Monday, April 30, 2012

Algebraic Expression Part-2

Addition and subtraction:


Let use consider on the following example:
1. Your are given 5 baskets A, B, C, D and E. Basket A contains 15 apples, basket B contains 21 apples, basket C contains 31 apples, basket D contains 8 apples and basket E contains 17 apples.

2. You are given 5 baskets. Basket A contains 12 apples, B contains 19 pineapples, C contains 17 oranges, D contains 24 apples and E contains 10 watermelons.

How many fruits are there in the baskets? What would be your answer to this question?
You may answer that in the first case, the baskets have 92 apples.

In the second case, the baskets contain 36 apples, 19 pineapples, 17 oranges and 10 watermelons.
So it is clear that there is need to add all articles. Hence, you have added the same type of fruit in the first case but written different types of fruits individually in the second case.

Similarly, while adding or subtracting of two algebraic expressions, you have to add or subtract only the like terms and write the unlike terms as they are.  



Addition: -


1. Add   6ax + 11by + 17 cz + 15
              8ax +               –   8cz  + 13
______________________________________
            14ax + 11by + 9cz + 28
-------------------------------------------------------------

As we have seen in the above example, for adding the two expressions, we first write them in separate rows. Like terms are written one below the other.

We then add the like terms and get (6 +8)ax = 14ax, (+ 17 – 80)cz = 9cz and the constants (15 + 13) = 28. We bring down 11by as it is, since there  is no like term, i.e., zero in the second expression.

2. Add 11a +15b – 8c + 7d, 10a + 9c – 13d and 9a + 8b – 11c – 17d.

Solution:   11a +15b – 8c + 7d
                     10a        + 9c – 13d
                     9a + 8b – 11c – 17d
-----------------------------------------------------------
                    30a + 23b – 10c – 23d
-----------------------------------------------------------



Subtraction:


3. Subtract 8sy – 5yz + 3xz + 21 from 6xy + 9yz – 7yz + 17

Solution:     6xy + 9yz – 7yz + 17
                    8sy – 5yz + 3xz + 21
                 (–)     (+)     (–)      (–)
        --------------------------------------------------------------
         2xy + 14 yz – 10 xz – 4                                                                          

As we have seen, for subtracting two expressions, we write them in two rows such that like terms are one below the other. The expression to be subtracted is written in the second row.

We know that subtraction of a number is the same as addition of its inverse that is subtracting – 5 is the same as adding + 5. Therefore, we change the sign of each term of the second row. The third row indicates the new sign of each term. With these new signs, we add the like terms.

4. From the sum of 11p + 9q – 7r and – 7p – 12q + 8r subtract the sum of 6p + 14q – 13r and 15p – q + 12r.

Solution: We are required to add the first two expressions and then from the result subtract the sum of the first two expressions.
Sum of the first two expressions
  11p + 9q – 7r
– 7p – 12q + 8r
   4p – 3q + r                                                              (i)

Sum of the last two expressions
6p + 14q – 13r
15p – q + 12r
21p + 13q – r                                                         (ii)


Now, we subtract (ii) from (i)
             4p –   3q + r
            21p + 13q – r
       (–)       (–)      (+)
– 17 p – 16q + 2r

5. The perimeter of a triangle is 14a – 3b + 13c. If two side or the triangle are 4a + 6b – 8c and 3a – 5b + 11c, find the third side.
Solution: Perimeter of a triangle is the sum of its three sides.
            Third side = Perimeter – Sum of two sides
Sum of two sides = (4a + 6b – 8c) + (3a – 5b + 11c)
                                = 7a + b + 3c
             Third side = (14a – 3b + 12c) – (7a + b + 3c)
                                = 7a – 4b +9c






Multiplication of Algebraic Expressions:



There are many situations where we need to multiply algebraic expressions. Take an example, the product of speed and time gives us the distance travelled, and the product of length and breadth gives us it’s area, as well as product of two side in square give us it’s are.

When we multiply two numbers, the product that we get is also a number. For example 5 ´ 8 = 40. However, in multiplication of algebraic expressions, we deal with both numbers and variables, and the answer is also mostly a combination of variables and numbers.

Suppose you go to the market to purchase 10 oranges and the cost of each orange is Rs. 20, then your total cost is Rs.( 10 ´ 20) = Rs.200

Now, again suppose you are asked to buy A oranges and the price of each orange is Rs. x, then your total expense is Rs Ax

Next day, suppose you purchase 4 oranges less because the price of each apple has increased by Rs. 5, then your total expense is Rs (A- 4) (x + 5).

Similarly, Suppose the length of a rectangle is p units and its breadth is q units, then their product p ´ q = A (say) denotes the area of the rectangle. Now, suppse the length of the rectangle is increased by 3 units and the breadth is decreased by 5 units, then

 Length of the new rectangle = (p + 3) units
Breadth of the new rectangle = (q – 5) units
     Area of the new rectangle = (p + 3) ´ ( q – 5) sq units

We many not know the exact value of the area but we know it in terms of the variables p and q.


Multiplying A Monomial By A Monomial:


We already know
´ x = x + x + x + x + x + x = 6x
´ y ´´ y  = y4

Therefore, you have

(7a ´ 3b) = (7 ´ 3) ´ (a ´ b) = 21ab
8y ´ 3y4 = (8 ´ 3) ´ (y ´ y4) = 24y5


Therefore, in the multiplication of two monomials, the product of their coefficients becomes the coefficient of their product. Similarly, the product of their variables becomes the variable part of the product.

Example: Find the product of (12x2y3´ (-3 x3y2)  

Solution: (12x2y3´ (-3 x3y2) = (12 ´ - 3) ´ ( x2y3)(x3y2)
                                                = - 36x5y5

[Hint: Property used in simplifying the exponents of the variables being multiplied in the algebraic expression is one of the basic laws of exponents. According to this laws, am ´ an = am+n]


Multiplying more than two monomials:


In this case, the procedure of multiplication remains the same as in the preceding topic. To get the answer, multiply the coefficients with coefficients and variables with variables of the given expressions. Observe the following examples.


Example: Find the product of (4a2b2´ (9b2c2´ (3c2d2´ (- 6a2c2)

Solution: (4a2b2´ (9b2c2´ (3c2d2´ (- 6a2c2) = ( 4 ´ 9 ´ 3 ´-6) ´ ( a2b2 ´b2c´ c2d2 ´a2c2)
                                                          = (-648) ´ ( a2 ´a2´ (b2 ´ b2´ (c2 ´ c2 ´ c2´ (d2)
                                                          = - 648a4b4c6d2



Multiplying a Monomial by a Polynomial:


To multiply a binomial by a monomial, we multiply both the terms of the binomial by the monomial and then add the products. In other words, we use the distributive property according to which if a, b, c are any three rational numbers,

We have  a ´ (b+ c) = ( a ´ b) + ( a ´ c)
And         a ´ (b – c) = (a ´ b) – (a ´ c)


Example: 4a ´ (3bc + 5cd) = ( 4a ´ 3bc) + (4a ´ 5cd)
                                           = 12abc + 20 acd

Example: Multiply (m2n2 + m2n + 8) by 6mn
6mn ´ (m2n2 + m2n + 8) = (6mn ´ m2n2) + ( 6mn ´ m2n) + ( 6mn ´ 8)
                                      = 6m3n3 + 6m3n2 + 48mn



Multiplying A Binomial By A Polynomial:


To multiply a binomial by a binomial, we multiply every term of one binomial by every term of the other binomial and add the products. And also combine like terms, if any. Observe the following numerical and algebraic examples to understand this multiplication.

You can apply the following method of numerical calculation in the multiplication of a binomial by a binomial.
37 ´ 108 = ( 30 + 7) ´ (100 +8) = 30 ´ (100 +8) + 7 ´ (100 + 8)
                                                  = (30 ´ 100) + (30 ´ 8) + (7 ´ 100) + (7 ´ 8)
                                                  = 3000 + 240 + 700 + 56
                                                  = 3996

Similarly, we will multiply a binomial by a binomial.


Example: Multiply (7a + 11b) by ( 13a + 9b)
( 7a + 11b) ´ (13a + 9b) = 7a ´ (13a + 9b) + 11b ´ ( 13a + 9b)
                                       =(7a ´ 13a) + ( 7a + 9b) + (11b ´ 13a) + (11b ´ 9b)
                                        = 91a2 + 63ab + 143ab + 99b2
                                        =91a2 + 206ab + 99b2
Since ab = ba, we add the like terms 63ab and 143ba.

Saturday, April 28, 2012

Algebraic Expressios



Algebraic expressions are formed by terms which contain both Variables and constants.
                         4x + 5, 8p – 9, 10ab – 2c and 6x2 + 7 
are examples of algebraic expressions.

In the algebraic expression 4x + 5, 4x and 5 are terms of the algebraic expression. A term may be a constant or a product of two or more constants and variables. For instant, the term 4x is the product of two factors, 4 and x, while the term 5 is a constant.

The numerical factor of a term is called its coefficient. The coefficient of x in 4x + 5 is 4. In expression 10ab – 2c, the coefficient of ab is 10 and coefficient of c is – 2.
What is coefficient if m in m + 2n + 4? Is it zero or 1? It is 1.

The value of the variable p in 7p + 3 can be anything. It can be 1, 2, 3, 4, 0, - 2 , - 3 etc. The value of the expression also changes depending on the value of p. If p = 1, the value of the expression is 10; if p = 6, the value of an expression changes with the value chosen for its variable.

ALGEBRAIC EXPRESSIONS ON THE NUMBER LINE
An algebraic expression does not have any specific position on the number line. However, if the variable of the expression is assigned a place on the number line, the position of the expression relative to the variable is well defined on the number line.

For instance; if the variable y of the expression y + 8 is represented by the point. Y on the number line as shown, then the expression y + 8 is at a distance of 8 units to the right of Y. Similarly, y – 3 at a distance of 4 units to the left of Y.


                                                     Y

  <-------------------------------|---|---|---|---|---|---|---|---|---|---|---|----------------------------->
                                   Y – 3          y                                + 8



We can also represent expressions, like 3y + 3 on the number line if we know the position of the variable y on the number line.
Assume variable y is represented by Y, then 3y is represented by P, which is at a distance equal to four times the distance of Y from O. The expression 3y + 3 is represented by Q which is at a distance of 3 units to the right of 3y.



                                           O   Y                  P          Q
          <---------------------------|---|---|---|---|---|---|---|---|---|--------------------------->-
                                                                    3y         3y + 3-  



Like and unlike Terms


Terms of an algebraic expression are said to be like terms if they have the same variable. The variable must also have the same power in the said terms. However, the coefficients of the variable may be different. Terms either not having the same variable or those with same variable but unequal powers are called unlike terms
Like Terms                                                           Unlike Terms
5a, 16a, - 6a                                                        12a, 10b, 23c,
2xy, 4xy, 8xy                                                       6x3, x2, x 6x0



Monomials, Binomials and Polynomials


An algebraic expression is said to be a monomial if it contains only one term. The following expressions are all monomials.
                                                      4xy, 11abc, 7 axby, - 6xyz, 2a2 b2
An algebraic expression is said to be a binomial if it contains two unlike terms. e.g.,
                                                       ax +b , 10a + 6b, 15q2 + 6
An algebraic expression is said to be a trinomial if it contains three unlike terms. eg.,
                                                      2x + 3y +4z, 2m2 + 5n2 + o2
In general, an algebraic expression, which contains one or more terms with nonzero coefficients and variables having nonnegative integral exponents, i.e., power of the variable, is called a polynomial. e.g.,
                                                     2x + 3y +4z + 9, 5abc, 4x2 + 6y2 +13

Wednesday, April 25, 2012

Linear Equation Part-4



Example 5: Five years ago, a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages.

Solution: Suppose that son’s age five years ago be x years
                                Then father’s age five years ago = (7x) years
                                                          Son’s present age = (x + 5) years
                                                       Father’s present age = (7x + 5) years
                                            Son’s age after five years = 7x + 5 + 5
                                                                                         = (7x + 10) years
According to the given condition,
Father’s age after five years = Three times son’s age after five years
Or,                                 7x + 10 = 3(x + 10)
Or,                                   7x + 10 =3x + 30
Or,                                   7x – 3x = 30 – 10
Or,                                            4x = 20
Or,                                                x = 20/4
                                                    X = 5
       Son’s age five years ago = 5 years
                     Son’s preset age = x + 5
                                                   = 5 + 5
                                                   = 10 years
             Father’s present age = 7x  ´  5  
                                                  = 7.5 ´5
                                                  = 35 + 5
                                                  = 40 years.

Check: Son’s age after five years = 10 + 5
                                                           = 15 years
         Father’s age after five years = 40 + 5
                                                            = 45 years
       Now,                                     45 = 3(15)  
Hence,                         father’s age = three times son’s age
Therefore, the solution is correct.

Example 6: The present ages of Isha and Raghu are in the ratio 5 : 7. Four year later, their ages wil be in the ratio 3 : 4. Find their present ages.
Solution: Suppose that present ages of Isha and Raghu is 5x years and 7x respectively.
Isha’s age after four years = (5x + 4) years
Raghu’s age after four years = (7x + 4) years.
According to the given condition,
                             5x + 4/7x + 4 = 3/4
Cross-multiplying , we have
                                   4(5x + 4) = 3(7x + 4)
Or,                              20x + 16 = 21x + 12
or                           – 21x + 20x = 12 – 16
or                                          – x =  – 4
Therefore,
           Isha’s present age = 5x = 5 ´ 4 = 20 years
Raghu’s present age = 7x = 7 ´ 4 = 28 years

Check: Isha’s ager after four years  = 20 + 4
                                                                 = 24 years
              Raghu’s age after four years = 28 + 4
                                                                   = 32 years
The ratio between Isha’s and Raghu’s age is 24 : 32 = 3 : 4
Hence, the solution is correct.


Example7: A motorboat covers a certain distanc downstream in a rive in five hours. It covers the same distance upstream in six hour. The speed of water is 2 km/h. Find the speed of the boat in still water.

Solution: Suppose that the speed of the boat in still water be x km/h
                                        Speed of water  = 2 km/h
             Speed of the boat downstream = (x – 2) km/h
                    Distance covered in 5 hours  =  Speed  ´ Time
                                                           = 5(x – 2) km
(The relative speed when the direction of the boat and the folw of water is the same = The sum of the speeds of the boat and water).
Distance covered in 6 hours = Speed ´ Distance = 6(x – 2) km
Speed of the boat upstream = (x – 2) km/h
(The relative speed when the boat travels opposite to the flow of water = The difference in speeds of the boat and water.
But the boat covers the same distance upstream and downstream.
Therefore,
                                      6(x – 2) = 5(x – 2)
                                     6x – 12 = 5x – 10
Or                                 6x – 5x = 12 + 10
Or                                            x = 22
Hence, the speed of the boat in still water = 22 km/h
Check: Distance covered in 5 hours (downstream) = Speed ´ Time
                                                                                    = 5(22 + 2) = 120 km
Distance covered in 6 hours (upstream) = Speed  ´ Time
                                                                = 6(22 – 2) = 120 km
In both cases, the distance is same. Hence the solution is correct.

Friday, April 20, 2012

Linear equation part 3


Some example of Linear equation:

Example 1: The sum of three consecutive multiples of 5 is 555. Find the numbers.

Solution: Alternative Method
We can also assume the three consecutive multiples of 5 to be 5x, 5x + 5 and 5x + 10
Now, we can frame the equation as
            5x + (5x + 5) + (5x + 10) = 555
Or,                               15x + 15 = 555
Or,                                          15x = 555 – 15 = 540
Or,                                              x = 540/15 
Or,                                           x = 36
             
             Therefore, the three required multiples of 5 are 5x = 5 ´ 36 = 180
                                                5x + 5 = 180 + 5 = 185
                                                5x + 10 = 180 + 10 = 190

Example 2: The length of a rectangle is 8 cm more that its breadth. If the perimeter of the rectangle is 108 cm, find the length and breadth of the rectangle.

Solution: Perimeter of the rectangle = 108 cm                                                          (i)

Let the breadth (b) of the rectangle = x cm  

Then the length (l) of the rectangle = (x + 8) cm

Perimeter of the rectangle = 2(l + b)
                                              = 2 (x + x + 8)
= 2(2x + 8)
= 4x + 16                                                                     (ii)
From (i) and (ii), we get
                               4x + 16 = 108
Or,                                  4x = 108 – 16
                                            = 92
 Or ,                                  x = 92/4
                                            = 23
Therefore, the breadth of the rectangle = x =23 cm and the length of the rectangle
= x + 8 = 23 + 8 = 31 cm.

Example 3: Two-third of a number is greater than three-fifths of the number by 6. Find the number.

Solution: Let the number be x
Two-thirds of the number = 2x/3
Three-fifths of the number = 3x/5
According to the given condition,
2x/3 – 3x/5 =6
Multiplying all the terms by 15, which is the LCM of 5 and 3, we get
                         10x – 9x = 90
Or                                  x = 90
Therefore, the required number is 90.
              Check 2/3 of 90 = 60
                           3/5 of 90= 54
And                      60 – 54 = 6
 Hence, the solution is correct.

Example 4: the sum of the digits of a two-digit number is 12. The number formed by interchanging the digits is greater than the original number by 54. Find the original number.

Solution: Let the units digit of the original number be x
Then the tens digit of the original number = 12 – x
The original number in expanded notation is
                              = 10(12 – x) + 1 ´ x
                              = 120 – 10x + x
                              = 120 – 9x

On interchanging the digits the unit digit is now
                               = 12 – x
And the tens digit is now
                                 = x
The number obtained by interchanging the digits in expanded notation
                                 = 10x + 1 ´ (12 – x)
                           = 10x + 12 – x
                           = 9x + 12
Applying the given condition,
The number obtained on interchanging the digits – The original number = 54
ð         9x + 12 – ( 120 – 9x) = 54
ð         9x + 12 – 120 + 9x =54
ð         18x – 108 = 54
ð         18x = 54 + 108 = 162
ð             x = 162/18 = 9

The units digit of the required number
                        = x
                        = 9
The tens digit
                       = 12 – x
                       = 12 – 9 = 3
So, the original number = 39

Check: The original number = 39
Number obtained by interchanging digits = 93
Their difference = 93 – 39 = 54
Hence, the solution is correct.