Class 8 RD Sharma Solutions- Chapter 6 Algebraic Expressions And Identities -Β Exercise 6.2

Last Updated : 11 Sep, 2024

Exercise 6.2 in Chapter 6 of RD Sharma's Class 8 mathematics textbook typically focuses on the multiplication of algebraic expressions. This exercise builds upon the foundational concepts introduced in Exercise 6.1 and helps students develop skills in multiplying monomials, binomials, and polynomials. Understanding these operations is crucial for manipulating more complex algebraic expressions and solving equations in future studies.

Question 1. Add the following algebraic expressions:

(i) 3a2b, -4a2b, 9a2b

Solution:

3a2b, -4a2b, 9a2b
Now we have add the given expression
= 3a2b + (-4a2b) + 9a2b
= 3a2b – 4a2b + 9a2b
= 8a2b

(ii) 2/3a, 3/5a, -6/5a

Solution:

We have to add the given expression
2/3a + 3/5a + (-6/5a)
2/3a + 3/5a – 6/5a
Now take LCM for 3 and 5 which will be 15
= (2Γ—5)/(3Γ—5)a + (3Γ—3)/(5Γ—3)a – (6Γ—3)/(5Γ—3)a
= 10/15a + 9/15a – 18/15a
= (10a+9a-18a)/15
= a/15

(iii) 4xy2 – 7x2y, 12x2y -6xy2, -3x2y + 5xy2

Solution:

We have to add the given expression
4xy2 – 7x2y + 12x2y – 6xy2 – 3x2y + 5xy2
Now rearrange the expression:
12x2y – 3x2y – 7x2y – 6xy2 + 5xy2 + 4xy
3xy2 + 2x2y

(iv) 3/2a – 5/4b + 2/5c, 2/3a – 7/2b + 7/2c, 5/3a + 5/2b – 5/4c

Solution:

3/2a – 5/4b + 2/5c, 2/3a – 7/2b + 7/2c, 5/3a + 5/2b – 5/4c
Now add the given expression
3/2a – 5/4b + 2/5c + 2/3a – 7/2b + 7/2c + 5/3a + 5/2b – 5/4c
rearrange 
3/2a + 2/3a + 5/3a – 5/4b – 7/2b + 5/2b + 2/5c + 7/2c – 5/4c
Now take LCM of (2 and 3 is 6), (4 and 2 is 4), (5,2 and 4 is 20)
(9a+4a+10a)/6 + (-5b-14b+10b)/4 + (8c+70c-25c)/20
23a/6 – 9b/4 + 53c/20

(v) 11/2xy + 12/5y + 13/7x, -11/2y – 12/5x – 137xy

Solution:

11/2xy + 12/5y + 13/7x, -11/2y – 12/5x – 13/7xy
Now add the given expression
11/2xy + 12/5y + 13/7x + -11/2y – 12/5x – 13/7xy
Now rearrange

11/2xy – 13/7xy + 13/7x – 12/5x + 12/5y -11/2y
Now take LCM for (2 and 7 is 14), (7 and 5 is 35), (5 and 2 is 10)
(11xy-12xy)/14 + (65x-84x)/35 + (24y-55y)/10
51xy/14 – 19x/35 – 31y/10

(vi) 7/2x3 – 1/2x3 + 5/3, 3/2x3 + 7/4x2 – x + 1/3, 3/2x2 -5/2x -2

Solution:

Now add the given expression
7/2x3 – 1/2x2 + 5/3 + 3/2x3 + 7/4x2 – x + 1/3 + 3/2x2 -5/2x – 2
Now rearrange
=7/2x3 + 3/2x3 – 1/2x2 + 7/4x2 + 3/2x2 – x – 5/2x + 5/3 + 1/3 – 2
=10/2x3 + 11/4x2 – 7/2x + 0/6
=5x3 + 11/4x2 -7/2x

Question 2. Subtract:

(i) -5xy from 12xy

Solution:

Subtract the given expression
= 12xy – (- 5xy)
= 5xy + 12xy
= 17xy

(ii) 2a2 from -7a2

Solution:

Subtract the given expression

= (-7a2) – 2a2
= -7a2 – 2a2
= -9a2

(iii) 2a-b from 3a-5b

Solution:

Subtract the given expression
=(3a – 5b) – (2a – b)
= 3a – 5b – 2a + b
= a – 4b

(iv) 2x3 – 4x2 + 3x + 5 from 4x3 + x2 + x + 6

Solution:

Subtract the given expression

(4x3 + x2 + x + 6) – (2x3 – 4x2 + 3x + 5)
4x3 + x2 + x + 6 – 2x3 + 4x2 – 3x – 5

2x3 + 5x2 – 2x + 1

(v) 2/3y3 – 2/7y2 – 5 from 1/3y3 + 5/7y2 + y – 2

Solution:

Subtract the given expression
1/3y3 + 5/7y2 + y – 2 – 2/3y3 + 2/7y2 + 5
On rearranging,
1/3y3 – 2/3y3 + 5/7y2 + 2/7y2 + y – 2 + 5
We will group similar expression:
= -1/3y3 + 7/7y2 + y + 3
= -1/3y3 + y2 + y + 3

(vi) 3/2x – 5/4y – 7/2z from 2/3x + 3/2y – 4/3z

Solution:

Subtract the given expression
2/3x + 3/2y – 4/3z – (3/2x – 5/4y – 7/2z)
On rearranging,
2/3x – 3/2x + 3/2y + 5/4y – 4/3z + 7/2z
We will group similar expression:
LCM of (3 and 2 is 6), (2 and 4 is 4), (3 and 2 is 6)
=(4x-9x)/6 + (6y+5y)/4 + (-8z+21z)/6
= -5x/6 + 11y/4 + 13z/6

(vii) x2y – 4/5xy2 + 4/3xy from 2/3x2y + 3/2xy2 – 1/3xy

Solution:

Subtract the given expression
2/3x2y + 3/2xy2 – 1/3xy – (x2y – 4/5xy2 + 4/3xy)
on rearrange
2/3x2y – x2y + 3/2xy2 + 4/5xy2 – 1/3xy – 4/3xy
We will group similar expression:
LCM of (3 and 1 is 3), (2 and 5 is 10), (3 and 3 is 3)
-1/3x2y + 23/10xy2 – 5/3xy

(viii) ab/7 - 35/3bc + 6/5ac from 3/5bc – 4/5ac

Solution:

Subtract the given expression
3/5bc – 4/5ac – (ab/7 – 35/3bc + 6/5ac)
On rearrange
3/5bc + 35/3bc – 4/5ac – 6/5ac – ab/7
We will group similar expression:
LCM of (5 and 3 is 15), (5 and 5 is 5)
(9bc+175bc)/15 + (-4ac-6ac)/5 – ab/7
184bc/15 + -10ac/5 – ab/7
– ab/7 + 184bc/15 – 2ac

Question 3. Take away:

(i) 6/5x2 – 4/5x3 + 5/6 + 3/2x from x3/3 – 5/2x2 + 3/5x + 1/4

Solution:

Subtract the given expression
1/3x3 – 5/2x2 + 3/5x + 1/4 – (6/5x2 – 4/5x3 + 5/6 + 3/2x)
On rearrange
1/3x3 + 4/5x3 – 5/2x2 – 6/5x2 + 3/5x – 3/2x + 1/4 – 5/6
By grouping similar expressions we get,
LCM of (3 and 5 is 15), (2 and 5 is 10), (5 and 2 is 10), (4 and 6 is 24)
17/15x3 – 37/10x2 – 9/10x – 14/24
17/15x3 – 37/10x2 – 9/10x – 7/12

(ii) 5a2/2 + 3a3/2 + a/3 – 6/5 from 1/3a3 – 3/4a2 – 5/2

Solution:

Subtract the given expression
1/3a3 – 3/4a2 – 5/2 – (5/2a2 + 3/2a3 + a/3 – 6/5)
On rearrange
1/3a5 – 3/2a3 – 3/4a2 – 5/2a2 – a/3 – 5/2 + 6/5
By grouping similar expressions we get,
LCM of (3 and 2 is 6), (4 and 2 is 4), (2 and 5 is 10)
= (2a3 – 9a3)/6 – (3a2 + 10a2)/4 – a/3 + (-25+12)/10
= -7/6a3 – 13/4a2 – a/3 – 13/10

(iii) 7/4x3 + 3/5x2 + 1/2x + 9/2 from 7/2 – x/3 – x2/5

Solution:

Subtract the given expression

7/2 β€“ x/3 β€“ 1/5x2 β€“ (7/4x3 + 3/5x2 + 1/2x + 9/2)

On rearranging,

-7/4x3 β€“ 1/5x2 β€“ 3/5x2 β€“ x/3 β€“ x/2 + 7/2 β€“ 9/2

By grouping similar expressions we get,

LCM of (3 and 2 is 6)

-7/4x3 β€“ 4/5x2– (2x-3x)/6 + (7-9)/2

-7/4x3 β€“ 4/5x2 β€“ 5/6x β€“ 1

(iv) y3/3 + 7/3y2 + 1/2y + 1/2 from 1/3 – 5/3y2

Solution:

Subtract the given expression
1/3 – 5/3y2 – (1/3y2 + 7/3y2 + 1/2y + 1/2)
On rearrange
-1/3y3 – 5/3y2 – 7/3y2 – 1/2y + 1/3 – 1/2
By grouping similar expressions we get,
LCM of (3 and 3 is 3), (3 and 2 is 6)
-1/3y3 + (-5y2 – 7y2)/3 – 1/2y + (2-3)/6 
-1/3y3 – 12/3y2 – 1/2y – 1/6

(v) 2/3ac – 5/7ab + 2/3bc from 3/2ab -7/4ac – 5/6bc

Solution:

Subtract the given expression
3/2ab – 7/4ac – 5/6bc – (2/3ac – 5/7ab + 2/3bc)
On rearrange
3/2ab + 5/7ab – 7/4ac – 2/3ac – 5/6bc – 2/3bc
By grouping similar expressions we get,
LCM of (2 and 7 is 14), (4 and 3 is 12), (6 and 3 is 6)
(21ab+10ab)/14 – (21ac-8ac)/12 – (5bc-4bc)/6
31/14ab – 29/12ac – 3/2bc

Question 4. Subtract 3x – 4y – 7z from the sum of x – 3y + 2z and -4x + 9y – 11z.  

Solution:

First we will find the sum:
The sum of x – 3y + 2z and -4x + 9y – 11z is
(x – 3y + 2z) + (-4x + 9y – 11z)
On rearrange
x – 4x – 3y + 9y + 2z – 11z
= -3x + 6y – 9z
Now Let's subtract it from -3x + 6y – 9z
(-3x + 6y – 9z) – (3x – 4y – 7z)
On rearranging again 
= -3x – 3x + 6y + 4y – 9z + 7z
= -6x + 10y – 2z

Question 5. Subtract the sum of 3l – 4m – 7n2 and 2l + 3m – 4n2 from the sum of 9l + 2m – 3n2 and -3l + m + 4n2.

Solution:

Sum of 3l – 4m – 7n2 and 2l + 5m – 4n2
3l – 4m – 7n2 + 2l + 3m – 4n2
On rearrange
3l + 2l – 4m + 3m – 7n2 – 4n2
5l – m – 11n2 ……………………..eq. (1)
Sum of 9l + 2m – 3n2 and -3l + m + 4n2
9l + 2m – 3n2 + (-3l + m + 4n2)
On rearrange
9l – 3l + 2m + m – 3n2 + 4n2
6l + 3m + n2 ……………………….eq. (2)
Let us subtract equ (i) from (ii), we get
6l + 3m + n2 – (5l – m – 11n2)
On rearrange
6l – 5l + 3m + m + n2 + 11n2
l + 4m + 12n2

Question 6. Subtract the sum of 2x – x2 + 5 and -4x – 3 + 7x2 from 5.

Solution:

Sum of 2x – x2 + 5 and -4x – 3 + 7x2 is
2x – x2 + 5 + (-4x – 3 + 7x2)
2x – x2 + 5 – 4x – 3 + 7x2
On rearrange
– x2 + 7x2 + 2x – 4x + 5 – 3
6x2 -2x + 2 …………eq (i)
Let subtract eq (i) from 5 we will get,
5 – (6x2 -2x + 2)
5 – 6x2 + 2x – 2
3 + 2x – 6x2

Question 7. Simplify each of the following:

(i) x2 – 3x + 5 – 1/2(3x2 – 5x + 7)

Solution:

x2 – 3x + 5 – 1/2(3x2 – 5x + 7)
On rearrange
x2 – 3/2x2 – 3x + 5/2x + 5 – 7/2
We will group similar expression:
LCM of (1 and 2 is 2)
= (2x2 – 3x2)/2 – (6x + 5x)/2 + (10-7)/2
= -1/2x2 – 1/2x + 3/2

(ii) [5 – 3x + 2y – (2x – y)] – (3x – 7y + 9)

Solution:

5 – 3x + 2y – 2x + y – 3x + 7y – 9
On rearrange
= – 3x – 2x – 3x + 2y + y + 7y + 5 – 9
We will group similar expression:
= -8x + 10y – 4

(iii) 11/2x2y – 9/4xy2 + 1/4xy – 1/14y2x + 1/15yx2 + 1/2xy

Solution:

On rearrange
11/2x2y + 1/15x2y – 9/4xy2 – 1/14xy2 + 1/4xy + 1/2xy
We will group similar expression:
LCM of (2 and 15 is 30), (4 and 14 is 56), (4 and 2 is 4)
= (165x2y + 2x2y)/30 + (-126xy2 – 4xy2)/56 + (xy + 2xy)/4 
= 167/30x2y – 130/56xy2 + 3/4xy
= 167/30x2y – 65/28xy2 + 3/4xy

(iv) (1/3y2 – 4/7y + 11) – (1/7y – 3 + 2y2) – (2/7y – 2/3y2 + 2)

Solution:

On rearrange
1/3y2 – 2y2 – 2/3y2 – 4/7y – 1/7y – 2/7y + 11 + 3 – 2
We will group similar expression:
LCM of (3, 1 and 3 is 3), (7, 7 and 7 is 7)
= (y2 – 6y2 + 2y2)/3 – (4y – y – 2y)/7 + 12
= -3/3y2 – 7/7y + 12
= -y2 – y + 12

(v) -1/2a2b2c + 1/3ab2c – 1/4abc2 – 1/5cb2a2 + 1/6cb2a – 1/7c2ab + 1/8ca2b

Solution:

On rearrange
-1/2a2b2c – 1/5a2b2c + 1/3ab2c + 1/6ab2c – 1/4abc2 – 1/7abc2 + 1/8a2bc
We will group similar expression:
LCM of (2 and 5 is 10), (3 and 6 is 6), (4 and 7 is 28)
-7/10a2b2c + 1/2ab2c – 11/28abc2 + 1/8a2bc

Summary

In this exercise, students learn to multiply various types of algebraic expressions. They practice multiplying monomials by monomials, monomials by polynomials, and polynomials by polynomials. The exercise emphasizes the use of the distributive property and the proper combination of like terms. Students also learn to recognize and apply special product formulas, such as the square of a binomial and the product of the sum and difference of two terms.

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