Review of Word Problems

Introduction on review of word problems are first used to identify the variables present in the given problem. Also the word problems are used for the unit numbers and unit variables. Word problems are used for finding the phrases. The consideration includes in this is, we have to analyse the problem present in the statements.

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Types of word problems:

There are many types review word problems present. They are,
1.Word problem for numbers
2.Word problem for mixtures
3.Word problem for Age
4.Word problem for time
5.Word problem for linear
Word problem for numbers:
          In word problem for numbers, we  first review the relationships are identified. The variables present in the word problems are reduced.
Word problem for mixtures:
       Different types of concentrations are included in this word problem mixtures.
Word problem for age:
       The relationship of ages are calculated on this word problem.
Word problem for time:
     In this word problem, we review  the various types of train problems are included. Time taken for a train to go and coming back are included in this word problem.
Word problem for linear:
      Linear word problem mostly review the cost problems. For example, the cost for an grapes and the apple is 14.etc..





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Example for review word problem:

Question: 1. The difference of twice a number 4 is 8. What is that number?
Solution:
Step 1: First we are changing the above said sentence as difference of twice a number 4 equals 8.
Step 2: Next step is to write the above said equation in the equation form.
     2V-4=8.
This is the solution for a word problem.
Question 2. The difference of twice a number 2 is 4. What is that number?
Solution:
Step 1: First we are changing the above said sentence as difference of twice a number 2 equals 4.
Step 2: Next step is to write the above said equation in the equation form.
     2V-2=4.
This is the solution for a word problem.
Practice to review word problem:
Practice 1.The difference of twice a number 12 is 14. What is that number?
Practice 2.The difference of twice a number 20 is 40. What is that number?

Factoring With Fractions

Factoring is the process of finding the factor of the given function. It is otherwise called as divide the given function as two separate part. We can factor the function more than two times. Fractions are also used in factoring process. Fraction means it has some numerator and denominator values. Both the numerator and denominator values are different. Now in this article, we see about ac factoring with fractions and their example problems.

Example problems for Ac factoring with fractions

Ac factoring with fractions example problem 1:

Factorize the given polynomial fraction equation `(1 / 6)` x2 - `(7 / 6)`x  - 20 = 0

Solution:

Given equation is `(1 / 6)` x2 - `(7 / 6)`x  - 20 = 0

For converting the above fraction equation in normal quadratic function,

Multiply the given equation by 6 on both the sides, we get

x2 - 7x - 120 = 0

Factorize the above equation, we get

x2 - 15x + 8x - 120 = 0

Grouping the first two terms and next two terms, we get

(x2 - 15x) + (8x - 120) = 0

x (x - 15) + 8 (x - 15) = 0

(x - 15) (x + 8) = 0

The factors are (x - 15) and (x + 8)

Answer:

The final answer is (x - 15) and (x + 8)

Ac factoring with fractions example problem 2:

Factorize the given polynomial fraction equation `(1 / 9)` x2 - `(4 / 3)` x + ` (11 / 9)` = 0

Solution:

Given equation is `(1 / 9)` x2 - `(4 / 3)` x + ` (11 / 9)` = 0

For converting the above fraction equation in normal quadratic function,

Multiply the given equation by 9 on both the sides, we get

x2 - 12x + 11 = 0

Factorize the above equation, we get

x2 - 11x - x + 11 = 0

Grouping the first two terms and next two terms, we get

(x2 - 11x) + (- x + 11) = 0

x (x - 11) - 1(x - 11) = 0

(x - 11) (x - 1) = 0

The factors are (x - 11) and (x - 1)

Answer:

The final answer is (x - 11) and (x - 1)

Ac factoring with fractions example problem 3:

Factorize the given polynomial fraction equation `(1 / 2)` x2 - `(11 / 2)` x + 14 = 0

Solution:

Given equation is `(1 / 2)` x2 - `(11 / 2)` x + 14 = 0

Multiply the given equation by 2 on both the sides, we get

x2 - 11x + 28 = 0

Factorize the above equation, we get

x2 - 7x - 4x + 28 = 0

Grouping the first two terms and next two terms, we get

(x2 - 7x) + (- 4x + 28) = 0

x (x - 7) - 4 (x - 7) = 0

(x - 4) (x - 7) = 0

The factors are (x - 4) and (x - 7)

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Answer:

The final answer is (x - 4) and (x - 7)

Adding Positive Integers

Let us see about adding positive integers. In mathematics the numbers or integers are very essential. The integers are either may be positive or it may be negative. The positive integers are represented as its normal numbers but the negative integers are represented as (-). Adding positive integers is the one of the operation in arithmetic operation. The integers are may be two digits or four digits or it may be more than that.

Definition:

Addition is the process of adding to integers. The positive integers are easy to add. It is also defined as, the group of objects mutually into a larger collection. It is signified by plus sign (+).Process of the addition is one of the easiest numerical tasks.

Examples:

Let us see about examples of positive integers.

Problem 1:

Adding the integers 24 and 35.

Solution:

The given integers are 24 and 35, both integers are positive integers. So, the addition is easy.

24

35  (+)

-------

59

----------.

This is the answer of the given problem.

Problem 2:

Adding then positive integers 12 and 55.

Solution:

The given integers are 12 and 55, both integers are positive integers. So, the addition is easy.

12

55  (+)

-------

67

----------.

This is the answer of the given problem.

Problem 3:

Adding the positive integers 61 and 76.

Solution:

The given integers are 61 and 76, both integers are positive integers. So, the addition is easy.

61

76  (+)

-------

137

----------.

This is the answer of the given problem.

Problem 4:

Adding the positive integers 90 and 40.

Solution:

The given integers are 90 and 40, both integers are positive integers. So, the addition is easy.

90

40  (+)

-------

130

----------.

This is the answer of the given problem.

Problem 5:

Adding the positive integers 35 and 49.

Solution:

The given integers are 35 and 49, both integers are positive integers. So, the addition is easy.

35

49  (+)

-------

84

----------.


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This is the answer of the given problem.

Problem 6:

Adding the positive integers 78 and 42.

Solution:

The given integers are 78 and 42, both integers are positive integers. So, the addition is easy.

78

42  (+)

-------

120

----------.

This is the answer of the given problem.

Non linear Differential Equations

An ordinary differential equation is an differential equation in For instance (i)dy/dx = x + 5 (ii) (y′)2 + (y′)3 + 3y = x2 which  is a single independent variable enters either explicitly or implicitly are all ordinary non linear differential equations And now we see about the non linear differential equations.

Non linear differential equations in First order

Order and degree of a non linear differential equations:

Definition of non linear differential equations :

The order of an differential equation is the order of the highest order derivative occurring in it. The degree of the differential equation is the highest order derivative which occurs in it, after the differential equation has been made free from radicals and fractions as far as the derivatives are concerned.The degree of a non linear differential equation does not require variables like r, s, t … to be free from radicals and fractions.

Find the order and degree of the following differential equations:

ii) y = 4dy/dx + 3xdx/dy

y = 4dy/dx + 3xdx/dy  y = 4(dy/dx) + 3x 1/(dy/dx )

Making the above equation free from fractions involving

dy/dx we get y .dy/dx = 4(dy/dx)2 + 3x 2Highest order = 1

Degree of Highest order = 2

(order, degree) = (1, 2)

Problem for non linear differential equations:

Formation of non linear differential equations :

Form the differential equation from the following equations.

y = e2x (A + Bx)

ye−2x = A+ Bx … (1)

Since the above equation contains two arbitrary constants, differentiating

twice, we get y′e−2x − 2y e−2x = B

{y′′e−2x − 2y′ e−2x} − 2{y′e−2x − 2y e−2x} = 0

e−2x {y′′ − 4y′ + 4y} = 0 [‡ e−2x ≠ 0]

y′′ − 4y′ + 4y = 0 is the differential equation.

Non linear Differential equations of first order and also first degree :

Solve: 3ex tan y dx + (1 + ex) sec2y dy = 0

`=>` 3 log (1 + ex) + log tan y = log c

`=>` log [tan y (1 + ex)3] = log c

`=>`   (1 + ex)3

required solution to the problem is tan y = c

Non linear Differential Equations

An ordinary differential equation is an differential equation in For instance (i)dy/dx = x + 5 (ii) (y′)2 + (y′)3 + 3y = x2 which  is a single independent variable enters either explicitly or implicitly are all ordinary non linear differential equations And now we see about the non linear differential equations.

Non linear differential equations in First order

Order and degree of a non linear differential equations:

Definition of non linear differential equations :

The order of an differential equation is the order of the highest order derivative occurring in it. The degree of the differential equation is the highest order derivative which occurs in it, after the differential equation has been made free from radicals and fractions as far as the derivatives are concerned.The degree of a non linear differential equation does not require variables like r, s, t … to be free from radicals and fractions.

Find the order and degree of the following differential equations:

ii) y = 4dy/dx + 3xdx/dy

y = 4dy/dx + 3xdx/dy  y = 4(dy/dx) + 3x 1/(dy/dx )

Making the above equation free from fractions involving

dy/dx we get y .dy/dx = 4(dy/dx)2 + 3x 2Highest order = 1

Degree of Highest order = 2

(order, degree) = (1, 2)

Problem for non linear differential equations:

Formation of non linear differential equations :

Form the differential equation from the following equations.

y = e2x (A + Bx)

ye−2x = A+ Bx … (1)

Since the above equation contains two arbitrary constants, differentiating

twice, we get y′e−2x − 2y e−2x = B

{y′′e−2x − 2y′ e−2x} − 2{y′e−2x − 2y e−2x} = 0

e−2x {y′′ − 4y′ + 4y} = 0 [‡ e−2x ≠ 0]

y′′ − 4y′ + 4y = 0 is the differential equation.

Non linear Differential equations of first order and also first degree :

Solve: 3ex tan y dx + (1 + ex) sec2y dy = 0

`=>` 3 log (1 + ex) + log tan y = log c

`=>` log [tan y (1 + ex)3] = log c

`=>`   (1 + ex)3

required solution to the problem is tan y = c

Radicals and Exponents

The opposite operation to the exponent is known as radical. A radical is an expression which contains the square roots, cube roots etc. For example the expression √49 can also be called as square root of 49 or root of 49. Radicals have the same property of the numbers. The simplification is to reduce the numbers and reduce the power of variables inside the roots.

Exponent of a number shows you how many times the number is to be used in a multiplication.  It is shown as a small number to the right and above the base number.  In this example: 22 = 2 × 2 = 4 ( Another name for exponent is index or power ).

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Examples for Radicals:

Example 1 for radical problem:

Simplify the radical:    √ (144) / (√36)

Step 1: Factors of 144 = 12 × 12

√ (144) = √ (12 × 12)

= √122

Step 2: Square root of 122 = 12

Step 3: Factors of 36 = 6 × 6

√ (36) = √ (6 × 6) = √62

Step 4:  √62 = 6

Step 5: so, √ (144) / √36) = 12 / 6

The answer of the given radical is 12 / 6
Example 2 for radical problem:

Simplify the radical:    3√8 × √12

Step 1: Find the factor of 8 = 2 × 2 × 2

Step 2: Take cubic root of 8 = 2

Step 3: Find the factor for 12 = 2 × 2 × 3

Step 4: Take square root of 12 = 2 √3

Step 5: so, 3√8 × √12 = 2 × 2 √3

The answer for the given radical is = 4√3

Examples for exponents:

EXAMPLE-1:

453 . 454 = 453+4 = 457

EXAMPLE-2:

148 / 145  = 14 8-5 = 143

EXAMPLE-3:

( 92 )3  = 9 2x3  = 96

EXAMPLE-4:

( 11 / 13 )2  = 112 / 122

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EXAMPLE-5:

70 = 1

EXAMPLE-6:

5-3 = 1/ 53

Inverse Variation

To understand the inverse variation,let is correlate it with our real life of inverse variation examples such as the total time taken with the speed traveled in a trip. Assuming distance of a trip would be fixed, let us say the distance of a trip would be 250 mile.

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Here  inverse variation equation will be used as T=250/r , according to the time will be  T be time of the trip will be equal to 250  divided by the Rate in miles per hour which will called r .so if rate is 50 miles per hour it would take,  5 hours  .

The next of the inverse example is total time taken to spread the soil and the number of people working on a part of land. so let us say if the total time of the job will take 12 hours the total time T to complete the job would be  12 hours divided by n, ( n is the number of workers).

Here the equation will be T= 12/n.  so if have 1 worker , we take 12 hours, 3 workers we  take 4 hours and so on.

Total amount of cost required by per person for petrol and the number of people in a cab. So if we know that it will cost 45 dollars’ worth of petrol to go for a picnic trip. The cost represents C= 45 dollars divided by number of people in the cab.

That is C= 45/no of people. So if n will increases the total cost decreases.
Here in inverse variation definition is as the ones quantityvalue gets bigger and the others value gets smaller in such that there product remains the same or proportional. Let us understand with one more example, Y varies inversely as X. where Y= 4 and when X=2.

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Find the value of the inverse-variation equation. Then determine Y when X= 16.  By performing substitution we can solve this question.  Y= k/x.  by doing cross multiplication , we get 4= k/2 which is k=8.  So our inverse-variation will be  Y=K/x that is 8/x. Now we can answer the second query on determining the y when x is 16. So, y= 8/x , substituting the value of x as 16. So y= 8/16. And we get  1/2 . y= ½ and x= 16.