TIME

& DISTANCE PROBLEMS

4WORK PROBLEMS

5COIN

PROBLEMS

6GEOMETRIC

PROBLEMS

NUMBER PROBLEMS1

The sum of two numbers is 99. The second is 3 more than the first. What are the numbers?

Solution:Let x = first numberX = 3 = second number

Equation:x + x + 3 = 99 2x + 3 = 99 2x = 99 – 3 2x = 96 x = 96/2

x = 48

Check: 48 + 51 = 99 99 = 99

 

MIXTURE PROBLEMS1 2

How many grams of 8% acid solution and how many grams of 15% acid solution must be mixed to obtain 10 gr4ams of a 12% acid solutions?

8% acid solution

8% x 0.08x

15% acid solution

15% 10-x 0.15(10-x)

Mixture 12% 10 0.12(10)

Equation:

0.08x + 0.015(10-x) = 0.12(10) 0.08 + 1.5 – 0.15x = 1.2 0.08x – 0.15x = 1.2-1.5

-0.07x = -0.30x = -0.30/-0.07x = 4.49 grams

RATE, TIME &

DISTANCE PROBLEMS1 3

John drove 4,000 meters in 5 minutes. Part of the trip was at 10 m/s and the r4est at 15 m/s. Find the time spent for the 10 m/s speed.

Solution:Time= distance traveled/ speed (T= d/s)

t= timed= distances= speed

Let t1= time spent with speed of 10 m/s

5min- t1 or 300s – t1 = time spent with speed of 15 m/s

Total distance traveled = 4,000 meters;D1 = distance traveled with lower speedD2 = distance traveled with higher speed

Equation:

Total distance = d1 +n d2

4,000 = (10m/s)(t1) + (15m/s)(300s – t1)

4000 = 10t1 + 4500 -15t1

-10t1 + 15t1 = 4500 – 4000

5t1 = 500

t1= 500/5

t1 = 100 seconds or 1 min and 40 seconds

WORK PROBLEMS

4

George can paint a house in 8 hours, John in 10 hours, and Paul in 12 hours. How long will it take to do the job if George and John work 2 hours and then George and Paul finish the job.

George 1/8 x/8John 1/10 2/10Paul 1/12 (x-2)/12

Solution:

Equation:

x/8 + 2/10 + (x-2)/12 = 1120(x/8) + 120(2/10) + 120(x-2)/12 = 120 (Multiplying by the led (120))

15x + 24 + 10x – 20 = 12025x + 4 = 120 - 4 25x = 116

x = 116/25 or 4 and 16/25 hours

 

COIN PROBLEMS

5

Jason has 50 coins, all in 5 and 1 peso coins, amounting to Php 130.00. How many 5 peso coin does he have?Solution:Let x = number of 5 peso coinsThen 50-x = number of 1 peso coins 

5x + 1(50-x) = 130 5x + 50 - x = 130 4x + 50 = 130

4x = 80 x = 20

He has 20 5-peso coins and 30 1-peso coins. (50-x = 50-20 = 30)

GEOMETRICPROBLEMS

6

If one side of a square is doubled in length and the adjacent side is decreased by two centimeters, the area of the resulting rectangle is 96 square centimeters larger than that of the original square. Find the dimensions of the rectangle.

2x2 – 4x = x2 + 96 x2 – 4x – 96 = 0 (x – 12)(x + 8) = 0 x = 12 or x = –8

Solution: 

Square’s side length: x One side is doubled: 2xNext side is decreased by two: x – 2 Square’s area: x2 Rectangle’s area: (2x)(x – 2) = 2x2 – 4x New area is 96 more than old area: 2x2 – 4x = x2 + 96

SOLVING QUADRATIC EQUATIONS

1SQUARE ROOT

PROPERTY

2COMPLETING THE

SQUARE

3QUADRATIC

FORMULA

4DISCRIMINANT

5FACTORING

SQUARE ROOT PROPERTY1

The square root property is one method that is used to find the solutions to a quadratic (second degree) equation.  This method involves taking the square roots of both sides of the equation. 

Steps:

1. Before taking the square root of each side, you must isolate the term that contains the squared variable.

2. Once this squared-variable term is fully isolated, you will take the square root of both sides and solve for the variable. 

3. We now introduce the possibility of two roots for every square root, one positive and one negative. 

4. Place a sign in front of the side containing the constant before you take the square root of that side.

Example:

√𝑥2=±√12

𝑥2=12

𝑥=√ 4(3)=±2√3

… the squared-variable term is isolated, so we will take the square root of each side

… notice the use of the sign, this will give us both a positive and  a negative root

… simplify both sides of the equation, here x is isolated so we have solved this equation 

So, the values of

Solution:

COMPLETINGTHE SQUARE1 2

An alternate method of solving quadratic equations is called completing the

square.  Actually completing the square involves using the square root property. 

However, we have a few steps to setting up the problem before we can use completing

the square.

Example:

Solution:

𝑥2−2 𝑥=8 … first isolate both terms containing the variable onto one side of the equation

𝑥2−2 𝑥+1=8+1 … next take  of the middle term now square this number and add to both sides

(𝑥−1)2=9 … the right-hand side will factor into the square of a binomial

     … now use the square root property and solve

            

QUADRATIC FORMULA1 3

In elementary algebra, the quadratic formula is the solution of the quadratic equation. There are other ways to solve the quadratic equation instead of using the quadratic formula, such as factoring, completing the square, or graphing. Using the quadratic formula is often the most convenient way.

Example:

Solution: By Quadratic Formula:  a = 1,  b = 2,  c = -8

Substitute the values and solve.

DISCRIMINANT 4

The discriminant is the expression b2 – 4ac. The value of the discriminant can be used to determine the number and type of roots of a quadratic equation.

We used the discriminant to determine whether a quadratic polynomial could be factored. If the value of the discriminant for a quadratic polynomial is a perfect square, the polynomial can be factored. The Discriminant can be negative, positive or zero.

REMEMBER:

If b2 – 4ac > 0 the equation will have two distinct real roots.

If b2 – 4ac = 0 the equation will have repeated roots.

If b2 – 4ac < 0 the equation will have no real roots.

1. 9x2+6x+1=0

a=9, b=6, c=1

b2-4ac=(6)2-4(9)(1) =36-36=0

1 real solution

Examples: (Just substitute the values of a, b, and c.)

2. 9x2+6x-4=0

a=9, b=6, c=-4

b2-4ac=(6)2-4(9)(-4) =36+144=180

2 real solutions

FACTORING 5

We use the Distributive Property to expand Algebraic Expressions. We sometimes need to reverse this process by factoring, an expression as a product of simpler ones.

Example:

-4= (x-2)(x+2)

A. FACTORING BY COMMON FACTORS

Example:

1. -6x The greatest common factor of the terms -6x is 3x, so we have

-6x= 3x(x-2)

B.FACTORING TRINOMIALS

To factor trinomial of the form + +, we note that (x+r)(x+s)= +(r+s)x+rs, so we need to choose numbers r and so that r+s=b and rs=c.

a. Trial and ErrorExample: – + 𝒙𝟐 𝟗𝒙 𝟐𝟎There is no common factor so we begin by setting up the first terms as follows: ( )( ) 𝑥

Next we get the signs right. Since the last term is positive, the signs will be the same as the middle term which is negative. ( − )( − ) 𝑥 𝑥

Finally we determine the last term by trial and error. The last terms multiplied together should equal 20 and when added equal −9.

Therefore our last terms will be −4 and −5. ( − 4)( − 5)𝑥 𝑥

b. Difference of Two SquareExample: Factor  x2 - 9

Both x2 and 9 are perfect squares.  Since subtraction is occurring between these squares, this expression is the Difference of Two Squares

Answer:  (x + 3) (x - 3)  or   (x - 3) (x + 3)  

c. Perfect Square TrinomialExample: Factor x2 + 2x + 1

Notice that x2 + 2x + 1 = x2 + 2x + 12

Using x2 + 2x + 12, we see that... the first term is x2 and the base is x; the last term is 12 and the base is 1

Put the bases inside parentheses with a plus between them    (x + 1)

Raise everything to the second power   (x + 1)2 

LINEAR INEQUALITIES

1SOLUTION SET

2SET BUILDER

NOTATION

3INTERVAL NOTATION

A linear inequality describes a region of the coordinate plane that has a boundary line.

Every point in the region is a solution set.

a < b and b > aFor two real numbers a and b, we say that a is less than b, and write a <b , if there is a positive real

number p so that a + p = b. the statement b > a, read b is greater than a, means exactly the same as a < b.

If you add a positive number to any number the sum is larger than the original number.

When we write a b we mean a < b or a = b and say a is than or equal to b. when we write a b we mean a > b or a = b and say a is greater than or equal to b.

SOLUTION SET1

A solution set is the set of values which satisfy a given inequality. It means, each and every value in the solution set will satisfy the inequality and no other value will satisfy the inequality.

Example:

Solve 2x + 3 ≤ 7, where x is a natural number.

Solution:2x + 3 ≤ 7 Subtracting 3 from both the sides2x ≤ 4 Dividing both sides by 2x ≤ 2 Since x is a natural number

Solution set = {1, 2}.

SET BUILDER NOTATION1 2

Shorthand used to write sets, often sets with an infinite number of elements.

Note: The set {x: x > 0} is read aloud, "the set of all x such that x is greater than 0." It is read aloud exactly the same way when the colon: is replaced by the vertical line | as in {x | x > 0}.

General Form:

{formula for elements: restrictions} or{formula for elements| restrictions}

Examples {x: x ≠ 3} the set of all real numbers except 3

{x | x < 5} the set of all real numbers less than 5

{x2 | x is a real number}

the set of all real numbers greater than or equal to 0

{2n + 1: n is an integer}

The set of all odd integers (e.g. ..., -3, -1, 1, 3, 5,...).

INTERVAL NOTATION1 3

INTERVAL NOTATION

INEQUALITYNOTATION

LINE GRAPH TYPE

Closed

Half – open

( Half – open

Open

Closed*

Open

Closed*

) Open

x

x

x

x

x

x

] x

x

PROPERTIES OF INEQUALITIES

1ADDITION PROPERTY OF INEQUALITY

2

3MULTIPLICATION PROPERTY OF INEQUALITY

4DIVISION PROPERTY OF INEQUALITY

SUBTRACTION PROPERTY OF INEQUALITY

ADDITION

PROPERTY OF INEQUALITY1

Using Addition to Solve Inequalities

The solutions of an inequality like x < 3 are easy to recognize. To solve some other inequalities, you may need to find a simpler equivalent inequality.

Equivalent inequalities have the same set of solutions.

Consider the inequality -4 < 1. The number line shows what happens when you add 2 to each side of the inequality.

-4 < 1-2 < 3-4 + 2 < 1 + 2

Notice that addition does not change the relationship between the numbers of the direction of the inequality symbol.

Addition Property of Inequality For all real numbers a, b and c, if a > b, then a + c > b + c. Example: 5 > -1, so 5 + 2 > -1 + 2 For all real numbers a, b, and c, if a < b, then a + c < b + c. Example: -4 < 1, so -4 + 2 < 1 + 2

EXAMPLE:

Solve x – 3 < 5. Graph the solutions on a number line.

x – 3 < 5x – 3 + 3< 5 + 3 add 3 to each side x < 8 combine like terms

The solutions are all numbers less than 8.Notation: SS: (-∞.8)

SUBTRACTION PROPERTY OF INEQUALITY1 2

Using Subtraction to Solve Inequality

 

Just as you can add the same number to each side of an inequality, you can subtract the same number from each side. The order, or direction, of the inequality is not changed.

Subtraction Property of Inequality

For all real numbers a, b, and c, if a > b, then a – c > b – c. Example: 3 > -1, so 3 – 2 > -1 – 2

For all real numbers a, b, and c, if a < b, then a – c < b – c. Example: -5 < 4, so -5 – 2 < 4 – 2

EXAMPLE:

x + 2 < -6x+ 2 – 2< -6 – 2subtract 2 from each side x < -8 combine like terms

Notation: SS: (

MULTIPLICACTION

PROPERTY OF INEQUALITY1 3

Solving Inequalities Using Multiplication

You can find equivalent inequality by multiplying or dividing each side of an inequality by the same number. If the number is positive, the order remains the same. If the number is negative, you reverse the order.

Multiplication Property of inequality

For all real numbers a and b, and for c > 0: If a > b, then ac > bc. If a, b, then ac < bc Example; 4 > -1, so 4(5) > -1 (5) -6 < 3, so -6 (5) < 3 (5)

For all real numbers a and b, and f or c < 0: If a > b, then ac <bc.Example: 2 > -1, so 4 (-2) < -1 (-2) -6 < 3, so -6(-2) > 3 (-2)

EXAMPLE:

Solve: < -1. Graph the solutions on a number line.

< -12 ( ) <2(-1) multiply each side by 2 x<-2 simplify each side 

The solutions are all numbers less than -2.Notation:SS :( -, -2)

DIVISION

PROPERTY OF INEQUALITY

4

Division Property of Inequality

For all real numbers a and b, and for c >0: If a>b, then > If a<b, then > . Ex.: 6> -4, so > -2 < 8, so < For all real numbers a and b, and for c<0. If a>b, then < . If a< b, then < . Ex.: 6>-4, so < -2<8, so >

EXAMPLE: 

4d ≤ -28 ≤ divide both side by 4 d≤ -7 simplify each side 

The solutions are all less than or equal to -7.Notation: SS: (-∞,

COMPOUND INEQUALITIES

1SOLVING COMPOUND INEQUALITIES JOINED BY “AND”

2 SOLVING COMPOUND INEQUALITIES JOINED BY “OR”

SOLVING COMPOUND INEQALITIES

JOINED BY “AND”1

1 2SOLVING

COMPOUND INEQALITIES

JOINED BY “OR”

Solving Compound Inequalities Joined By “Or”

A solution of a compound inequality joined by or is any number that makes either inequality true.

EXAMPLE: Solve 4x = 3 < -5 or -2x + 7 < 1. Graph the solutions.(4x) <(-8) (-2x) >(-6) x < -2 or x >3

The solutions are all numbers that are less than -2 or are greater than 3.SS: (−∞, -2 ) µ (3,∞)

ABSOLUTE VALUE EQUATIONS AND INEQUALITIES

Absolute Value Equations

Use the following property to solve equations that involve absolute value.

|x| = c is equivalent to |x| =± c

This property says that to solve an absolute value equation, we must solve two separate equations. For example, the equation |x| = 5 is equivalent to the two equations x= 5 and |x| = -5.

EXAMPLESolve the equation |2x-5| = 3 Solution: the equation |2x-5| = 3 is equivalent to two equations:

2x-5 = 3 or 2x-5 = -3 2x=8 2x=2 transpose 5 to

the right side then simplify

x= 4 x=1 divide by 2  The solutions are 1 and 4.

Absolute Value Inequalities 

We use the following properties to solve inequalities that involve absolute value.

These properties can be proved using the definition of absolute value. To prove the property one, that the inequality |x| < c says that the distance from x to 0 is less than c and based from the figure in left side you can see that this is true if and only if x is between c and –c.

ExampleIn solving an Absolute Value inequality 

Sole the inequality |x-5|<2Solution 1: The inequality |x-5|<2 is equivalent to  -2<x - 5<2 property 1 3 < x < 7 add 5

The solution set is the open interval (3,7)

Solution 2 Geometrically, the solution set consists of all numbers x whose distance from 5 is less than 2 and according to the figure on the left side the interval is (3,7)

POLYNOMIAL INEQUALITIES

Polynomial Inequalities A polynomial inequality is a mathematical statement that relates a polynomial expression as either less than or greater than another.Example 1 – Graph: 

Step 1: Write the polynomial in the correct form. The polynomial must be written in descending order and must be less than, greater than, less than or equal to, or greater than or equal to zero.

Step 2: Find the key or critical values. To find the key/critical values, set the equation equal to zero and solve.

Step 3: Make a sign analysis chart. To make a sign analysis chart, use the key/critical values found in Step 2 to divide the number line into sections.

Step 4: Perform the sign analysis. To do the sign analysis, pick one number from each of the sections created in Step 3 and plug that number into the polynomial to determine the sign of the resulting answer. In this case, you can choose x = –3 which results in +7, x = 0 which results in –8, and x = 5 which results in +7.

Step 5: Use the sign analysis chart to determine which sections satisfy the inequality. In this case, we have greater than or equal to zero, so we want all of the positive sections.

Step 6: Use interval notation to write the final answer.

FINDING THE DOMAIN OF THE FUNCTION

If a function is defined by an equation and the domain is not indicated, then we assume that the domain is not the set of all real number replacements of the independent variable that produce real value for the independent variables.

The domain of a function is the set of numbers that can go in to a given function. In other words, it is the set of x-values that you can put in to any given equation. The set of possible y-values is called the range.

EXAMPLEFind the domain of f(x) =  

Solution:The fraction represents a real number for all

replacements of x by real numbers except x =3, since division by 0 is not defined. Thus, f(3) does not exist and the domain of f is the set of all real numbers except 3. We often indicate this by writing f(x) =

x is not equal to 3.

EVALUATING FUNCTIONS

To evaluate a function, simply replace (substitute) the function's variable with the indicated number or expression. Example:f(x) = 2x+4 for x=5Just replace the variable "x" with "5":f(5) = 2(5) + 4 = 14

10 + 4 = 14

Answer: f(5) = 14

1. If , find . The notation means that everywhere we see x in the function, we are going to replace it with 2. This gives . Our notation is • Notice that it is not absolutely necessary to enclose the “2” in parentheses, but if functions are more complicated it is helpful to avoid confusion.

OPERATIONS OF FUNCTIONS

Each function is defined for all x in the domains of both f and h.

A. Sum of f and h is defined as:

The domain of f + h consists of the numbers x that are in the domain of f and in the domain of h.

B. Difference of f and h is defined as:

The domain of f-h consists of the numbers x that are in the domain of f and in the domain oh h.

C. Product of f and h is defined as:

The domain of f·h consists of the numbers x that are in the domain of f and in the domain of h.

D. Quotient of f and h is defined as:

The domain of f/h consists of the numbers x for which h(x) not equal to 0 that are in the domain of f and in the domain of h.

Illustrative example: 

Let f(x)=x+1 and h(x)=2x2

1) The sum = (f + h)(x) = f(x) + h(x) = 2x2 x+1

2)  The difference = (f - h)(x) = f(x) - h(x) =( x+1)- 2x2=  -2x2 + x+1

3)  The product = (f . h)(x) = f(x) . h(x) =( x+1)( 2x2) = 2x3+2x2

4)  The quotient = (f/h)(x) = f(x)/h(x) =  x+1/2x2; h(x)not equal to 0

COMPOSITION OF FUNCTIONS

If f and g are functions, then the composite function, or composition, of g and f is defined by

The domain of the composition function f o g is the set of all x such that

• x is in the domain of g and• g(x) is in the domain of f.

Let f(x)=2x-1 and g(x)

A. Find (f o g)(2) Solution find g(2) Since g(x)

g(2)=42−1 = 41 =4

41x

41x

EXAMPLE

Find (g o f) (-3)

Solution:

(g o f) (-3)= g(f(-3))= g(-7)

= 4/(-7-1)= 4/(-8)

= -1/2

INVERSE FUNCTIONS

Definition: An inverse function is a function that undoes the action of the other function.

Remember: 

The inverse of a function may not always be a function!

The original function must be a one-to-one function to guarantee that its inverse will also be a function. 

a. Given function f, find the inverse relation. Is the inverse relation also a function?

Answer:

Function f is a one-to-one function since the x and y values are used only once. Since function f is a one-to-one function, the inverse relation is also a function.

Therefore, the inverse function is:

b. Determine the inverse of this function.  Is the inverse also a function?

x 1 -2 -1 0 2 3 4 -3 f (x) 2 0 3 -1 1 -2 5 1

Answer:    Swap the x and y variables to create the inverse relation.  The inverse relation will be the set of ordered pairs:

{(2,1), (0,-2), (3,-1), (-1,0), (1,2), (-2,3), (5,4),(1,-3)} 

Since function f was not a one-to-one function (the y value of 1 was used twice), the inverse relation will NOT be a function (because the x value of 1 now gets mapped to two separate y values which is not possible for functions).

Finding the Inverse of a FunctionExample:Let f be defined by f(x)=3x+1. Find the inverse function f−1(x).

f(x) = 3x + 1 Steps: y = 3x + 1 Replace f(x) by y. x – 1 = 3y Interchange x and y.Solve for y.

f-1(x) Replace y with f-1(x).

Therefore, we found out that x=y/3−1/3, so we can write the inverse function as f-1(x) =

Check:First, we apply f followed by f−1.

f−1 f)(x) =f−1(f(x))∘ =f−1(3x+1) =(3x+1)/3−1/3 =x+1/3−1/3 =x

Second, we apply f−1 followed by f.

(f f−1)(x)=f(f−1(x))∘ =f(x/3−1/3) =3(x/3−1/3)+1 =x−1+1 =x

In both cases, applying both f and f−1 to x gave us back x. Indeed, f−1(x)=x/3−1/3.

GRAPHING LINEAR EQUATIONS

A linear equation is an equation with two variables whose graph is a line. The graph of the linear equation is a set of points in the coordinate plane that all are solutions to the equation. If all variables represent real numbers one can graph the equation by plotting enough points to recognize a pattern and then connect the points to include all points.

If you want to graph a linear equation you have to have at least two points, but it's usually a good idea to use more than two points. When choosing your points try to include both positive and negative values as well as zero.

Graphing Linear Equations Using the Values of x and y A. Example: Graph y = x + 2 Steps in Graphing :

1. Begin by choosing a couple of values for x e.g. -2, -1, 0, 1 and 2 and calculate the corresponding y values.

2. Evaluate y = 5 – 3x for different values of x, and create a table of corresponding x and y values.

X Y = x + 2 Ordered pair

-2 -2 + 2 = 0 (-2, 0)-1 -1 + 2 = 1 (-1, 1)0 0 + 2 = 2 (0, 2)1 1 + 2 = 3 (1, 3)2 2 + 2 = 4 (2, 4)

3. Plot the resulting ordered pairs in the Cartesian plane.

4. Connect the points by drawing a line through them.

Graphing Linear Equations Using Intercepts Just remember:To find the X Intercept: Let y = 0To find the Y Intercept: Let x = 0 

Example:  Graph the equation y = 3x + 2 by using the intercepts method.

Steps:

1. Substitute y= 0 and solve for x.

2. Substitute x= 0 and solve for y.

3. Plot the two points and draw the graph.

GRAPHING FUNCTIONS USING SLOPE-INTERCEPT FORM

Slope of a line- measures of its steepness

Slope =

SLOPE-INTERCEPT FORM: 

y=mx+b

Slope y-intercept

HOW TO GRAPH:

1.) Plot the y-intercept on the y-axis. This is the point (0, b).

2.) Obtain the second point using the slope. Write m as fraction and use rise over run, starting at the point containing the y-intercept, to plot this point.

3.) Use a straightedge to draw a line through the two points. Draw the arrowheads at the ends of the line to show that the line continue indefinitely in both directions.

y=𝟑𝟓x-2 m=

b=-4

(0,-2)

Graph: 

move 3 units up (rise) move 5 units right (run)

GRAPHING f(x), f-1(x), y=x

Example:

1.)f(x)=2x-5 Given equation y=2x-5 Change f(x) to y m=2 b=-5 (0,-5) x=2y-5 Substitute y to x and vice versa

-2y=-x-5 Isolate y-term -12(2y=-x-5) Divide both sides by 13 y= −𝑥2 +52𝐹−1(x) (inverse function)

m=−12 b=52 (0,52 )

How to Graph:

1. Graph the f(x) by plotting the y-intercept on the y-axis and obtaining the second point using the slope.

2. Use straightedge to draw a line through the

two points. Draw the arrowheads at the ends of the line to show that the line continue s indefinitely in both directions.

3.) Graph the inverse function or the f-1(x).

4.) Use straightedge to draw a line through the two points. Draw the arrowheads at the ends of the line to show that the line continues indefinitely in both directions.

5.) From the origin, draw a line in which the two lines are intersect.

Graph: y=x

z 𝐹−1(x)=−𝑥2 -52 f(x)=2x-5

Submitted by:Allado, SwedeniaArtizona, AnamaeBolivar, Ma. Cris Annabelle Callocallo, April Mae

Gabilagon, Liezl Lerado, Aira GraceMatalubos, Lyra Sorenio, Shiela MaeTabuga, NatalieT0rda, Marjocel

Submitted to:Prof. Danilo Parreño

ADVANCE ALGEBRA