Equations with Variables on both Sides Calculator

Introduction for equations with variables on both sides calculator:

Equation with variable are the important part of algebra, generally the equation with same variable are said to be equation with  variables on both sides. Fro example x+2x+4 = x+8 here x is the variable in this equation we need to find the value of variable using calculator. In this article we will discuss about equation with variables on both sides calculator with suitable example problem.

Problem on Equations with Variables on both Sides Calculator:

Simplify the following equations with variables on both sides calculator

Problem(i) : The equation  is 2x+4 = 4x+6+8.

Solution

The given equation  is 2x+4 = 4x+6+8.

Step 1: In this given equation collect the variable

2x-4x+4 = 6+8

Step2: Collecting the constant

2x-4x+4 = 14

2x-4x = 14-4.

Step 3: simplifying the equation

-2x = 10

Step 4: Dividing both sides by -2

`(-2x)/-2` =`10/-2`

x= -5

Step 5: The variable of x is -5.

Problem(ii) : The equation  is 3x= 4x+6+1.

Solution

The given equation  is 3x = 4x+6+1.

Step 1: In this given equation collect the variable

3x-4x = 6+1

Step 2: Collecting the constant

3x-4x = 7

3x-4x = 7.

Step 3: simplifying the equation

-x = 7

Step 4: The variable of x is -7.

Problem on Equations with Variables on both Sides Calculator:

Simplify the following equations with variables on both sides calculator

Problem (i) The equation is -5x+7+8 = -9x +15+10

Solution

The given equation  is -5x+7+8 = -9x +15+10

Step 1: In this given equation collect the variable

-5x+9x+15 = 15+10

Step2: Collecting the constant

-5x+9x+15 = 25

-5x+9x = 25-15.

Step 3: simplifying the equation

4x = 10

Step 4: Dividing both sides by 4

`(4x)/4` = `10/4`

x = `5/2`

Step 5: The variable of x is `5/2` .

Problem (ii) The equation is -2x+7+13 = 4x +14

Solution

The given equation  is -2x+7+13 = 4x +14

Step 1: In this given equation collect the variable

-2x-4x+7+13 = 14

Step2: Collecting the constant

-2x-4x+20 = 14

-6x = 14-20

Step 3: simplifying the equation

-6x = -6

Step 4: Dividing both sides by -6

`(-6x)/-6` =`(-6)/(-6)`

x = 1

Step 5: The variable of x is 1.

Simple Regression Formula

Introduction to simple regression formula:

In mathematics, one of the most important topics in statistics is regression. Regression is determining the relationship between two variables. Regression math are used to analysis the several variables. Regression is one of the statistical analysis methods that can be used to assessing the association between the two different variables.

Example of Simple Regression Formula:
Here we study about the simple regression formula are,

Formula for regression analysis:

Regression Equation (y) = a + bx

Slope `(b) = (NsumXY-(sumX)(sumY))/(NsumX^2-(NsumX)^2)`

Intercept`(a) = (sumY-b(sumX))/N`


Where,
x and y are the variables.
b = the slope of the regression line is also defined as regression coefficient
a = intercept point of the regression line where is in the y-axis.
N = Number of values or elements
X = First Score
Y = Second Score
`(sumXY)` = Sum of the product of the first scores and Second Scores
`(sumX)` = Sum of First Scores
`(sumY)` = Sum of Second Scores
`(sumX^2)` = Sum of square First Scores.

Example Problem for Simple Regression Formula:

Problem for simple regression formula:

Example 1:

Find the regression slope coefficient, intercept value and create a regression equation by using the given table.

X Values   Y Values

10            11

20            22

30            33

40            44

50            55


For the given data set of data, solve the regression slope and intercept values.

Solution:

Let us count the number of values.
N = 5
Determine the values for XY, X2

X Value  Y Value    X*Y      X*X

10        11        110      100

20        22        440      400

30        33        990      900

40        44       1760     1600

50        55       2750     2500


Determine the following values `(sumX), (sumY), (sumXY), (sumX^2).`
`(sumX) = 150`
`(sumY)= 165`
`(sumXY)= 6050`
`(sumX^2) = 5500`



Plug values in the slope formula,


Slope `(b) = (NsumXY-(sumX)(sumY))/(NsumX^2-(NsumX)^2)`


`= (5 xx(6050)-(150)xx(165))/((5)xx(5500)-(150)^2)`


`= (30250 - 24750)/(27500-22500)`


`= 5500/5000`

`b= 1.1`

Plug the values in the intercept formula,


Intercept `(a) = (sumY- b(sumX))/N`


`= (165-(1.1xx150))/5`


`= (165 - 165)/5`


`= 0/5`


`a = 0`

Plug the Regression coefficient values and intercept values in the regression equation,
Regression Equation(y) = a + bx
= 0 + 1.1x

Answer:

Slope (or) Regression coefficient (b) = 1.1

Intercept (y) = 0

Regression equation y = 0 + 1.1x

Number of Factors of an Integer

Introduction to number of factors of an integer:

In math, the natural numbers are form the integer and another name of an integer is whole number. The factor is divisor of a given number. This divisor is divides the given integer without any remainder. The factors may be two or more in an integer. Now we are going to see about number of factors of an integer.

Explanation for Number of Factors of an Integer

Some notes about integer and factors in math:

An integer may be positive and negative. We can’t calculate the number of factors of decimal or fraction.

The integer has one or more divisors. We can classify the factors into two. The names of factors are prime and composite. The prime factor is defining the prime number that is prime number has only two factors. The composite factor is defining the composite number that is the composite number has more than two factors.

We can count the number of factors of integer by listing method. We can separate the normal factors and prime factors.

More about Number of Factors of an Integer

Example problems for number of an integer in math:

Problem 1: Count how much number of factors and prime factors are present in given integer.

94

Answer:

Given integer are 94.

The given integer 94 has 4 factors as 1, 2, 47, 94 and 2 prime factors are present as 2 x 47.

Problem 2: Count how many number of factors present in given integer?

112

Answer:

Given integer are 112.

The given integer has 10 factors. They are 1, 2, 4, 7, 8, 14, 16, 28, 56 and 112.

It has 5 prime factors. They are 2 x 2 x 2 x 2 x 7.

Exercise problems for number of factors of an integer:

1. How many numbers of factors in integer 75?

Answer: The integer 75 has 6 factors as 1, 3, 5, 15, 25, and 75.

2. How many numbers of factors in integer 19?

Answer: The integer 19 has 2 factors.

The Various Types of Sets in Set Theory

One of the important theories in modern mathematics is the set theory. This theory has been present now for a very long time. This was developed during the 1870’s itself. There are various operations in the set theory. Set is basically a collection of objects. When it comes to sets in mathematics, the objects must be related to mathematics. There is the presence of a universal set in set theory. This acts as the reference set. Other sets are compared with the same. Set theory is best explained with the help of Venn diagrams. The operation on sets like union, intersection, and compliment and so on can be well represented with the help of Venn diagrams. The subset R contains some of the elements present in R and not all the elements.’ R’ acts as the universal set in this case. The concepts sets and subsets are closely related. Without the presence of one the other doesn’t exist. The definition of subset is that it is a set which contains some of the elements present in the original set.

An example can be used to explain the concept. A set has elements {x, y, z} and another one contains {x). The latter set is the subset of the former one. There can be a number of subsets of the same set. If there was another set containing the element {y}, then it also becomes the subset of the given set.  The subset notation is used to convey that a particular set is the subset of the other set. The other set is called the super set. It contains all the elements present in the subset and the subset contains some of the elements of the super set. This is nothing but the subset is a part of the super set.

Set theory can be very helpful in solving mathematical problems. The Venn diagrams give a clear picture on the sets and their operations. By the Venn diagrams problems can be solved. So, lot of arithmetical calculations can be avoided. The process also becomes very simple and easy to understand. Once one is thorough with the concepts of the set theory, the problems can be very easily solved. One of the important operations in set theory is that of intersection of two sets. The intersection of two sets yields a new set which contains the common elements of both the sets.

Number Cube Probability

Number Cube Probability

Probability is the chance of the outcome of an event of a particular experiment. Probabilities are occurs always numbers between 0(impossible) and 1(possible). The set of all possible outcomes of a particular experiment is called as sample space. For example probability of getting a 3 when rolling a dice is 1/6. In this article we will discuss about probability problems using number cube.

Number Cube Probability - Example Problems

Example 1: If rolling a number cube, what is the probability of getting prime number?

Solution:

Let S be the sample space, n(S) = 6.

A be the event of getting prime number.

A = {2,3, 5}, n(A) = 3

P(A) = `(n(A))/(n(S))` = `3/6` = `1/2`

Example 2: If rolling two number cubes, what is the probability of getting a sum of odd?

Solution:

Let S be the sample space, n(S) = 6 * 6 = 36.

A be the event of getting a sum of odd.

n(A) = {{1, 2), (1, 4), (1, 6)....(6, 1), (6, 3), (6, 5)} = 18

P(A) = `(n(A))/(n(S))` = `18/36` = `1/2`

Therefore probability of getting a sum of odd is `1/2`

Example 3: If rolling two number cubes, what is the probability of getting a sum of 10 or 7?
Solution:

Let S be the sample space, n(S) = 6 * 6 = 36.

A be the event of getting 10.

n(A) = {{4, 6), (5, 5), (6, 4)} = 3

P(A) = `(n(A))/(n(S))` = `3/36` = `1/12`

Let B be the event of getting sum of 7.

n(B) = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} = 6

P(B) = `(n(B))/(n(S))` = `6/36` = `1/6`

P(A or B) = P(A) + P(B) = `1/12` + `1/6` = `1/4`

P(A or B) = `1/4`

Therefore probability of getting a sum of 10 or 7 is `1/4`


Number Cube Probability - Practice Problems

Problem 1: If rolling a number cube, what is the probability of getting composite number?

Problem 2: If rolling two number cubes, what is the probability of getting 9 or 6?

Answer: 1) `2/3` 2) `5/18`

Linear Approximation Table

Introduction to linear approximation table:

In mathematics, the procedure for determining the straight line is that closely fit curvature at various sites. Linear approximation equation is given as y = ax + b, the values of a and b are chosen that the line meet the curve at the preferred site, or value of x, and the evaluation of the line equals the rate of change of the curve at that site.

The majority curvature, linear approximation are first-rate only particularly close to the preferred x. The table consist both rows and columns. We can put the values in the table. It should be very easy for the student to study.

Formula-linear Approximation Table:

Linear approximation of the domain function represented as  f(x). Digression line to the table of f(x) at the point (`x_0, y_0` ) where `y_0` is given as `y_0` = f(`x_0` ). It is represented as

` y-y_0= f(x_0)(x-x_0)`

If x is closed the `x_0` , the formula can be written as `x_1 = x_0Deltax `

In mathematics, a linear approximation is an approximation of a universal function. The linear approximation uses a linear function. The linear function is more accurate and affine functions .They are widely used in the method of finite differences to create first order methods for solving or determining solutions to equations. It is given by,

f(x )= f(a) + f'(a)(x-a) + R2
f(x) = f(a) + f'(a)(x-a)

Linear approximation function:

A linear approximation functions to a function at a point can be computed by striking the primary expression in the Taylor series,

`F(x_0+Delta)`  = `f(x_0)+f(x_0)Delta+....`

The Newton’s method linear approximation can be estimated using linear approximation.

Example Problems- Linear Approximation Table:

Example 1- Using linear approximation table

Consider the linear function y = f(x) = `5x^2`

Solution:

Let` Deltax ` is an increment of x.

Therefore, `Deltay ` is also an increment of y.

Hence, we have

= f(x + `Deltax` )-f(x)

= 5(x + `Deltax` )`^2` – `5x^2`

= 25x( `Deltax` ) + 4 `(Deltax)^2`

The differential equation can be determined by using the given linear function.

Therefore `(dy)/(dx)` =25x

Hence, dy = 10x dx

Example 2- Using linear approximation table

Consider the linear function y = f(x) = `3x^2`

Solution:

Let` Deltax ` is an increment of x.

Therefore, `Deltay` is also an increment of y.

Hence, we have

= f(x + `Deltax` )-f(x)

= 3(x + `Deltax` )`^2` – `3x^2`

= 9x( `Deltax` ) + 2` (Deltax)^2`

The differential equation can be determined by using the given linear function.

Therefore `(dy)/(dx)` =9x

Hence, dy = 6x dx.

These all are the above explanation and examples make clear about this linear approximation.

Solve Math Fractions Problems

Solve math fractions problems:

In this article we are discussing  the solve math fractions problems. A fraction is a number that can represent part of a whole. The earliest fractions were reciprocals of integers: ancient symbols representing one part of two, one part of three, one part of four, and so on. A much later development were the common or "vulgar" fractions which are still used today (`1/2` , `5/8` , `3/4` etc.) and which consist of a numerator and a denominator. (Source – Wikipedia)

Solve math fractions problems is simple addition fraction, multiplication fraction, subtraction fraction and dividing fractions.

Solve Math Fractions Problems-example Problems:

Example 1:

Add the fractions for given two fraction, `3/9` + ` 3/9`

Solution:

The given two fractions are `3/9` + `3/9`

The same denominators of the two fractions, so

= `3/9 ` + `3/9`

Add the numerators the 3 and 3 = 3+3 = 6.

The same denominator is 9.

=` 6/9`

The addition fraction solution is `2/3.`

Example 2:

Subtract the fractions for given two fractions `4/6` - `6/5`

Solution:

The denominator is different so we take a (lcd) least common denominator

LCD = 6 x 5 = 30

So multiply and divide by 5 in first term we get

` (4 xx 5) / (6 xx 5)`

=`20/30`

Multiply and divide by 6 in second terms

= `(6 xx 6) / (5 xx 6)`

= `36/30`

The denominators are equals

So subtracting the numerator directly = `(20-36)/30`

Simplify the above equation we get = `-16/30`

Therefore the final answer is `-8/15`

Example 3:

Multiply the fractions for given two fractions, `4/5` x `5/4`

Solution:

The given two fractions are `4/5` x `5/4`

The multiply numerator and denominators of the two fractions, so

= `4/5` x `5/4`

Multiply the numerators the 4 and 5 = 4 x 5 = 20.

Multiply the denominators the 5 and 4 = 5 x 4= 20

= `20/20`

The multiply fraction solution is 1

Example 4:

Dividing fraction:

`4/3` divides `2/4`

Solution:

First we have to take the reciprocal of the second number

Reciprocal of `2/4 ` = `4/2`

Now we multiply with first term we get

`4/3` x `4/2`

Multiply the numerator and denominator

`(4 xx 4) / (3 xx 2)`

Simplify the above equation we get

= `16/6`

Therefore the final answer is `8/3`


Solve Math Fractions Problems-practice Problems:

Problem 1: Add the two fraction `2/8` +`2/8`

Solution: `1/2`

Problem 2: Subtract two fractions `10/10` – ` 6/10`

Solution: `2/5`

Problem 3: multiply two fractions `3/5` x  `5/5`

Solution: `3/5`

Problem 4: Dividing two fractions `6/3` and `2/4`

Solution: 4