2) Subtract the sum of x2 - 5xy + 2y2 and y2" 5xy - 2x2 from the sum of 6x2 - 4xy
ya and 4xy + 3y2– 2x?.
Answers
Answer:
first you add this 6x2 - 4xy
ya and 4xy + 3y2– 2x?.
and subtract answer from x2 - 5xy + 2y2
Chapter 12
NCERT Book with important points highlighted and links to past board questions
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Algebraic Expressions
12.1 Introduction
We have already come across simple algebraic expressions like x + 3, y – 5, 4x + 5,
10y – 5 and so on. In Class VI, we have seen how these expressions are useful in formulating puzzles and problems. We have also seen examples of several expressions in the chapter on simple equations.
Expressions are a central concept in algebra. This Chapter is devoted to algebraic expressions. When you have studied this Chapter, you will know how algebraic expressions are formed, how they can be combined, how we can find their values and how they can be used.
12.2 How are Expressions Formed?
We now know very well what a variable is. We use letters x, y, l, m, ... etc. to denote variables. A variable can take various values. Its value is not fixed. On the other hand, a constant has a fixed value. Examples of constants are: 4, 100, –17, etc.
We combine variables and constants to make algebraic expressions. For this, we use the operations of addition, subtraction, multiplication and division. We have already come across expressions like 4x + 5, 10y – 20. The expression 4x + 5 is obtained from the variable x, first by multiplying x by the constant 4 and then adding the constant 5 to the product. Similarly, 10y – 20 is obtained by first multiplying y by 10 and then subtracting 20 from the product.
The above expressions were obtained by combining variables with constants. We can also obtain expressions by combining variables with themselves or with other variables.
Look at how the following expressions are obtained:
x2, 2y2, 3x2 – 5, xy, 4xy + 7
(i) The expression x2 is obtained by multiplying the variable x by itself;
x × x = x2
Just as 4 × 4 is written as 42, we write x × x = x2. It is commonly read as x squared.
(Later, when you study the chapter ‘Exponents and Powers’ you will realise that x2 may also be read as x raised to the power 2).
In the same manner, we can write x × x × x = x3
Commonly, x3 is read as ‘x cubed’. Later, you will realise that x3 may also be read as x raised to the power 3.
x, x2, x3, ... are all algebraic expressions obtained from x.
(ii) The expression 2y2 is obtained from y:
2y2 = 2 × y × y
Here by multiplying y with y we obtain y2 and then we multiply y2 by the constant 2.
(iii) In (3x2 – 5) we first obtain x2, and multiply it by 3 to get 3x2.
From 3x2, we subtract 5 to finally arrive at 3x2 – 5.
(iv) In xy, we multiply the variable x with another variable y. Thus, x × y = xy.
(v) In 4xy + 7, we first obtain xy, multiply it by 4 to get 4xy and add 7 to 4xy to get the expression.
Try These
TryThese2L
Describe how the following expressions are obtained:
7xy + 5, x2y, 4x2 – 5x
between two like terms is a like term with a numerical coefficient equal to the difference between the numerical coefficients of the two like terms.
Note, unlike terms cannot be added or subtracted the way like terms are added or subtracted. We have already seen examples of this, when 5 is added to x, we write the result as (x + 5). Observe that in (x + 5) both the terms 5 and x are retained.
Similarly, if we add the unlike terms 3xy and 7, the sum is 3xy + 7.
If we subtract 7 from 3xy, the result is 3xy – 7
Adding and subtracting general algebraic expressions
Let us take some examples:
Add 3x + 11 and 7x – 5
The sum = 3x + 11 + 7x – 5
Now, we know that the terms 3x and 7x are like terms and so also are 11 and – 5.
Further 3x + 7x = 10 x and 11 + (– 5) = 6. We can, therefore, simplify the sum as:
The sum = 3x + 11 + 7x – 5
= 3x + 7x + 11 – 5 (rearranging terms)
= 10x + 6
Hence, 3x + 11 + 7x – 5 = 10x + 6
Add 3x + 11 + 8z and 7x – 5.
The sum = 3x + 11 + 8z + 7x – 5
= 3x + 7x + 11 – 5 + 8z (rearranging terms)
Note we have put like terms together; the single unlike term 8z will remain as it is.
Therefore, the sum = 10x + 6 + 8z
Subtract a – b from 3a – b + 4
The difference = 3a – b + 4 – (a – b)
= 3a – b + 4 – a + b