Solve Algebra Fraction

Algebra is the branch of 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. Here we are going to study about How to solve algebra fraction and its example problems. (Source from Wikipedia)

Algebra fraction:

In algebra fraction is a simple fraction it contains algebraic expression in either numerator or denominator.

Example:
`(3x+3) / 2` = 3

Example problems:

Example: 1

Solve the following algebraic equation 

`(2x+4) / 2` + `(3x+3) / 4` = 6

Solution:

Here we have two fraction term and both have different denominator so we take Common denominator

Least common denominator = 2 * 4 = 8

Multiply first term with 4 we get

`(8x+16) / 8` + `(6x+6) / 8` = 6

Now both denominators is equal so take away

`1/ 8` {(8x+16) + (6x+6)} = 6

Multiply both sides 8 we get

8 * `1/ 8 ` {(8x+16) + (6x+6)} = 6 * 8

In left hand side 8 will be canceling

(8x+16) + (6x+6) = 48

Combine the like terms we get

14 x + 22 = 48

Add both sides -22

14x + 22 -22 = 48 -22

14x = 26

Divide both sides 14 we get

x = `26 /14`

The simplest fraction is `13 / 7`

Therefore the value of x = `13 / 7`


Example : 2

Solve the following algebraic equation

`(3x+3) / 2` = 3

Solution:

Here the denominator is 2

So multiply both sides 2 we get

2 `(3x+3) / 2` = 3 *2

In left hand side numerator 2 and denominator 2 will be canceling

3x +3 = 6

Add both sides -3 we get

3x+3-3 = 6 -3

3x = 3

Now we get without fraction equation.it is simple algebraic equation.

Divide both sides 3 we get

x =` 3 / 3`

Therefore the value of x = 1


Example: 3

Solve the following algebraic equation

`4 / (2x+3)` = 3

Solution:

Multiply both sides (2x+3)

(2x+3) * `4 / (2x+3)` = 3 (2x+3)

Numerator (2x+3) and denominator (2x+3) will be cancelling

4 = 3 (2x+3)

4 = 6x+9

Add both sides -9 we get

4-9 = 6x+9-9

-5 = 6x

Therefore x = - `(5/6)`

The value of x – `(5/6)`

Solve Second Derivative Test

Solve second derivative test involves the process differentiating the given algebraic function twice with respect to the given variable. Generally the derivative is discussed in calculus whereas it is mainly used to find the rate of change of the given function with respect to the change in the input. The following are the solved example problems with detailed step by step solution in second derivative to study for the test.


Second derivative test example problems to solve:


Example 1:

Solve the test function to find second derivative.

f(z) = 2z6 + 2 z5 + 3 z4 + 3z

Solution:

The given equation is

f(z) = 2z6 + 2 z5 + 3 z4 + 3z

The above function is differentiated with respect to z to find the first derivative

f '(z) =  2(6z 5)  +2 (5 z4 ) +3(4 z3) + 3

By solving above terms

f '(z) =  12z 5  +  10z4  + 12 z3 – 3

The above function is again differentiated with respect to z to find second derivative

f ''(z) =  12(5z 4 ) – 10(4z3)  + 12(3z2)

f ''(z) =  60z 4 – 40z3 +36z2  is the answer.

Example 2:
Solve the test function to find second derivative.

f(z) = 5z 2 +5z 4  + 12

Solution:

The given function is

f(z) = 5z 2 +5z 4  + 12

The above function is differentiated with respect to z to find the first derivative

f '(z) = 5(2z  )+5(4 z 3 ) + 0

By solving above terms

f '(z) = 10z +20z3

The above function is again differentiated with respect to z to find second derivative

f ''(z) =  10(1 ) +20(3z2)

f ''(z) =  10 + 60z2 is the answer.

Example 3:

Solve the test function to find second derivative.

f(z) = 4z4 +5z 5 +6z 6  + 2z

Solution:

The given function is

f(z) = 4z4 +5z 5 +6z 6  + 2z

The above function is differentiated with respect to z to find the first derivative

f '(z) = 4(4z 3 )+5(5z 4 ) +6( 6z 5) +2

By solving above terms

f '(z) = 16z 3 +25z 4 +36 z 5 + 2

The above function is again differentiated with respect to z to find second derivative

f ''(z)= 16(3z 2) +25(4z 3) +36 (5z 4)

f ''(z)= 48z 2 +100z 3 +180z 4 is the answer.


Second derivative test practice problems to solve:


1) Solve the test function to find second derivative.

f(z) = z 3 + z 4 + z 5

Answer: f ''(z) = 6z +12z2+ 20z 3

2) Solve the test function to find second derivative.

f(z) = 2z 3+3z5 + 4z 6

Answer: f ''(z) = 12z + 60z3 + 120 z 4

Learning Intermediate Algebra

Algebra executes most common four basic types of operations. That includes the addition, subtraction, multiplication and division operations. Algebraic problems contains the variables, constant, coefficients, exponents, terms and expressions. Balanced equations on both of the sides are considered as the fundamental concept of algebra. Important properties such as commutative, associative, identities and inverse are also involved in the algebraic problems.


Learning common term in intermediate algebra:


Variables:

Algebraic variables are use for assigning the variables. Alphabetical characters alone used as variables. There are possibilities for changes of variable values while solving the algebraic problems.  x, y and z are the three commonly used variables in algebra.

Constant:

Algebraic constants are values that never change during the process of solving the algebraic equation. Consider the algebraic form of equation 6y + 5, in which the value 5 is the constant.

Expressions:

Algebraic expressions are considered as the combinations of variables, constant, coefficients, exponents and terms.Addition, subtraction, multiplication and division expressions proves this. The example of an algebraic expression is given below

15y + 6.

Term:

Terms are used to form the algebraic expressions by some of the arithmetic operations such as addition, subtraction, multiplication and division. Consider the following example 5n^2 + 6n in which the terms 5n^2, 6n are combined to form the algebraic expression 3n^2 + 2n by using the addition operator ( + ) to perform addition operation.

Coefficient:

The coefficients are the values that are present in front of the terms of an algebraic expression. Consider the following example, 6n2 + 4n in which the coefficient of 6n2 is 6 and 4n is 4.

Equations:

Algebraic equations are the combinations of the numbers or expressions. Mostly the algebraic equations are used for evaluating the the value of the variables.


Learning Examples of intermediate algebra:


Example 1:

4x - 6 = 2x – 4

4x-6+6 =2x-4+6 (adding common value(6) in both sides)

4x      = 2x +2 (we get)

4x -2x = 2x -2x +2 (adding common value(-2x) in both sides)

2x   = 2 (we get)

2x / 2  =  2/2 (dividing common value (2) in both sides)

X  =  1 (We get x value)

Example 2:

6n +15 = 150

6n+15-15 =150-15 (adding common value(-15) in both sides)

6n = 135 (we get )

6n / 6 = 135 / 6 ( dividing common value (6) in both sides)

n = 22.5 (We get n value).

Scale Factor in Algebra

A scale factor is a number which scales, or multiplies, some quantity. In the equation y=Cx, C is the scale factor for x. C is also the coefficient of x, and may be called the constant of proportionality of y to x. For example, doubling distances corresponds to a scale factor of 2 for distance, while cutting a cake in half results in pieces with a scale factor of ½


Explanation of Scale factor in algebra:


the following table will explain the scale factor in algebra some example given below

                            LENGTH           AREA              VOLUME

Scale Factor X      X times            X2 times              X3 times

Scale Factor 2      2 times             4 times                8 times

Scale Factor 3      3 times             9 times                27 times

Scale Factor 4      4 times             16 times              64 times

Scale Factor 5      5 times             25 times              125 times

For example in a square having 6cm all its sides. If the same squares length is reduced to 3cm then the scale factor of the square is 1/2.


Finding Scale factor in algebra:


Some examples for scale factor in algebra.

Ex : Consider the equation C= YZ, Y was the scale factor or Z, Y is also a coefficient of Z and may be called as constant of proportionality of a to Z. By using scale factor, you are creation a smaller model of something that is quite big.

In these algebra examples we discuss the scale factor

For the given problem Y = (4) (3).

4 was the scale factor of 3

4 is also a coefficient of 3

Ex :

                        

                    square1                                            square2

In the example  square1 value of each side is 4 cm. the square2 value of each side is 8 cm. so we find the scale factor of this problem?

Sol : This example we find the scale factor of two square figures.

The second square value by first square value

Scale factor =8/4

                   =2

Ex :  The two triangle are in the same ratio. The first triangle area is 9 sq. units and  the second triangle area is 18 sq. units. Find the scale factor of the two triangle.

Sol : For the first triangle the  area is 9 sq. units.

 For the second triangle the area is 18 sq. units.

The scale factor = second triangle area /first triangle area

        = 18/9

         =2

answer for this problem = 2

Equation

Equation is one of the major concepts in mathematics. Taught in middle school, equation plays an important role throughout the mathematical years. Equation can be classified into algebraic equation, linear equation, quadratic equation and so on. Let’s have a basic overview of equation and formulas in this post.

Equation in simple terms defines the equality of two things. Equation includes the equal sign to define this equality. For example: X + 2 child phone for kids = 7 child phone for kids. Here, the equation is demonstrating that the things in left side are equal to the things in right side. Solving equations in mathematics is like solving a puzzle. A number is missing and one needs to find out the missing value.

For example: 2X + 6 crib toys = 12 crib toys = 2X = 12 – 6 = X = 6 / 2 = 3

Therefore, X is equal to 3.
Formula on the other hand in mathematics means a type of equation that defines the relationship between two variables. There are different formulas used in mathematics to solve equations. For example: Formula for finding volume of a box, area of a circle, radius of a circle and so on.

Example 1: Maya bought a box that includes the best flashlight available in the market. The height of the box is 4, width is 5 and length is 10. Now find the volume of the box including the best flashlight:

Applying the formula of finding the volume of a box i.e, V = h*w*l, we get
Volume of box = 4 * 5 *10 = 200.

Example 2: Arpita got a rectangular box. The length of the box is 4 and the breadth of the box is 8. Find the volume of the rectangle.

Applying the formula of finding the volume of a rectangle i.e, V = 2 (l + b),
we get Volume of rectangle = 2 (4 + 8) = 24
These are the basics about equations and formulas in mathematics.

Study Midpoint Formula

A midpoint formula is a mathematical form halfway between two stellar bodies that tells an interpretative picture for the individual. Direct and indirect are the types of mid point. Sometimes you need to find the point that exactly between two other points. For other instance, you might need to find a line that bisects in a given line segment. These middle point is called the "midpoint".



Discussion on study midpoint formula


If the co-ordinates of the end points of a segment are (x1, y1) and (x2, y2), then the co-ordinates of the mid point of this segment is given by the mid point formula :

M=[(x1+x2)/2,(y1+y2)/2]


Examples on study midpoint formula


Study midpoint formula Example 1:

To find the mid point formula between A(1,2)and B(3,6)

solution:
Let the coordinates of  A(1,2) andB(3,6)

=((1+3)/2,(2+6)/2)

=(4/2,8/2)

=(2,4)

Study midpoint formula Example 2:

M(3, 8) is the midpoint formula of the line AB. B has the coordinates (-2, 3), Find the coordinates of A.

Solution:
Let the coordinates of B be (x, y)

[(-2+x)/2,(3-y)/2]=(3,8)

[(-2+x)/2]=3==>x=8

[(3+y)/2]=3==>x=13

Coordinates of B = (8, 13)

Study midpoint formula Example 3:

What is the midpoint formula between the points (5, 6) and (– 12, 40)?

Solution:

(m1,m2)=((5+(-12)/2),(6+40)/2)

=(-7/2,46/2)

=(-7/2,23)

Study midpoint formula Example 4:

If the midpoint formula between (1, 4) and (x, 10) is (–4, 7), what is the value of x?

solution:

1+x=-8 (or)(1+x/2)=-4
which gives the value x=-9

Study midpoint formula Example 5:

Find the mid point formula of a segment if A is (6,8) and B is (-2, 4).

Solution :

Let (6, 8) be (x1, y1) and    (-2, 4) be (x2, y2).

Therefore their mid point M is given as :

M=[(x1+x2)/2,(y1+y2)/2]

=((6-2)/ 2,(8+4)/2)

= (4/2,12/2)

=(2,6)

Example 6:

If the mid point of segment. AB is (-2, 8) and A is (12, -1), find co-ordinates of B.

Solution :

Let the co-ordinates of B be (x, y). According to the mid-point formula

(-2,8)=[(12+x/2,-1+y/2)]

-2=(12+x)/2, 8=(-1+y)/2

-4=12+x, 16=-1+y

(x,y)=(-16,17)

Therefore the co-ordinates of B are (-16, 17).

linear probability model

Definition:

Linear probability model is one of the econometric model where the dependent variables having the probability between 0 and 1. There are two things related to Linear probability model

I) Logit

II) Probit

Logit is one of the important part in linear probability model. The Logit function is nothing but the inverse function of logistic which is used in mathematics.
Probit model is a popular method in linear probability model. This model is established using standard maximum likelihood procedure.

Explanation for Linear probability model:


The declining model places no limitation on the values that the independent variables take on. They may be continuous, period level, they may be only positive or zero or they may be dichotomous variable (1=male, 0= female)

The dependent variable is implicit to be continuous. There is no constraint on the IVs; the DVs must be free to range in value from negative infinity to positive infinity

We put into practice, only a small variety of Y values will be observed. Because it is also the case that only a small range of X values will be observed. The best guess on continuous interval measurement is frequently not problematic, That is, even though degeneration assumes that Y can range from negative infinity to positive infinity. It regularly won’t be too much of a disaster if. It really only ranges from 1 to 17

Y can only take two values if Y can equal to 0 or 1 then

E (Yi)= 1 x P (Yi = 1) + 0 x P (Yi = 0) =P (Yi = 1)

Final equation is: E (Yi) = P (Yi) = α + Σ β k X k


Examples for linear probability model:


The yield of apple in an acre of apple plantation depends on various types of agriculture practice (treatments). An experiment may be planned where various ploys are subjected to one  out of two possible treatment over a period of time .The yield of tea before the  application of treatment is also recorded .A  Model for post treatment yield(y) is

y = `beta` 0+ `beta`1 x 1+ `beta` 2 x 2 + `in`

Where the binary variables x1 represent the treatment type and the real valued variable x2 is the pre treatment yield. The error term mainly consists of unaccounted factors such as soil type or the inherent differences in apple bushes