Practice Algebra Notes

Introduction :

Algebra is the most important branch of the mathematics concerning the study of the rules of operations and relations, and the constructions and concepts arising from them, including terms, polynomials, equations and algebraic structures. Together with the geometry and analysis, topology, combinatorics, and number theory, algebra is one of the main branches of pure mathematics. And now let us see about the practice algebra notes (source in Wikipedia)


Practice Algebraic Notes:

Practice Notes for Variable:

A term or measure, which takes as the unusual numerical values, is called a variable. Generally we represent the variables and have some letters, says x, y, z, etc.

Finding the value of an Expression:

The method of replacing the variables by numerical values are called substitution. So the value of an expression depends on the value of its variables.

Practice algebra notes for Algebraic Equation:

An algebraic equation is the report given of equality, which contains unknown quantities or variables.

Practice algebra notes for Simple equation:

An equation, which has only one variable whose power is one, is called a simple equation

Example Algebraic Problem in Practice Algebra Notes:
Ex 1:

This problem illustrates that the correct use of the distributive property.

Solve 3× (x + 6 + 9) = 66.

Sol:

•    Grouping like terms, the left side of the equation becomes

3 × (x + 6 + 9) `=>` 3 × (x + 15)

•    Using the distributive property,

3 × (x + 15) `=>` 3 × x + 3 × 15

•    Carrying out multiplications,

3 × x + 3 × 15  `=>` to 3x + 45

•    The equation now becomes

3 x + 45 = 66.

•    Subtracting a 45 (adding a -45) to each side gives us

3x + 45 + (-45) = 66 + (-45)

3x + (45 + (-45)) = 66 - 45

3x + 0 = 21

3x = 21

•    Since the x is multiplied by 3, we divide both sides by 3 to solve for x:

3x = 21

3x ÷ 3 = 21 ÷ 3

(3x)/3 = 7

x = 7.

We get the answer value is X = 7

•    We can check this answer in solution of the original equation:

3 × (7 + 6 + 9) = 66

3 × 22 = 66

66 = 66 so our answer is correct.

Practice Algebra Notes Problem:

Solve 3× (X + 3 + 6) = 54.

Answer: 9

Solve 4× (X + 5 + 7) = 60.

Answer: 3

Practice Counting Numbers

Practice counting Numbers:

A number is a numerical thing, which is used for calculating and counting. It is also referred as numeral and involves zero, negative numbers, rational numbers, irrational numbers, etc,.

The procedure of numerical function involves one or more numerical as income and generate its relevant numerical outcome. This function includes arithmetic operation such as addition, subtraction, multiplication, division, and exponentiation.

The numbers are classified as follows,

Natural numbers
Integers
Rational numbers
Real numbers
Complex numbers
Computable numbers
Prime number

Examples for Practice Counting Numbers:
Counting number is a positive number with countable of one, two, three, and so on. Sometimes it contains the integer zero (0) in the table of that counting numbers. So, it is also referred to as Natural number. The sign of natural number is ‘N’. Some case it is called as skip count because of each count must be done by skipping.

General Rules:

Counting numbers include some general rules. Let take ‘B’ is counting number then it follows

Add ‘B’ to the earlier number
Skipping and counting each further number by ‘B’.
Note: Here ‘B’ may be a number or symbol or object

Practice Counting Problems:

1solve the practice problem) Counting downwards from number 13 to 2 with numbers:

Solution:

Starting from 13 to 2

N = ( 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2)

solve the practice problem2) How many heart symbols are there in following set (♥, ♥, ♥, ♥)

Solution:

There are 4 heart symbols in the set

solve the pratice problem 3) Counting the number from 18 to 42 by 6’s

Solution:

Thus, the count by 6’s may be adding number 6 to the earlier number (Rule1)

So Start from 18= 18+6, 18+6+6.

N = (18,24, 30, 36, 42)

4) How many square numbers are there in 100

Solution:

Square numbers in 100 = (1, 4, 9, 16, 25, 36, 49, 64, 81, 100)

Therefore, total square numbers are 10 in the given set.

5) How many club suits are there in following set (♣, ♣, ♣, ♣, ♣)

Solution:

There are 5 club suit in given set

6) How many six letter words are there in the following set (tennis, ball, four, pen, hundred, street, head, man)

Solution:

Six letter words are tennis, street. Therefore, there are 2 six letter words.

Inverse Operations Addition and Subtraction

Introduction :

The addition is the basic process in arithmetic’s. The addition   process is to add the two numbers and give the combination of that number. Summation the arithmetic process of summing calculating the sum of two or more numbers. A part that is added to something to improve it; the subtraction process is inverse of the addition A component that is subtracted to something to reduce it.

Explanation of Inverse Operations Addition and Subtraction

Addition:

The addition process is the main function in the arithmetic. The simple and effective process. Every function is based on the addition.

We will learn the addition process in simple way.

Here will explain the concepts.

Ex:

The two apple is added to four apple finally what will be the out put of the problem.

Sol:

Step1: First, add   two apples to four apples.

Step2: 2 + 4

Step3: the final out put 6. The output will changed for the reason combination of two quantities.

\Subtraction:

The subtraction is the inverse of the addition process. This process is reducing the total quantities.

We will learn the basic process of subtraction method.

Ex:

The six apples in the one bag. Ram already took two apples in the bag. What will be the remaining apples in the bag?

Sol:

The subtraction method is opposite of the addition process. This process reduce the total quantities,

Step1: the total quantities are 6 apples.

Step2: the ram took apples are 2

Step3: So, total apples – ram took apples= final quantities

Step4:6-2=4. Final apples are 4.

Practice Problems on Inverse Operations Addition and Subtraction

Ex:1 Ravi had 3 oranges and raja have 4 apples and 2 oranges. What are the total apples and oranges both have solve this problem?

Sol: 5 oranges and 4 apples

Ex:2 Ravi had 4 pens and 3 books, raja have 4 pencil and 2books. What are the total books, pens and pencil both have solved this problem?

Sol :5 books,4 pens,4 pencils

Factoring Fractional Exponents

Introduction to factoring:

In mathematics, a rational function is any function which can be written as the ratio of two polynomial functions. Factorization (also factorization in British English) or factoring is the decomposition of an object (for example, a number, a polynomial, or a matrix) into a product of other objects, or factors, which when multiplied together give the original.

Example Problems for Factoring Fractional Exponents

Factoring fractional exponents example problem 1:

Factoring the given fractional expression `(1 / 3)` x^2 - `(11 / 3)` x - 144 = 0

Solution:

Given fractional expression is `(1 / 3)` x^2 - `(11 / 3)` x - 144 = 0

Multiply the given expression by 3 on both the sides, we get

x^2 - 11x - 432 = 0

Factorize the above equation, we get

x^2 - 27x + 16x - 432 = 0

Grouping the first two terms and second terms, we get

(x^2 - 27x) + (16x - 432) = 0

Take common terms, we get

x (x - 27) + 16 (x - 27) = 0

(x - 27) (x + 16) = 0

The factors of the given fractional expression is (x - 27) and (x + 16)

Answer:

The final answer is (x - 27) and (x + 16)

Factoring fractional exponents example problem 2:

Factoring the given fractional expression `(1 / 5)` x^2 - `(27 / 5)` x - 74 = 0

Solution:

Given fractional expression is `(1 / 5)` x^2 - `(27 / 5)`x - 74 = 0

Multiply the given expression by 5 on both the sides, we get

x^2 - 27x - 370 = 0

Factorize the above equation, we get

x^2 - 37x + 10x - 370 = 0

Grouping the first two terms and second terms, we get

(x^2 - 37x) + (10x - 370) = 0

Take common terms, we get

x (x - 37) + 10 (x - 37) = 0

(x - 37) (x + 10) = 0

The factors of the given fractional expression is (x - 37) and (x + 10)

Answer:

The final answer is (x - 37) and (x + 10)

Factoring fractional exponents example problem 3:

Factoring the given fractional expression `(1 / 7)`x^2 - `(12 / 7)`x + 5 = 0

Solution:

Given fractional expression is `(1 / 7)`x^2 - `(12 / 7)`x + 5 = 0

Multiply the given expression by 7 on both the sides, we get

x^2 - 12x + 35 = 0

Factorize the above equation, we get

x^2 - 7x - 5x + 35 = 0

Grouping the first two terms and second terms, we get

(x^2 - 7x) - (5x - 35) = 0

Take common terms, we get

x (x - 7) - 5 (x - 7) = 0

(x - 7) (x - 5) = 0

The factors of the given fractional expression is (x - 7) and (x - 5)

Answer:

The final answer is (x - 7) and (x - 5)

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Practice Problems for Factoring Fractional Exponents

Factoring fractional exponents practice problem 1:

Factoring the given fractional expression `(1 / 24)`x^2 + `(23 / 24)`x - 12 = 0

Answer:

The final answer is (x + 32) and (x - 9)

Factoring fractional exponents practice problem 2:

Factoring the given fractional expression `(10 / 7)`x^2 - `(59 / 7)`x + 7 = 0

Answer:

The final answer is (10x - 49) and (10x - 10)

Factoring Polynomials by Grouping

Introduction of factoring polynomials by grouping:

• In general, the sum of a countable number of monomials is referred as a polynomials. The way of writing a polynomials as a product of two or more simpler polynomials is calledfactorization. The process of factorization is also known as the resolution into factors or factoring polynomials. Each simpler polynomial in the product is called a factor of the given polynomial.   
• Factoring polynomial by grouping method is used for the expressions containing more than three terms.

Factoring Polynomials Using Grouping:

Factoring polynomials by grouping involving four or more than four terms can be grouped using the following steps:

Step 1:  The terms having common factors should be grouped.

Step 2: The Greatest Common Factor (GCF) is taken out.

Step 3:  In the third step, we have to use the distributive law to find the factors.

                                               The distributive law is given by

                                                         a (b + c) = a b + a c

Example Problems on Factoring Polynomials by Grouping:

Ex 1:  Factoring the given polynomial, 3x² + 9x³ + 7x⁷ + 21x⁸, by grouping:     

Sol :   Step 1:  The terms having common factors should be grouped.

                                     (3x² + 9x³⁾ + (7x⁷ + 21x⁸).

Step 2: The Greatest Common Factor (GCF) is taken out.

                            3x² + 9x³ and 7x⁷ + 21x⁸ both can have a GCF.

               Taking out the GCF outside, we get

                         (3x² + 9x³⁾+ (7x⁷ + 21x⁸) = 3x² (1 + 3x) + 7x⁷ (1 + 3x)

Step 3: Use  distributive law to find the factors.

                Note that there is a common factor, 1 + 3x. So, factor out 1+3x.

                  By distributive law, we get

                                3x² + 9x³ + 7x⁷ + 21x⁸ = (3x² + 7x⁷) (1 + 3x).

Ex  2:   Factoring the given polynomial by grouping:  2x³+6x²−6x−18

Sol :   Step 1: The terms having common factors should be grouped.

                                               (2x³+6x²) + (−6x−18)

Step 2: The Greatest Common Factor (GCF) is taken out.

                                  (2x³+6x²) and  (−6x−18) both have a GCF

               Taking out the GCF outside, we get

                                   (2x³+6x²) + (−6x−18) = 2x²(x+3)-6(x+3)

Step 3:  Use distributive law to find the factors.

              Note that there is a common factor, x+3. So, Factor out (x+3).

               By distributive law, we get

                                             2x³+6x²−6x−18=(x +3) (2x²-6).

Ex 3 :  Factoring the given expression by grouping: 2x² - 3x + 10x - 15.

Sol :  Step 1:  The terms having common factors should be grouped.

                                       (2x² + 10x) + (- 3x – 15)

Step 2: The Greatest Common Factor (GCF) is taken out.

                            2x²+10x) and  (−3x−15) both have a GCF

              Taking out the GCF outside, we get

                              (2x²+10x) + (−3x−15) = 2x(x+5)-3(x+5)

Step 3:  Use distributive law to find the factors.

              Note that there is a common factor, x+5. So, Factor out (x+5).

               By distributive law,we get

                                             2x² - 3x + 10x – 15=(x+5) (2x-3).

These are the examples for the factoring polynomials by grouping.

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Complex Matrix Inverse

Introduction

In algebra, the determinant is a particular number related with some square matrix. The essential geometric significance of a determinant is a scale factor for compute when the matrix is considered as a linear transformation. Thus a 3× 3 matrix with determinant 3 when useful to a place of points with fixed area will convert those points into a place with double the area. Determinants are significant together in calculus, where they go into the replacement rule for various variables, and in multi linear algebra.

`A = [[A,B],[C,D]] If AD - BD != 0` , then A has an inverse, Denoted` A^-1`

`A^-1 = 1/(AD-BC) [[D,-B],[-C,A]]`

Complex Matrix Inverse - Examples:

Complex matrix inverse - Example 1:

Find the Complex inverse matrix of the matrix below?

`A = [[4,2],[3, 8]]`

Solution:

Inverse A =` 1/((8*4)-(3*2)) [[8,-3],[-2,4]]`

= `1/(32-6) [[8,-3],[-2,4]]`

=` 1/26[[8,-3],[-2,4]]`

= `1/26 * [[8,-3],[-2,4]] = [[8/26, -3/26],[-2/26,4/26]]`

= `[[4/13, -3/26], [-1/13, 2/13]]`

Complex matrix inverse - Example 2:

What is the inverse of the matrix below?

`[[9, 1], [6, 3]]`

Solution:

Inverse of the matrix is the

Inverse A =` 1/((9*3)-(6*1)) [[3,-1],[-6,9]]`

= `1/(27-6) [[3,-1],[-6,9]]`

= `1/21[[3,-1],[-6,9]]`

= `1/21 * [[3,-1],[-6,9]] = [[3/21, -1/21],[-6/21,9/21]]`

` [[3/21, -1/21],[-6/21,9/21]] = [[1/7, -1/21],[-2/7,3/7]]`

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Complex matrix inverse - Example 3:

What is the inverse of the matrix below?

A =  ` [[7, 1], [5, 3]]`

Solution:

Inverse of the matrix is the

Inverse A =` 1/((7*3)-(5*1)) [[3,-1],[-5,7]]`

= `1/(21-5) [[3,-1],[-5,7]]`

= `1/16[[3,-1],[-5,7]]`

= `1/16 * [[3,-1],[-5,7]]`

= `[[3/16, -1/16],[-6/16,9/16]]`

`[[3/16, -1/16],[-6/16,9/16]] = [[3/16, -1/16],[-3/8,9/16]]`

Complex Matrix Inverse - more Examples:

Complex matrix inverse - Example 1:

What is the inverse of the matrix below?

` [[8, 2], [6, 4]]`

Solution:

Inverse of the matrix is the

Inverse A = `1/((8*4)-(6*2)) [[4,-2],[-6,8]]`

`= 1/(32-12) [[4,-2],[-6,8]]`

`= 1/20[[4,-2],[-6,8]]`

` = 1/20* [[4,-2],[-6,8]]`

`= [[4/20, -2/20],[-6/20,8/20]]`

`[[4/20, -2/20],[-6/20,8/20]]= [[1/5, -1/10],[-3/10,2/5]]`

Complex matrix inverse - Example 2:

Find the inverse Complex matrix of the matrix below?

`[[9, 3], [7, 5]]`

Solution:

Inverse of the matrix is the

Inverse `A = 1/(9*5)-(7*3) [[5,-3],[-7,9]]`

`= 1/(45-21) [[5,-3],[-7,9]]`

`= 1/24[[5,-3],[-7,9]]`

`= 1/24* [[5,-3],[-7,9]]`

`= [[5/24, -3/24],[-7/24,9/24]]`

`[[5/24, -3/24],[-7/24,9/24]] = [[5/24, -1/8],[-7/24,9/24]]`

Complex matrix inverse - Example 3:

What is the inverse of the matrix below?

`[[10, 4], [8, 6]]`

Solution:

Inverse of the matrix is the

Inverse `A = 1/(10*6)-(8*4) [[6,-4],[-8,10]]`

`= 1/(60-32) [[6,-4],[-8,10]]`

`= 1/28 [[6,-4],[-8,10]]`

`= 1/28* [[6,-4],[-8,10]]`

`= [[6/28, -4/28],[-8/28,10/28]]`

`[[6/28, -4/28],[-8/28,10/28]] = [[3/14, -1/7],[-2/7,5/14]]`

Factoring Quadratic Trinomials

Introduction

In elementary algebra, a trinomial is a polynomial consisting of three terms or monomials. In mathematics, factorization (also factorisation in British English) or factoring is the decomposition of an object (for example, a number, a polynomial, or a matrix) into a product of other objects, or factors, which when multiplied together give the original. In this article we shall discuss about factoring quadratic trinomial.

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Example Problem on Factoring Trinomials

Problem 1:

Factoring trinomials

15 – 2x – x2.

Solution: Writing in the standard form,

15 – 2x – x2 = –x2 – 2x + 15

= (–1) (x2 + 2x – 15).

Here, we find –15 = 5 × –3, 5 + (–3) = 2

Hence, we get 15 – 2x – x2 = (–1) [(x+5) {x + (–3)}]

= (–1) (x +5)(x – 3)

= (x + 5) ( 3 – x).

Problem 2:

Factoring  trinomials

x2 – x – 132.

Solution: We find –132 = (–12) × (11), (–12) + 11 = –1.

Hence we get x2 – x – 132 = [x + (–12)] (x + 11) = (x – 12) (x + 11).

Next, we consider the quadratic polynomial ax2 + bx + c where a, b, c are integers and a ≠0,1. If we are able to find two integers p and q such that pq = ac and p + q = b. Then

ax2 + bx + c =`1/a ` (a2x2 + abx + ac)

=`1/a` [a2x2 + a(p+q)x + pq]= `1/a` [a2x2 + apx + aqx + pq]=`1/a` [ax (ax + p) + q(ax + p)]

=`1/a` (ax + p) (ax + q)

Thus, we are able to factorize the expression

Problem 3:

Factoring trinomial

2x2+ 7x + 3.

Solution: Here a = coefficient of x2 = 2

b = coefficient of x = 7

c = constant term = 3

We find a × c = 2 × 3 = 6 = 6 × 1, 6 + 1 = 7 = b. Hence

2x2 + 7x + 3 = 21 (2x + 6) (2x + 1) =(x+3)(2x+1).

Instead of applying the final result of the rule, we can also do the factorization by splitting the middle term and grouping as follows:

2x2 + 7x + 3 = 2x2 + (6 + 1)x + 3

= 2x2 + 6x + x + 3

= 2x(x + 3) + (1)(x+3) = (2x+1) (x+3).

Practice Problem on Factoring Quadratic Trinomials

Problem 1:

Factorize 8a2 + 2a – 3.

Answer:

(4a + 3) (2a – 1)

Problem 2:

Factorize 6 + 11/2 x + x2.

Answer:

1/2 (x + 4) (2x + 3).