Concept and Definition of Markup

In mathematics the concept of cost price and selling price are studied. These are very important not only in mathematics but also in the field of business. Both the cost price and the selling price will help in determining the profit for the business. Cost price is the price that is incurred in producing the product and the selling price is the price at which the product is sold. There should be a difference between the two.

Basically the selling price must be greater than the cost price in order to earn a profit otherwise the business cannot run forward and it has to be shut down. So, one has to be very careful in fixing the selling price of the product otherwise the business might run into losses and must be closed down. The markup definition can be got from the concepts of the cost price and the selling price. Once both these concepts are clear. Then the concept is easy to understand.

So, basically it is nothing but part of the selling price. This is added to the cost price in order to find the selling price. This is done in order to get profit for the business. If markup is not added then no profit can be obtained.

Profit is the main motive for any business. No business can move forward without the help of profit. The percentage value can also be calculated for the same.  For this purpose the cost price is used. The cost price comes in the denominator in the calculation.

The final answer is multiplied with ‘100’. Multiplication with ‘100’ is done to obtain the figure in percentage. In retail markup also the same concept is used. The retail price is decided in a manner so that the component of profit is always present in the selling price. The retailer cannot sell the product with losses incurred. Otherwise the retailer has to shut shop.

To calculate markup both the cost price and the selling price is required. This is calculated by subtracting the cost price from the selling price. In this case the selling price must be greater than the cost price.

This is because the profit component is also included in the total price. The total price in this case is nothing but the selling price of the product.  Once mark-up is calculated it helps in knowing the margin that is required in finding selling price.

Factoring polynomials

When factoring out polynomials,  we find   the  polynomial  that divide out evenly from the original polynomial  .   How to factor  Polynomials:  For this we have to find all the terms that if multiplied together we get the  original polynomial.  This is continued to all the terms until this cannot be simplified any more.  If the polynomial cannot be factored any more then the polynomial is said to be completely factored.

A factor of a polynomial is any polynomial which divides evenly into   the given polynomial   For example, x + 2 is a factor of the polynomial x^2 – 4.
The factorization of a polynomial is its representation as a product its factors. For example, the factorization of x^2 – 4 is (x – 2)(x + 2).

In mathematics, factorization or factoring is the breaking apart of a polynomial into a product of other smaller polynomials. If you choose, you could then multiply these factors together, and you should get the original polynomial.
This example shows how to factoring polynomials.    Take 3x^2 – 12x + 9.   The common term in these is 3.  So take 3 out and divide each term by 3.  We get   3 (x 2-4x + 3).

Step by step explanation as how to Factor out Polynomials is given below
Factor 4a2 + 20a – 3a - 15

The first two terms have a common factor in 4a.  The last two terms have a   common factor in 3.
We need to factor those terms out.

4a ( a + 5)  -3(a+5)

Now you have a binomial.  Each term     has a factor of (a + 5).

(4a -3) (a+5) is the factored terms.

Another example is given below on factoring of polynomials
x^2 -8x + 15

start by looking at the factor pairs of 15.  We are looking for a pair of factors which add up to equal -8.  Look for the factor pairs of  +15 so that they add up to -8    The negative factor pairs of 15 are:
-15  and  -1
 -5 and  -3
Since -5 + -3 = -8 this is the pair we are looking for and we can factor the
original expression into:  (x-5)(x-3)

Addition of matrices

Addition of Matrices is an operation done on two or more matrics according to their corresponding entries. Matrix of same order can only be added. If there orders are different then addition operation cannot be applied on them.

Adding Matrices requires addition of corresponding elements of them. For example if two Matrices are added then elements in first row and first column of the two matrices will be added together to get the element in the first row and first column of the resultant matrix.

Two Matrices of order m × n (m is number of rows and n is number of columns) can be added:

Let sum of these = C = A+B then
C11 = a11 + b11, C12 = a12 + b12….,  C21 = a21 + b21, C22 = a22 + b22 and so on.

A + B = C =  ( )
Order of matrix C will also be m × n.
If two matrix are equal but opposite in sign, then there addition will be null matrix (whose all elements are zero). For example if A = ((1@2@5)) and B = ((-1@-2@-5)) then A+B = 0 as A+B = ((1+(-1)@2+(-2)@5+(-5)))= ((0@0@0))

Example 1) Add Matrices A and B if A = ((1&9@5&2))  and B = ((2&4@5&8))
Solution) Let sum of A and B be C. order of C will be same as that of A and B i.e. 2x2.
Elements of matrix C will be: C11, C 12, C 21, C22.
C11 = A11 + B11 = 1+2,
C12 = A12 + B12 = 9+4,
C21 = A21 + B21 = 5+5,
C22 = A22 + B22 = 2+8,
C = ((C11 = A11 + B11 &C12 = A12 + B12 @C21 = A21 + B21&C22 = A22 + B22))
C =   ((1 + 2&9 + 4@5 + 5 &2 + 8))
C = ((3&13@10&10))
Example 2) Perform addition of Matrices P and Q given as:
P =   ((2&0&- 3@5&3&7@3&- 7&1)), Q =  ((- 4&6&2@5&2&5@21&- 15&- 4))
Solution) let P+Q = R
R11 = P11 + Q11 = 2 +(-4) =-2,
R12 = P12 + Q12 = 0+6 = 6,
R13 = P13 + Q12 = -3+2 =-1
R21 = P21 + Q21 = 5+5 =10,
R22 = P22 + Q22 = 3+2 =5,
R23 = P23 + Q23 = 7+5 = 12
R31 = P31 + Q31 = 3+21=24
R32 = P32 + Q32 = -7+(-15) =-22
R33 = P33 + Q33 = 1+(-4)= -3
R = ((R_11&R_12&R_13@R_21&R_22&R_23@R_31&R_32&R_33 ))
R = ((-2&6&-1@10&5&12@24&-22&-3))

Gaussjordan elimination method

With the concept of matrices one can solve the system of linear equations apart from usual algebraic methods. Even in matrices topic, a system there are different methods to solve and one such method is called as Gauss Jordan elimination method named after the two mathematicians who introduced this elimination method. It is also called in short as Gauss elimination method or as Jordan elimination.
Let us first study the concept behind the Gauss elimination method. A system of equations is expressed in the matrix form as, AX = B, where A is the matrix of the coefficients, X is the matrix of the variables and B is the matrix of the constants on the right side of the system of equations. If the augmented matrix [A¦B] can be expressed in reduced row echelon form, by the method of Gauss-Jordan, that is, in the form [I¦C] (I being the identity matrix), then X = C, which gives us the solution to the system.
Now let us do Gauss-Jordan elimination step by step, with an example for clear understanding. Let a system of equations is,


-3x + 4y + z = -5
x +3y -2z = -6
2x –y + 3z = 9

   Then,A is,

-3	4	1
1	3	-2
2	-1	3

x

y

z

X is, 

-5

-6

9

B is

The following are the steps of elimination.

-3	4	1	-5
1	3	-2	-6
2	-1	3	9

[A│B]

1	-4/3	-1/3	5/3
1	3	-2	-6
2	-1	3	9

[R₁│-3]

1	-4/3	-1/3	5/3
0	13/3	-5/3	-23/3
2	-1	3	9

[R₂│-R₁]

1	-4/3	-1/3	5/3
0	13/3	-5/3	-23/3
0	5/3	11/3	17/3

[R₃-2R₁]

1	-4/3	-1/3	5/3
0	1	-5/13	-23/13
0	5/3	11/3	17/3

[R₂/(13/3]

1	-4/3	-1/3	5/3
0	1	-5/13	-23/13
0	0	168/39	336/39

[R₃│-(5/3)R₂]

1	-4/3	-1/3	5/3
0	1	-5/13	-23/13
0	0	1	2

[R₃/(168/39)]

1	-4/3	0	7/3
0	1	-5/13	-23/13
0	0	1	2

[R₁+(R₃/3)]

1	-4/3	0	7/3
0	1	0	-1
0	0	1	2

[R₂+(5R₃/13)]

1	0	0	1
0	1	0	-1
0	0	1	2

[R₁+(4R₂/3)]

Now the finally reduced matrix is in the form [I│C], where,

1

-1

2

           C and therefore X = 

Hence, the solution to the given system of equations is, x = 1, y = -1 and z = 2. 

Thus, we had seen the used of the elimination method introduced by Gauss and Jordon.

Learning to find the determinant value of a 3x3 matrix

Consider a matrix A. If A is a square matrix then there is some value that can be assigned to A. This is called the determinant of matrix A. It is denoted by the symbol |A| or det(A). The entries of the matrices are used and an arithmetic expression is formed out of them to come to a value of the matrix. Such square matrices are most commonly used in solving system of linear equations. If the number of variables in the system is 2, then the square matrix would be a 2x2 matrix. Similarly if the number of variables in the system is n, then the square matrix would be an nXn matrix.
Determinant of a 3x3 matrix:
The determinant of a 3x3 matrix is also called a third order determinant. If we have to solve a system of three simultaneous linear equations in three variables, we shall have to deal with third order determinants or 3x3 determinant. For real numbers a1, a2, a3, b1, b2, b3, c1, c2, c3, the symbol
|a1 a2 a3|
|b1 b2 b3|
|c1 c2 c3|

Represents the determinant of 3x3 matrix.

The numbers a1, a2, a3, b1, b2, b3, c1, c2, c3 are the entries or elements of the det.  The first, the second and the third rows of the det are respectively: a1, a2, a3; b1, b2, b3 and c1, c2, c3. The first, the second and the third columns are respectively;
a1 a2 a3
b1; b2; b3;
c1 c2 c3

The value of this determinant of 3x3 matrix is defined as:

|a1 a2 a3|
|b1 b2 b3|
|c1 c2 c3|

= a1*|b2 b3| - a2 * |b1 b3| + a3 * |b1 b2|
          |c2 c3|            |c1 c3|             |c1 c2|

= a1 * (b2c3 – b3c2) – a2 * (b1c3 – b3c2) + a3 * (b1c2 – b2c1)

The above formula can be used to find the value of any 3x3 determinant. This is not the only method used to evaluate a 3x3 determinant. We can also use the Sarrus’ expansion method. It is as follows:
Suppose D
= |a b c|
   |d e f|
   |g h i|
We write five columns like this:

a b c a b
d e f d e
g h I g h
Then we multiply the terms long the red and blue lines as shown in the picture below:

The red diagonals are aei, bfg and cdh. The blue diagonals are gec, hfa and idb.
Then D = aei + bfg + cdh – gec – hfa - idb

Solution of a quadratic equation by completing the square

One method of obtaining the roots of a quadratic equation is the factoring method. Another method that is used is completing the square method.

Consider the following situation:
The product of Sara’s age (in years) two years ago and her age four years from now is one more than twice her present age. What is her present age?
To answer this, let her present age (in years) be = x. Then the product of her ages two years ago and four years from now is (x-2)*(x+4).
Therefore, (x-2)*(x+4) = 2x+1
I.e, x^2 + 2x – 8 = 2x + 1
i.e, x^2 – 9 = 0

So Sara’s present age satisfies the quadratic equation x^2 – 9 = 0
We can write this as x^2 = 9. Taking square roots, we get x = 3 or x = -3. Since age is always a positive number, x = 3 would be our answer.

Now consider the quadratic equation (x+2)^2 – 9 = 0. To solve this we can write it as (x+2)^2 = 9. Taking roots we have x+2 = 3 and x+2 = -3. Therefore x = 1 or x = -5. These are the roots.

In both the examples above, the term containing x is completely inside a square, and we found the roots easily by taking square roots. But what if we are given the equation x^2 + 4x – 5 = 0 to solve? We would probably apply factorization unless we (somehow!) realize that x^+4x-5 = (x+2)^2 – 9.

So solving x^2+4x-5 = 0 is equivalent to solving (x+2)^2-9 = 0, which we have seen is faster. In fact, we can convert any quadratic equation to the form (x+a)^2 – b^2 = 0 and then we can easily find its roots. This process is called completing a square.

Suppose we were to complete the square for x^2+4x. The process is as follows:
X^2+4x
=(x^2 + (4/2)x) + (4/2)x
= x^2+2x+2x
= (x+2)x + 2*x
= (x+2)x + 2*x + 2*2 – 2*2
= (x+2)x + (x+2)2 – 2*2
= (x+2)(x+2) – 2^2
= (x+2)^2 – 4
So, x^2+4x-5 = (x+2)^2 -4 -5 = (x+2)^2 – 9.
So turning x^2+4x-5 to (x+2)^2-9 makes a complete square of the expression.

In brief, this can be shown as follows:
X^2+4x = (x+(4/2))^2 – (4/2)^2 = (x+(4/2))^2 – 4
So, x^2+4x-5 = 0 can be rewritten as
(x+(4/2))^2 -4-5 = 0
i.e., (x+2)^2 – 9 = 0. This was the completing square method in brief.

Addition of like fractions

We know that whole numbers can be added, subtracted, divided or multiplied. Similarly fractions can also be added, subtracted, divided or multiplied. A problem of addition of fractions can be of either of the two types:
(a) Adding fractions like denominators
(b) Adding fractions – unlike denominators

Adding like denominators fractions:

Adding fractions with like denominators word problems may be worded using different terms such as: “How to add fractions with common denominators?’ or ‘adding fractions with same denominators’.  These words, “common denominators” or “ same denominators” mean the same thing as “like denominators”. By definition, if two fractions have the same or equal denominators they are called, like fractions.

For being able to add fractions, the denominators have to be equal or same. In other words, for adding fractions it is important that all the addends are like fractions and have the same common denominator. For adding fractions with unlike denominators, we first make the denominators equal by taking the lowest common multiple of the denominators and converting each of the fractions to its corresponding like fraction that has denominator equal to the lowest common multiple found.

For adding like fraction, the following steps are to be followed:

Step 1: Make sure that the fractions are like fractions. That means, ensure that the denominators of all the fractions to be added are equal.
Step 2: Add up all the numbers on the numerators of all the fractions.
Step 3: Put the result found in step 2 over the common denominator that we confirmed in step 1.
Step 4: Simplify the fraction if needed. That is, reduce the fraction to its lowest terms if needed.

Let us illustrate the above process with an example:

Example 1: Add 1/10, 3/10, 1/10

Solution:
Step 1: The fractions to be added are 1/10, 3/10 and 1/10. Note that the denominators of all the three fractions are same and equal to 10. So we can move on to step 2.

Step 2: The numerators of the fractions are 1,3 and 1. Adding them we have,
1 + 3 + 1 = 5.

Step 3: The result found in step 2 was the number 5. We put this number over the common denominator that we confirmed in step 1 which was 10. So we have,
= 5/10

Step 4: Now we reduce the fraction thus found to its lowest terms. So,
5/10 = (5*1)/(5*2), the common factor 5 cancels out and we are left with,
= ½ <- answer.="answer." p="p">