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

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            14ax + 11by + 9cz + 28

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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

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                    30a + 23b – 10c – 23d

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Subtraction:

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

Solution:     6xy + 9yz – 7yz + 17

                    8sy – 5yz + 3xz + 21

                 (–)     (+)     (–)      (–)

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–         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

6 ´ x = x + x + x + x + x + x = 6x

y ´ y ´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, a^m ´ aⁿ = a^m+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 ´b²c² ´ 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.