Glossary of Algebraic Symbols

Algebra is a branch of mathematics. Algebra plays an important role in our day to day life. The glossary of algebraic symbols involved in the four basic operations such as addition, subtraction, multiplication and division. The most important terms with the glossary of algebraic symbols are variables, constant, coefficients, exponents, terms and expressions. In Algebra, besides numerals we use symbols and alphabets in place of unknown numbers to make a statement. Hence, glossary of algebraic symbols may be regarded as an extension of Arithmetic.  

Glossary of algebraic symbols:

Most important terms for glossary of algebraic symbols:

Expressions

          An algebraic Expression is the combination of variables, constant, coefficients, exponents, terms which are combined by the following arithmetic operations Addition, subtraction, multiplication and division. The example of an algebraic expression is given below

                           2y + 5
Term

          Terms of the algebraic expression is concatenated to form the algebraic expression by the arithmetic operations such as addition, subtraction, multiplication and division. In the following example 3n² + 2n the terms 3n², 2n are combined to form the algebraic expression 3n² + 2n by the addition operation ( + )

Coefficient

          The coefficient of an algebraic expression is the value is present just before the terms. From the following example, 3n² + 2n the coefficient of 3n² is 3 and 2n is 2

Equations

          An algebraic equation equals the numbers or expressions. Most probably algebraic equation is used for the value of the variable. The example of the equation is given below

                          2y + 5

Examples for glossary of algebraic symbols:

Example 1:

8x - 3 = 2x

Solution:

8x - 3 = 2x

8x – 3 + 3 =2x + 3 (Add 3 on both sides)

8x =2x +3

8x – 2x =2x -2x + 3 (Add -2x on both sides)

6x = 3

6x /6 = 3 / 6  (both sides by divided 6)

X = 1/2

Example 2:

Solve the equation |-3x + 3| -8 = -6

Solution:

|-3x + 3| -8 = -6

|-3x + 3| -8 + 8 = -6 + 8 (Add 8 on both sides)

|-3x + 3| = 2

Case (i)

+ (-3x+3) = 2

-3x + 3 = 2 

Subtract 3 on both sides,

-3x + 3 -3 = 2-3

-3x=-1

x =1/3

Case (ii)

- (-3x+3) = 2

3x-3 =2

Add 3 on both sides

3x-3+3=2+3

3x=5

x=5/3

What is Absolute Value in Math

Absolute value in math is nothing but if we represent a number as absolute value the result will be positive. If the number is positive or negative the result of the absolute value is math is positive. For example |-2| = +2 and |+ 2| = +2. Here we are going to learn what absolute value in math is and operations using the absolute values in math. It is help us to understand what the absolute value in math is.

Examples for what is absolute value in math:

Example 1:
           What is the absolute value of |-5|?
Solution:
            We know in math absolute value of any value is its positive  value. Here we won’t consider the sign.
             So |-5| = +5

Example 2:
             What is the absolute value of |10|?
Solution:
            We know in math absolute value of any value is its positive value. Here we won’t consider the sign.
             So |10| = +10

Example 3:
             What is the absolute value of |-9|?
Solution:
             We know in math absolute value of any value is its positive value. Here we won’t consider the sign.
               So |-9| = +9

Operations using absolute value in math:

 Addition operation:
            Perform the following operation. |-3| + |-2|
Solution:
             Given numbers are |-3| + |-2|
              Absolute value of |-3| = +3
              Absolute value of |-2| = +2
              So |-3| + |-2| = +3 + 2 = +5

Subtraction operation:
               Perform the following operation. |-8| - |-2|
Solution:
               Given numbers are |-8| - |-2|
                Absolute value of |-8| = +8
                Absolute value of |-2| = +2
                So |-8| - |-2| = 8 – 2 = +6


Multiplication operation:
               Perform the following operation. |6| `xx` |-5|
Solution:
            Given numbers are |6| `xx` |-5|
            Absolute value of |6| = +6
            Absolute value of |-5| = +5
            So |6| `xx` |-5| = 6 `xx` 5 = 30

       These are some of the examples for absolute value in math. It is better to understand the absolute values.

Singular Matrices

The term determinant was first introduced by Gauss in 1801 while discussing quadratic forms. He used the term because the determinant determines the properties of the quadratic forms. We know that the area of a triangle by means of vertices (x1, y1) (x2, y2) and (x3, y3) is

1/2 [x1(y2-y3) + x2 (y3-y1) + (y1-y2) ]

Similarity the condition for a second degree equation in x and y to represent a pair of straight lines is

abc + 2fgh – af2 – bg2 –ch2 =0.

Definition of determinant:

To every square matrix A of order n with entries as real or complex numbers, we can associate a number called determinant of matrix A and it is denoted by | A | or det (A) or Δ.

Thus determinant formed by the elements of A is said to be the determinant of matrix A.

If A = `[[a_(11),a_(12)],[a_(21),a_(22)]]` then its | A | =

Minors:

Let | A | = |[aij]| be a determinant of order n. The minor of an arbitrary element aij is the determinant obtained by deleting the ithjth column in which the element aij stands. The minor of aij is denoted by Mij. row and

Cofactors:

The cofactor is a signed minor. The cofactor of aij is denoted by Aij and is defined as

Aij = (− 1)i + j Mij.

Singular and non-singular matrices:

A square matrix A is said to be singular if | A | = 0 (i.e) determinant zero

A square matrix A is said to be non-singular matrix, if | A | ≠ 0. (i.e) determinant is not zero


Examples of determinant zero:

Let us see some examples of determinant zero:

Example 1:

A = `[[1,2, 3],[4,5,6],[7,8,9]]`

Solution:



= 1(45 - 48) - 2(36 -42) +(32 -35)

= -3 +12-9 = 0

|A| = 0

So, A is singular  matrix.


Example 2:

Solve for x if

Solution:

`rArr` (x2 - 35) + (1-2) = 0

`rArr` x2 -35 -1 = 0

`rArr` x2 - 36 =0

`rArr` x2 = 36

`rArr` x = `+-` 6.

These are examples of determinant zero.

Isosceles Acute Triangle

 An Isosceles triangles has

• All angles add up to 180 degrees
• Has two equal sides.
• only one unequal side called the base.
• Base angles of isosceles are all equal.        

An acute triangle is a triangle which has all the angles less than 90°. We know all the internal angles in triangle is equal to 180°. In acute triangle the sum of all the internal angles is equal to 180° but not even the one angle is equal to or greater than 90°.  acute triangle drawn below  L, M and N are all acute angles.

    

Example Problems for Isosceles acute triangle:

Example 1:
   Find the area of the Isosceles triangle of b = 13 cm, h = 7 cm..
Solution:
   Given side of the Isosceles triangle, b = 13 cm, h = 7 cm.
        Area of the Isosceles triangle   = `1/2` * b * h
                                                      = `1/2 ` * 13 * 7
                                                      = 0.5 * 91
                                                      = 45.5 cm²
  Area of  Isosceles triangle   = 45.5 cm²
Example 2:
Find the area of the Isosceles triangle of b = 15 cm, h = 8 cm..
Solution:
Given side of the Isosceles triangle, b = 15 cm, h = 8 cm.
 Area of the Isosceles triangle   = `1/2` * b * h
                                                      = `1/2 ` * 15 * 8
                                                      = 0.5 * 120
                                                      = 60 cm²
  Area of  Isosceles triangle   = 60 cm²
Example 3:
Find the perimeter of Isosceles triangle that has side S₁ =16 cm,S₂ = 16,S₃ = 10 cm.
Solution:
 Given, S₁ =16 cm,S₂ = 16,S₃ = 10 cm.
  perimeter of Isosceles triangle   = S₁+S₂+S₃
                                                         = 16+16+10
  perimeter of Isosceles  triangle  =  42 cm

I am planning to write more post on cbse sample papers class 10 and cbse syllabus for class 11. Keep checking my blog.

Some other examples regarding Isosceles acute triangle:

Example 1:

A triangle has angle 43º, 64º and 73º. What type of triangle is this?

Solution:

  Here all the given angles are less than 90°, hence the given triangle angles are an acute triangle.

Example 2:

A triangle has angle 39º, 65º and 76º. What type of triangle is this?

Solution:

  Here all the given angles are less than 90°, hence the given triangle angles are an acute triangle.

Example 3:

A triangle has angle 45º, 56º and 79º. What type of triangle is this?

Solution:

  Here all the given angles are less than 90°, hence the given triangle angles are an acute triangle.

Associative Property of Addition

Associatively is the property of addition is nothing but in an expression containing two or more occurrences in a row of the addition operator, the order of operations that are performed will not be a matter it can be performed in any order of operation, and rearranging the brackets so that the values will not be changed. Here we are going to see about the associative property of addition

Associative property of Addition is

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

Examples for associative property of addition

Example problem 1 for associative property of addition:

Prove that (1 + 2) + 3 = 1+ (2 + 3) by using associative property of addition.

Solution:

Given (1 + 2) + 3 = 1+ (2 + 3) it is in the form (a + b) + c = a + (b + c)

Left hand side:   (1 + 2) + 3 = 3 + 3

= 6

Right hand side: 1 + (2 + 3) = 1 + 5

= 6

Left hand side is equal to right hand side. Therefore it is proved using associative property of addition.
Example problem 2 for associative property of addition:

Prove that (3 + 6) + 9 = 3+ (6 + 9) by using associative property of addition.

Solution:

Given (3 + 6) + 9 = 3+ (6 + 9) it is in the form (a + b) + c = a + (b + c)

Left hand side:   (3 + 6) + 9 = 9 + 9

= 18

Right hand side: 3 + (6 + 9) = 3 + 15

= 18

Left hand side is equal to right hand side. Therefore it is proved using associative property of addition.


Example problem 3 for associative property of addition:

Prove that (2 + 5) + 8 = 2+ (5 + 8) by using associative property of addition.

Solution:

Given (2 + 5) + 8 = 2+ (5 + 8) it is in the form (a + b) + c = a + (b + c)

Left hand side:   (2 + 5) + 8 = 7 + 8

= 15

Right hand side: 2 + (5 + 8) = 2 + 13

= 15

Left hand side is equal to right hand side. Therefore it is proved using associative property of addition.

Nonlinear Equation Algorithms

Nonlinear equation is the form of the multi variable equations and functions. In the equation there will be having more number of terms available. In the nonlinear equations the variables are not dependent to each other in the equation. Nonlinear equation has the more number of different orders of degrees. The graph of the nonlinear equation is not a straight line. It includes the quadratic function of equation and cubic function of equation. Here we are showing about the nonlinear equation algorithms and example problems in it .




Step by step algorithm for solving nonlinear equation:

Nonlinear equation algorithms:

Find whether the given equation is linear or nonlinear.

If the equation is equal to y = m x + b then the equation is linear equation
If the equation is not equal to y = mx + b then the given equation is nonlinear equation.

Find the equations of the order equal to any one of their variable and substitute the value in another equation.
Calculate the value for that equation and find two values.
Substitute the two values of the first equation in the second equation
Now we get each variable having two values.


Nonlinear equation algorithms - Example Problems:

Nonlinear equation algorithms - Problem 1:

Solve the nonlinear equations and find the value of x and y.

x2 - 8y = - 32

- x + y = 4

Solution:

Given equations

x2 - 8y = - 32   ------> Equation 1

- x + y = 4         ------> Equation 2

From the equation 2 rearrange and equal to y

y = 4 + x

Substitute the y value in the equation 1

x2 - 8(4 + x) = - 32

x2 - 32 -8x = -32

x2 - 8x -32+32 = 0

x2 -8x = 0

Solve the above equation by using factorization, we get

x(x - 8) = 0

Therefore the x value will be,

x=0            x - 8 = 0

x = 0 and x = 8

Substitute the x values in equation 2, we get

For x = 0,

y = 4 -x

y = 4 - 0

y = 4

For x = 8

y = 4 - 8

y = - 4

The value of y is 4, - 4


Nonlinear equation algorithms - Practice Problems:

1, Solve the nonlinear equations and find the value of x and y.

x2 - 7y = - 31

- x + y = 3

Answer:

x = 2 , 5 and y = 5, 8

2.

Find the nonlinear equations and find the value of x and y.

x = 2y - 3

y2 + 3x = - 2

Answer:

x = - 1, - 17 and y = 1, - 7

Learn Points and Lines

To learn about points and lines, A point used to represent a place in a plane with a help of pencil, a point is nothing but the dot , it has no dimension or no width, it’s only a simple black dot. In geometry co ordinates of a point which shows the particular place in a segment for representation.Line has two end points is called segment. Line segment is denoted with a connected piece of line.line segments names  has two endpoints and it is named by its endpoints.


learn about points and lines:

To learn about the geometric points and lines we have to know the classification of a points and lines.points and lines classification are as follows.
Collinear points:
When three or more points lies on the same line is said to be collinear points.

Midpoint:
A halfway point where line segment divides into two equal parts are called midpoint.

Equidistant point:
A point which is said to be equidistant in a line segment where point is equal length from other points which are in congruent then the point is equidistant point.

Parallel line segment:
Two lines which does not touch each other are called parallel lines.

Perpendicular line segment:
Two line segment  that form a L shape are called perpendicular lines.


learn problems in points and lines:

Example 1:
Find the distance between the points A(5,2) and B (7,3).


Solution:
Let assume "d" be the distance between A and B.           (x1,y1)= (5,2), (x2,y2)= (7,3).

Then d (A, B) =`sqrt((x2-x1)^2+(y2-y1)^2)`

= `sqrt((7-5)^2 +(3-2))^2)`

= `sqrt(2^2+1^2)`

= `sqrt(4+1)`

=`sqrt5`

Example 2:
Find co-ordinate of the mid point of the line segment joining given points A(-1,1) and B(3,4)

Solution:
The required mid point is
Formul a   `((x_1+x_2)/2 ,(y_1+y_2)/2)` here,  (x1, y1) = (-1,1),(x2, y2) = (3,4)

=  `((-1+3)/(2))``((1 +4)/(2)) `

= `(2/2) ` ,  ` (5/2)`

=`(1,5/2)`

Example 3:
Find the slope of the lines given (2,-1) and (1,3)

Solution:
(x1,y1)= (2,-1), (x2,y2)= (1,3).
We know to find slope of line,m=` (y_2-y_1) /(x_2-x_1)`

=`(3+1)/(1-2)`

m =`4/-1` = -4
Example 4:

Find the equation of the line having slope `1/2` and y-intercept −3.
Solution:
Applying the equation of the line is y = mx + c
Given,       m = `1/2` ,c = −3
y = `1/2` x + (−3)

or  2y = x − 6
or  x− 2y − 6 = 0.