Solve Inconsistent System of Equations

Introduction:

Algebra is a section in math, which comprises of infinite number of operations on equations, polynomials, radicals, rational numbers, logarithms, etc. Solving a set of linear equation is also an element of algebra. While solving, if there is no solution it is called as inconsistent set of equations else they are said to be consistent.

Examples to Solve Inconsistent System of Equations:
1. Find whether the following set of equation is consistent or inconsistent?

2x+3y = 1

4x+6y =2

Solution:

Elimination method:

In this method we try to eliminate one variable either ‘x’ or ‘y’ and then find out the remaining variable’s value.

2x+3y = 1-----------> (1)

4x+6y =2 -----------> (2)

Multiply equation (1) with ‘2’and subtract it from (2),

Therefore, 2(2x+3y – 1) =0

4x+6y -2 =0

4x+6y-1 – (4x+6y -2) =0’

4x +6y-1-4x-6y+2 =0,

4x-4x =0 similarly 6y-6y =0,

Therefore, -1+2=0

Since both the ‘x’ and ‘y’ terms have vanished, it is not possible to get a solution.

Hence, this set of equations is inconsistent.

2. Find whether the following set of equation is consistent or inconsistent?

x+4y = 1

3x+12y =2

Solution:

Substitution method:

x+3y = 1-----------> (1)

3x+9y =2 -----------> (2)

Solve the first equation for ‘x’

Subtract by ‘3y’ on both sides,

Therefore x+3y -3y =1-3y,

Since 3y-3y =0, x =1-3y,

Substitute x =1-3y in (2)

3(1-3y) + (9y) =2,

3-9y+9y =2,

3=2.

Since both the ‘x’ and ‘y’ terms have vanished, it is not possible to get a solution.

Hence, this set of equations is inconsistent.

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Solve Practice Problems: Inconsistant System of Equations

1. Check whether the system of equations are inconsistent?
x +y=1
x +y=2
2. Solve and comment on the system solution,
5x +3y =1
15x+9y =4
3. Check whether the set of equations are inconsistent?
3x+2y=5
4x+6y=3
4. Solve and comment on the system solution,
2x+2y=5
4x+6y=6
5. Solve and comment on the solution,
7x+2y=5
5x+6y=7
6. Solve and comment on the system  solution?
8x+6y=5
8x+6y=3
7. Solve and comment on the system solution?
x -y=1
x -y=5

Correct Scientific Notation

Introduction to correct scientific notation: 

Generally, we approach across extremely big as well as extremely little numbers. Writing this number such as extremely difficult. The use of correct scientific notation helps us to state them in a suitable form. Exponential notation is also identified as scientific conversion which is nothing but writing the numbers in the shape of m x 10 n.

Correct Scientific Notation Terms:

The correct scientific notation help terms are given below that,

Coefficient

Exponent

In the term m x 10 n, m is the co-efficient and n is the exponent.  The co-efficient term m can be of some real number and the exponent n have to be an integer. The coefficient expression is as well recognized as mantissa.

Example: 

3. 4 x 108

3.4 = coefficient or mantissa

8 = exponent.

Steps to find correct scientific notation:

Steps to find correct scientific notation is given below that:

The correct scientific notations have one number to the left of the decimal point.

We surround to set the decimal point subsequent the first digit.

How many places the decimal position was stimulated point out through the power of ten (10).

A fact in correct Scientific notation is writing as an ordinary form:

X×10y  (where X is the coefficient that is in among 1 and 10. And ‘y’ is the exponents)

Coefficient ×10exponents

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Examples for Correct Scientific Notation:

Example 1:

4876 change into correct scientific notation.

Solution:

Specified that 6581

Specified integer is not a big number, other than it is easy to convert scientific conversion.

Initial note down 6.581this is not equivalent to specified integer

(6.581)(1000) = 6581

At the present 1000 = 10 3          

Following that it can be note down in scientific conversion = 6.581 x 10 3

Example 2:

654123897 change into correct scientific notation.

Solution:

Specified that 654123897

Specified integer is a large number, other than it is easy to convert scientific conversion.

Initial note down 6.54123897 this is not equivalent to specified integer

Estimate how many digits available subsequent the decimal point in previous step

There are 8 digits subsequent the decimal point so you be able to place 8 zero

6.54123897 (100000000)

So the conversion determination be 10 8

The scientific conversion can be note down as 6.54 x 10 8

Example 3:

0.00753 change into correct scientific notation.

Solution:

Specified that 0.00753

Specified integer is not a large number, other than it is simple to convert scientific conversion.

Initial note down the integer as 7.53

Present are 3 digit subsequent the number is there so we can write 10 -3

The scientific conversion can be note down as 7.53 x 10 -3

Practice problems for correct scientific notation:

Problem 1:

91000000 change into correct scientific notation.

Answer: 9.1 x 107

Problem 2:

591000 change into correct scientific notation.

Answer: 5.91 x 105

Problem 3:

35759182 change into correct scientific notation. 

Answer: 3.57 x 107

Sector of a Circle

Introduction

The two radii of the central angle θ and the formation of the arc in between the points on the circle which forms the sector of a circle .The sector of a circle is nothing but the calculation of a part of the whole area of the circle. The angles in the circle which can be measured in degrees or radians.

Let us take the angle θ of the circle to be the central angle and the total area circle is given as πr2.In this the 2π is the angle for the whole part of the circle.

Formula for the Sector of Circle:

Formula:  The area of a sector of circle:

Area = `(theta pi R^2)/360`

Where, C is the central angle in degrees,

R is the radius of the circle

π is Pi, approximately 3.142

The formula is obtained by taking the area of a whole circle, and then a fraction of it which depends on the central angle is the area of sector. In the sector of the circle, if the central angle is 90°, then the sector of the circle will have some portion which is equal to one quarter of the whole circle..

When the central angle is in radians, then the formula for area of the sector can be given as,

Area=R2C/2

Where, C is the central angle in radians and

R is the radius of the circle of which the sector is part.          

Example for Sector of Circle:

Ex 1: The sector of a circle consists of radius 8 cm and angle 40°.Find the area of the sector of a circle and also the major part of the circle.

Sol:

Area of sector = `theta/360` × πr2

= `40/360`  × 3.14 × 64

= 22.34 cm2

Thus, the area of the major circle = area of circle – area of minor circle.

= 50.2 – 22.34

= 27.86 cm2

Ex 2: If the radius of the circle is equal to the value of 12 inches and the central angle is 1500.  Solve the area of the sector of a circle.

Sol:

The given values are r = 12 inches and `theta` = 1500

Area:

Area of a sector is solved through the following formula,

A=` (theta)/(360^0)` x π r2

That is A= `(150^0) / (360^0) ` x 3.14 x 12 x 12

Simplifying this we have to get

A= 10800 square inches.

This is the area of the arc of a sector.

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Ex 3: If the radius of the circle is  5 cm and the central angle is  1470.  Solve the area of the sector of a circle.

Sol:

The given values are r = 5 cm and theta = 1470

Area:

Area of a sector is solved through the following formula,

A= `(theta)/(360^0)` x π r2

That is A= `(147^0) / (360^0)` x 3.14 x 5 x 5

Simplifying this we have to get

A= 1837.5 square cm.

This is the area of the arc of a sector.

Minimum Deviation of a Prism

Introduction to minimum deviation of a prism:
A prism is a transparent medium bounded by two plane surfaces inclined to each other at a suitable angle. The angle between two surfaces is known as refracting angle or angle of prism. In a prism, a ray of light suffers two refraction and the result is deviation. In other words, we say that after passing through prism the ray of light deviates through a certain angle from its original path.

Minimum Deviation of a Prism:

Look at the given diagram, here ABC is the principal section of the prism and the angle of prism is A.

A ray of light KL is incident on face AB of the prism at Ði1. It bends towards the normal NO and refracted along LM at Ðr1. The refracted ray LM is incident at Ðr2 on face AC of the prism. It bends away from the normal PO and emerges MS at angle Ði2. In passing through the prism, the ray KL suffers two refraction and finally turned through an ÐQTM = d, this is the angle of deviation.


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Calculation for the Minimum Deviation of a Prism:
In  TLM, `delta` = TLM + ÐTML

`delta` = (i1 – r1) + (i2 – r2)

`delta` = (i1 + i2) - (r1 + r2)                       ….(1)

In D OLM, ÐO + r1 + r2 = 180°           ….(2)

In quadrilateral ALOM,

ÐL + ÐM = 180°

ÐA + ÐO = 180°

Put this value in equation (2), we get

ÐO + r1 + r2 = ÐA + ÐO

r1 + r2 = ÐA                                                     ….(3)

Put this value in equation (1)

`delta` = (i1 + i2) – A                                                ….(4)

Let n be the refractive index of the medium of prism with respect to air

n = Sin i1 / Sin r1 = i1 /r1                        (when angles are small)

i1 = n r1 , Similarly i2 = n r2

Put this value in equation (4), we get

d = n(r1 + r2) – A

d = n A – A

`delta` = (n – 1)A

In case of minimum deviation of a prism, i1 = i2 = i so that r1 = r2 = r, so put these values in equation (3) and in equation (4), we get

r + r = A

r = A/2

`delta`m = (i + i) – A = 2i – A   where dm is the minimum deviation angle

i = ( `delta`m + A)/2

n =  

Plotting Ordered Pairs

Introduction to plotting ordered pairs:

Plotting ordered pairs mean we have the x and y value in a graph.  It is denoted by the open and close parenthesis. Let us take an example for plotting the ordered pairs (3, 4). In this 3 is the x axis value and 4 is the y axis value. X axis value it represent the x component length and the y value mean it represent the y component length. If we placed an ordered pair in a grid it is called coordinates.

Method to Plotting Ordered Pairs:

We will plot the ordered pairs on the graph based on the origin. In a graph the origin is also a ordered pair which is (0, 0). Here the first 0 indicates the x coordinate and the second 0 indicates the y coordinate. Always the x coordinate is indicated first. So if we want to place any coordinates we have to horizontally and vertically from the origin. If we want to locate the positive x value we have to place it left to the origin. If we want to place the negative x value we have to move right side of the origin. Likewise if we want to place the positive y value we have to move upward and for negative y value we have to move downwards of the origin. Let us see some examples for plotting ordered pairs on a graph.

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Examples for Plotting Ordered Pairs:

Ex:1 Plot the given ordered pairs (2, 2), (2, 3), (5, 6)

Sol:

Let us take the first ordered pair. To plot the ordered pairs we have to move 2 units horizontally and 2 units vertically.

Here the x component value is 2 and the y component value is 2.

Likewise we have to plot the second and the third ordered pairs. For the second ordered pair we have to move 2 units horizontally and 3 units vertically. For the third ordered pair we have to move 5 units horizontally and 6 units vertically.

Solving discrete variables

Introduction:

      In our article let us discuss the solving discrete variables. Solving discrete variables is the important one in math and it explains how to solve discrete variable. Discrete variables are mainly  based on two types of variables and we will study about solving discrete variables with its example.

Example for discrete variable: In an event of the nearest millisecond to the time.

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Solving Discrete Variable:

 Description about solving discrete variable:

Solving discrete variable:

    Solving discrete variable is the important one in mathematics. The discrete variable only takes the values under in its boundary itself. The limits of discrete value are completed. The best example for discrete variable is 'Real world quantities approximation'.

Example for solving discrete variable:

Solution:

            If the student flips the coin, he counts how much possible number of heads as outcomes in an  event. The possible number of head values is present between 0 and other values such as x=1, 2, 3 like that.

             The above examples are used to solve the discrete variable with its discrete probability function.

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Solving Discrete Variable Probability Distribution:-

DISCRETE PROBABILITY DISTRIBUTION:-

Definition:

If the variable is said to be discrete variable then its probability distribution of the variable is known as discrete probability distribution.

The following example clearly explains the concept of discretr random variable with its probability function.

For example:

solution:

If a coin flips in twice, we get  4 possible values of outcomes such as HH, HT, TH, and TT.

The random variable X represents number of heads in an event.



                                  No. of heads                                     Probability

                                        0                                                        `1/4`

                                        1                                                        `1/2`

                                        2                                                        `1/2`

Another example:

An event of flipping a coin three times then find the probability of possible number of tails as outcomes and x=1,2,3 ?

soluton:

If you flip the coin 3 times, we shall get possible values of 8 outcomes such as HHH, HHT, HTH, HTT, THH, THT, TTH and TTT

Here the possible number of values for the above event is showed below.

                                  Number of tails                                  Probability

                                             0                                              `1/8`

                                             1                                               `7/8`

                                             2                                               `5/8`

                                             3                                              `1/8`

so the above examples clears the concept of solving discrete variable.

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Introduction to adding and subtracting of matrices

Introduction to adding and subtracting of matrices:
In this article let us learn what is a matrix and how to add and subtract matrices.

What is a matrix:

Consider a set of entries or elements. When they are arranged in rows and columns enclosing them with a square bracket [ ] or parenthesis ( ) is called as matrix.

We can denote matrix using capital letters like A, B, C...

The set of entries or elements in the matrix could be real or complex numbers, algebraic expressions.

In matrix, Counting the horizontal segment from top to bottom is termed as number of rows. 

                Counting the vertical segment from left to right is termed as number of columns.

Example:

R1,R2,R3 represent rows

C1,C2,C3 represent columns.

Rules to be Followed while Adding and Subtracting Matrices:

 When the matrices have same number of rows and columns, then we can add them.
       For example:

        Any two matrices of 5 rows and 3 columns could be added.

        When a  matrix with 2 rows and 3 columns and a matrix with 3 rows and 4 columns could not be added.

 Add the corresponding rows and columns of the given two matrix and the resulatat matrix is the added matrix.
      Example:

      We could add or subtract the element in the third row and second column of the first matrix  with the element in the third row and second column of the second matrix and not with any other entries. that is , corresponding entries are added or subtracted. For example  the element in the first row first column of the first matrix can be  added to the element in the first row first column in the second matrix.

Subtraction is nothing but  negative addition.
Additive inverse of matrix A is - A
             A + ( -A ) = ( -A ) + A =  zero matrix

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Problems on Adding and Subtracting Matrices:

Let us  add and subtract the given matrices

A = `[[2,5,7],[0,8,1]]` : B = `[[6,3,6],[0,3,5]]`

A+B = `[[2+6,5+3,7+6],[0+0,8+3,1+5]]`

         = `[[8,8,13],[0,11,6]]`

A - B = `[[2-6,5-3,7-6],[0-0,8-3,1-5]]`

         = `[[-4,2,1],[0,5,-4]]`