RD Sharma Solutions for Class 8 Maths Exercise 6.2 of Chapter 6 Algebraic Expressions and Identities are available here in simple PDF. Students can download the RD Sharma Solutions PDF from the links provided below. BYJU’S expert team have prepared the solutions in an easily understandable manner which helps students score good marks in the final exam. In Exercise 6.2 of Chapter 6, Algebraic Expressions and Identities, we will discuss the problems based on the addition and subtraction of algebraic expressions.
RD Sharma Solutions for Class 8 Maths Exercise 6.2 Chapter 6 Algebraic Expressions and Identities
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EXERCISE 6.2 PAGE NO: 6.5
1. Add the following algebraic expressions:
(i) 3a2b, -4a2b, 9a2b
(ii) 2/3a, 3/5a, -6/5a
(iii) 4xy2 – 7x2y, 12x2y -6xy2, -3x2y + 5xy2
(iv) 3/2a – 5/4b + 2/5c, 2/3a – 7/2b + 7/2c, 5/3a + 5/2b – 5/4c
(v) 11/2xy + 12/5y + 13/7x, -11/2y – 12/5x – 137xy
(vi) 7/2x3 – 1/2x2 + 5/3, 3/2x3 + 7/4x2 – x + 1/3, 3/2x2 -5/2x -2
Solution:
(i) 3a2b, -4a2b, 9a2b
Let us add the given expression
3a2b + (-4a2b) + 9a2b
3a2b – 4a2b + 9a2b
8a2b
(ii) 2/3a, 3/5a, -6/5a
Let us add the given expression
2/3a + 3/5a + (-6/5a)
2/3a + 3/5a – 6/5a
Let us take LCM for 3 and 5, which is 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
Let us add the given expression
4xy2 – 7x2y + 12x2y – 6xy2 – 3x2y + 5xy2
Upon rearranging
12x2y – 3x2y – 7x2y – 6xy2 + 5xy2 + 4xy2
3xy2 + 2x2y
(iv) 3/2a – 5/4b + 2/5c, 2/3a – 7/2b + 7/2c, 5/3a + 5/2b – 5/4c
Let us add the given expression
3/2a – 5/4b + 2/5c + 2/3a – 7/2b + 7/2c + 5/3a + 5/2b – 5/4c
Upon rearranging
3/2a + 2/3a + 5/3a – 5/4b – 7/2b + 5/2b + 2/5c + 7/2c – 5/4c
By taking LCM for (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 – 13/7xy
Let us add the given expression
11/2xy + 12/5y + 13/7x + -11/2y – 12/5x – 13/7xy
Upon rearranging
11/2xy – 13/7xy + 13/7x – 12/5x + 12/5y -11/2y
By taking 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 – 1/2x2 + 5/3, 3/2x3 + 7/4x2 – x + 1/3, 3/2x2 -5/2x – 2
Let us add the given expression
7/2x3 – 1/2x2 + 5/3 + 3/2x3 + 7/4x2 – x + 1/3 + 3/2x2 -5/2x – 2
Upon rearranging
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
2. Subtract:
(i) -5xy from 12xy
(ii) 2a2 from -7a2
(iii) 2a-b from 3a-5b
(iv) 2x3 – 4x2 + 3x + 5 from 4x3 + x2 + x + 6
(v) 2/3y3 – 2/7y2 – 5 from 1/3y3 + 5/7y2 + y – 2
(vi) 3/2x – 5/4y – 7/2z from 2/3x + 3/2y – 4/3z
(vii) x2y – 4/5xy2 + 4/3xy from 2/3x2y + 3/2xy2 – 1/3xy
(viii) ab/7 -35/3bc + 6/5ac from 3/5bc – 4/5ac
Solution:
(i) -5xy from 12xy
Let us subtract the given expression
12xy – (- 5xy)
5xy + 12xy
17xy
(ii) 2a2 from -7a2
Let us subtract the given expression
(-7a2) – 2a2
-7a2 – 2a2
-9a2
(iii) 2a-b from 3a-5b
Let us subtract the given expression
(3a – 5b) – (2a – b)
3a – 5b – 2a + b
a – 4b
(iv) 2x3 – 4x2 + 3x + 5 from 4x3 + x2 + x + 6
Let us 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
Let us subtract the given expression
1/3y3 + 5/7y2 + y – 2 – 2/3y3 + 2/7y2 + 5
Upon rearranging
1/3y3 – 2/3y3 + 5/7y2 + 2/7y2 + y – 2 + 5
By grouping similar expressions, we get,
-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
Let us subtract the given expression
2/3x + 3/2y – 4/3z – (3/2x – 5/4y – 7/2z)
Upon rearranging
2/3x – 3/2x + 3/2y + 5/4y – 4/3z + 7/2z
By grouping similar expressions, we get,
LCM for (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
Let us subtract the given expression
2/3x2y + 3/2xy2 – 1/3xy – (x2y – 4/5xy2 + 4/3xy)
Upon rearranging
2/3x2y – x2y + 3/2xy2 + 4/5xy2 – 1/3xy – 4/3xy
By grouping similar expressions, we get,
LCM for (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
Let us subtract the given expression
3/5bc – 4/5ac – (ab/7 – 35/3bc + 6/5ac)
Upon rearranging
3/5bc + 35/3bc – 4/5ac – 6/5ac – ab/7
By grouping similar expressions, we get,
LCM for (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
3. Take away:
(i) 6/5x2 – 4/5x3 + 5/6 + 3/2x from x3/3 – 5/2x2 + 3/5x + 1/4
(ii) 5a2/2 + 3a3/2 + a/3 – 6/5 from 1/3a3 – 3/4a2 – 5/2
(iii) 7/4x3 + 3/5x2 + 1/2x + 9/2 from 7/2 – x/3 – x2/5
(iv) y3/3 + 7/3y2 + 1/2y + 1/2 from 1/3 – 5/3y2
(v) 2/3ac – 5/7ab + 2/3bc from 3/2ab -7/4ac – 5/6bc
Solution:
(i) 6/5x2 – 4/5x3 + 5/6 + 3/2x from x3/3 – 5/2x2 + 3/5x + 1/4
Let us subtract the given expression
1/3x3 – 5/2x2 + 3/5x + 1/4 – (6/5x2 – 4/5x3 + 5/6 + 3/2x)
Upon rearranging
1/3x3 + 4/5x3 – 5/2x2 – 6/5x2 + 3/5x – 3/2x + 1/4 – 5/6
By grouping similar expressions, we get,
LCM for (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
Let us subtract the given expression
1/3a3 – 3/4a2 – 5/2 – (5/2a2 + 3/2a3 + a/3 – 6/5)
Upon rearranging
1/3a5 – 3/2a3 – 3/4a2 – 5/2a2 – a/3 – 5/2 + 6/5
By grouping similar expressions, we get,
LCM for (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
Let us subtract the given expression
7/2 – x/3 – 1/5x2 – (7/4x3 + 3/5x2 + 1/2x + 9/2)
Upon rearranging
-7/4x3 – 1/5x2 – 3/5x2 – x/3 – x/2 + 7/2 – 9/2
By grouping similar expressions, we get,
LCM for (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
Let us subtract the given expression
1/3 – 5/3y2 – (1/3y3 + 7/3y2 + 1/2y + 1/2)
Upon rearranging
-1/3y3 – 5/3y2 – 7/3y2 – 1/2y + 1/3 – 1/2
By grouping similar expressions, we get,
LCM for (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
Let us subtract the given expression
3/2ab – 7/4ac – 5/6bc – (2/3ac – 5/7ab + 2/3bc)
Upon rearranging
3/2ab + 5/7ab – 7/4ac – 2/3ac – 5/6bc – 2/3bc
By grouping similar expressions, we get,
LCM for (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
4. Subtract 3x – 4y – 7z from the sum of x – 3y + 2z and -4x + 9y – 11z.
Solution:
The sum of x – 3y + 2z and -4x + 9y – 11z is
(x – 3y + 2z) + (-4x + 9y – 11z)
Upon rearranging
x – 4x – 3y + 9y + 2z – 11z
-3x + 6y – 9z
Now, Let us subtract the given expression from -3x + 6y – 9z
(-3x + 6y – 9z) – (3x – 4y – 7z)
Upon rearranging
-3x – 3x + 6y + 4y – 9z + 7z
-6x + 10y – 2z
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
Upon rearranging
3l + 2l – 4m + 3m – 7n2 – 4n2
5l – m – 11n2 ……………………..equation (1)
Sum of 9l + 2m – 3n2 and -3l + m + 4n2
9l + 2m – 3n2 + (-3l + m + 4n2)
Upon rearranging
9l – 3l + 2m + m – 3n2 + 4n2
6l + 3m + n2 ……………………….equation (2)
Let us subtract equation (i) from (ii), and we get
6l + 3m + n2 – (5l – m – 11n2)
Upon rearranging
6l – 5l + 3m + m + n2 + 11n2
l + 4m + 12n2
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
Upon rearranging
– x2 + 7x2 + 2x – 4x + 5 – 3
6x2 -2x + 2 ………….equation (i)
Let us subtract equation (i) from 5 we get,
5 – (6x2 -2x + 2)
5 – 6x2 + 2x – 2
3 + 2x – 6x2
7. Simplify each of the following:
(i) x2 – 3x + 5 – 1/2(3x2 – 5x + 7)
(ii) [5 – 3x + 2y – (2x – y)] – (3x – 7y + 9)
(iii) 11/2x2y – 9/4xy2 + 1/4xy – 1/14y2x + 1/15yx2 + 1/2xy
(iv) (1/3y2 – 4/7y + 11) – (1/7y – 3 + 2y2) – (2/7y – 2/3y2 + 2)
(v) -1/2a2b2c + 1/3ab2c – 1/4abc2 – 1/5cb2a2 + 1/6cb2a – 1/7c2ab + 1/8ca2b
Solution:
(i) x2 – 3x + 5 – 1/2(3x2 – 5x + 7)
Upon rearranging
x2 – 3/2x2 – 3x + 5/2x + 5 – 7/2
By grouping similar expressions, we get,
LCM for (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)
5 – 3x + 2y – 2x + y – 3x + 7y – 9
Upon rearranging
– 3x – 2x – 3x + 2y + y + 7y + 5 – 9
By grouping similar expressions, we get,
-8x + 10y – 4
(iii) 11/2x2y – 9/4xy2 + 1/4xy – 1/14y2x + 1/15yx2 + 1/2xy
Upon rearranging
11/2x2y + 1/15x2y – 9/4xy2 – 1/14xy2 + 1/4xy + 1/2xy
By grouping similar expressions, we get,
LCM for (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)
Upon rearranging
1/3y2 – 2y2 – 2/3y2 – 4/7y – 1/7y – 2/7y + 11 + 3 – 2
By grouping similar expressions, we get,
LCM for (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
Upon rearranging
-1/2a2b2c – 1/5a2b2c + 1/3ab2c + 1/6ab2c – 1/4abc2 – 1/7abc2 + 1/8a2bc
By grouping similar expressions, we get,
LCM for (2 and 5 is 10), (3 and 6 is 6), and (4 and 7 is 28)
-7/10a2b2c + 1/2ab2c – 11/28abc2 + 1/8a2bc
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