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• RD Sharma Class 8 Solutions for Maths

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

(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
= (2×5)/(3×5)a + (3×3)/(5×3)a – (6×3)/(5×3)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
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
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:

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

(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

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

On rearranging,

-7/4x3 – 1/5x2 – 3/5x2 – x/3 – x/2 + 7/2 – 9/2

By grouping similar expressions we get,

LCM of (3 and 2 is 6)

-7/4x3 – 4/5x2– (2x-3x)/6 + (7-9)/2

-7/4x3 – 4/5x2 – 5/6x – 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

(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

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