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).