Rd Sharma Solutions Class 7 Chapter 7 Algebraic Expressions
Welcome to NCTB Solution. Here with this post we are going to help 7th class students for the Solutions of Rd Sharma Class 7 Mathematics, Chapter 7, Algebraic Expressions. Here students can easily find Exercise wise solution for chapter 7, Algebraic Expressions. Students will find proper solutions for Exercise 7.1, 7.2, 7.3 and 7.4. Our teachers solved every problem with easily understandable methods so that every students can understand easily. Here all solutions are based on the CBSE latest curriculum.
Algebraic Expressions Exercise 7.1 Solution
Question no – (1)
Solution :
(i) a2
= Monomial
(ii) a2 – b2
= Binomial
(iii) x3 + y3 + z3
= Trinomial
(iv) x3 + y3 + z3 + 3xyz
= Quadrinomial
(v) 7 + 5
= Monomial
(vi) abc + 1
= Binomial
(vii) 3x – 2 + 5
= Binomial
(viii) 2x – 3y + 4
= Trinomial
(ix) xy + yz + zx
= Trinomial
(x) ax3 + bx2 + cx + d
= Quadrinomial
Question no – (3)
Solution :
(i) 4xy, – 5x2y, – 3yx, 2xy2
Like term = 4x, -3yx
Coefficient = 4, -3
(ii) 7a2bc, – 3ca2b, – 5/2 abc2, 3/2 abc2, – 4/3 cba2
Like term = 7a2bc, – 3ca2b, – 4/3 cba2, -5/2 abc2, 3/2 abc2
Coefficient = 7, -3, -4/3, -5/2, 3/2
Question no – (4)
Solution :
(i) a2 + b2 – 2a2 + c2 + 4a
∴ like terms = a2, -2a2
(ii) 3x + 4xy – 2yz + 5/2 zy
∴ like terms = – 2yz, 5/2 zy
(iii) abc + ab2c + 2acb2 + 3c2ab + b2ac – 2a2bc + 3cab2
∴ like terms = ab2c, 2acb2, 3cab2
Question no – (5)
Solution :
(i) Given, -12x
Here, coefficient of x is -12.
(ii) Given, -7xy
So, the the coefficient of x is -7y
(iii) Given, xyz
So, the the coefficient of x is yz
(iv) Given, -7ax
So, the the coefficient of x is -7a
Question no – (6)
Solution :
(i) Given, -3x²
∴ The the coefficient of x² is -3
(ii) In the question, 5x²yz
Hence, the the coefficient of x² is 5y
(iii) In the question, 5/7x²z
Thus, the the coefficient of x² is 5/7z
(iv) In the question, -3/2ax² + yx
So, the the coefficient of x² is -3/2a.
Question no – (7)
Solution :
(i) Given, -3y
Hence, the the coefficient of y is -3.
(ii) In the question, 2ab
Thus, the the coefficient of a is 2b
(iii) In the question, -7xyz
So, the the coefficient of z is -7xy
(iv) In the question, -3pqr
Therefore, the the coefficient of p is -3qr
(v) In the question, 9xy²z
So, the the coefficient of y² is 9xz
(vi) In the question, x³ + 1
Hence, the the coefficient of x³ is 1
(vii) In the question, -x²
Thus, the the coefficient of x² is -1
Question no – (8)
Solution :
(i) In the question, -x²
Hence, the numerical coefficient of x² is -1
(ii) In the question, -x²
Thus, the numerical coefficient of x² is -1
(iii) In the question, -x²
Therefore, the numerical coefficient of x² is -1.
Question no – (9)
Solution :
(i) In the given question,
= 4x2y – 3/2 xy + 5/2 xy2
∴ Numerical coefficient,
= 4, -3/2, 5/2
(ii) As per the question,
= -5/3 x2y + 7/4xyz + 3
∴ Numerical coefficient,
= -5/3, 7/4, +3
Question no – (10)
Solution :
(i) x2y – xy2 + 7xy – 3
Here, the constant term is -3
(ii) -a3 – 3a2 + 7a + 5
Hare, the constant term is 5.
Question no – (11)
Solution :
(i) x/y + y/z + z/x
= -2/- 1 + -1/3 + -3/2
= 2 – 1/3 – 3/2
= 12 – 2- 9/6
= 1/6
Hence the answer will be 1/6
(ii) x2 + y2 + z2 – xy – yz – zx
= (- 2)2 + (-1)2 + (3)2 – (- 2) × (- 1) – (- 1) × 3 – 3 × (- 2)
= 4 + 1 + 9 – 2 + 3 + 6
= 21
Thus, the answer will be 21.
Question no – (12)
Solution :
(i) ax + by + cz
= 1 × (-2) + 1 (-1) + 2 × (-2)
= -2 -1 -4
= -7
(ii) ax² + by² -cz²
= (2) (1)² + (1) (-2)² -(-2) (2)²
= -2 + 1 + 2 × 4
= -2 + 1 + 8
= 7
Algebraic Expressions Exercise 7.2 Solution
Question no – (1)
Solution :
(i) 3x and 7x
= 3x + 7x
= 10x
So, the answer will be 10x
(ii) -5xy and 9xy
= -5xy + 9xy
= 4xy
Hence, the answer will be 4xy.
Question no – (2)
Solution :
(i) 7x3y + 9yx3
= 16x3y…(Simplified)
(ii) 12a2b + 3ba2
= 15ab2…(Simplified)
Question no – (5)
Solution :
(i) As per the question,
= 8a – 6ab + 5b, – 6a- ab – 8b and – 4a + 2ab + 3b
∴ The sum is ,
8a – 6ab + 5b, – 6a- ab – 8b and – 4a + 2ab + 3b
= a (8 – 6 – 4) + b (- 5 8 + 3) + ab (- 6 + 2)
= 2a – 5ab
(ii) According to the question,
5x3 + 7 + 6x – 5x2, 2x2 – 8 – 9x, 4x – 2x2 + 3x3, 3x3 – 9x – x2 and x – x2 – x3 – 4
∴ The sum is,
= 5x3 + 7 + 6x – 5x2, 2x2 – 8 – 9x, 4x – 2x2 + 3x3, 3x3 – 9x – x2 and x – x2 – x3 – 4
= x2 + x – x2 + x3 – 4
= 10x3 – 7x2 – 7x – 5
Question no – (6)
Solution :
(i) In the given question,
x – 3y – 2x; 5x + 7y – 8z; 3x – 2y + 5z
∴ (x – 3y – 2x) + (5x + 7y – 8z) + (3x – 2y + 5z)
= 9x + 2y – 5z
For more better understanding,
+ x – 3y – 2x
+ 5x + 7y – 8z
+ 3x – 2y + 5z
———————————–
9x + 2y – 5z
(ii) In the given question,
= 4ab – 5bc + 7ca; -3ab + 2bc – 3ca; 5ab – 3bc + 4ca
= (4ab – 5bc + 7ca) + (- 3ab + 2bc – 3ca) + (5ab – 3bc + 4ca)
= 6ab – 6bc + 8ca
For more better understanding,
4ab – 5bc + 7ca
– 3ab + 2bc – 3ca
5ab – 3bc + 4ca
———————————–
6ab – 6bc + 8ca
Question no – (7)
Solution :
As per the question,
= 2x2 – 3x + 1 to the sum of 3x2 – 2x and 3x + 7.
∴ (3x2 – 2x + 3x + 7) + 2x2 – 3x + 1
= 5x2 – 2x + 8
Question no – (10)
Solution :
(i) In the given question,
7a2b from 3a2b
∴ 3a2b – 7a2b
= 4a2b
(ii) From question we get,
= 4xy from -3xy
∴ – 3x – 4xy
= – 7xy
Question no – (11)
Solution :
(i) Given, -4x from 3y
∴ 3y – (- 4x)
= 3y + 4x
(ii) In the question,
= -2x from -5y
∴ -5y – (-2x)
= -5y + 2x
Question no – (12)
Solution :
(i) According to the question,
= 6x3 – 7x2 + 5x – 3 from 4 – 5x + 6x2 – 8x3
∴ (4 – 5x + 6x2 – 8x3) – (6x3 – 7x2 + 5x – 3)
= 4 – 5x + 6x2 – 8x3 – 6x3+ 7x2 – 5x + 3
= -14x3 + 13x2 – 10x + 7
(ii) From the question we get,
= -x2 – 3z from 5x2 – y + z + 7
∴ (5x2 – y + z + 7) – (- x2 – 3z)
= 5x2 – y + z + 7 + x2 + 3z
= 6x2 – y + 4z + 7
Question no – (13)
Solution :
(i) In the given question,
= p3 – 4 + 3p2, take away 5p2 – 3p3 + p – 6
∴ (p3 – 4 + 3p2) – (5p2 – 3p3 + p – 6)
= p3 – 4 + 3p2 – 5p2 + 3p3 – p + 6
= 4p3 – 2p2 – p + 2
(ii) Given in the question,
7 + x – x2, take away 9 + x + 3x2 + 7x3
∴ (7 + x – x2) – (9 – x – 3x2 – 7x3)
= -7x3 – 4x2 – 2
(iii) In the question,
= 1 – 5y2, take away y3 + 7y2 + y + 1
∴ (1 – 5y2) – (y3 – 7y2 – y – 1)
= -y3 – 12y2 – y
Question no – (14)
Solution :
Sum : 3x2 – 5x + 2 – 5x2 – 8x + 9
= -2x2 – 13x + 11
Then :
= -2x2 – 13x + 11 – 4x2 + 7x – 9
= -6x2 – 6x + 2
Question no – (15)
Solution :
As per the question,
= [6x – 4y – 4z + 2x + 4y – 7] – [13x – 4y + 7z – 6z + 6x + 3y]
= 8x – 4z – 7 – 19x + y – z
= – 11x – 5z + y – 7
Question no – (16)
Solution :
As per the question,
= [x2 + 3y2 – 6xy + 2x2 – y2 + 8xy + y2 + 8 + x2 – 3xy – [- 3 3x2 + 4y2 – xy + x – y + 3]
= 4x2 + 3y2 – xy + 3x2 – 4y2 + xy – x + y – 3 + 8
= 7x2 – y2 – x + y + 5
Question no – (17)
Solution :
According to the question,
= 4xy – 3xz + 4yz + 7 – (xy – 3yz + yzx)
= 4xy – 3xz + 4yz + 7 – xy + 3yz – yzx
= 3xy + 7yz – 7zx + 7
Hence, 3xy + 7yz – 7zx + 7 should be added to get 4xy – 3zx + 4yz + 7.
Question no – (18)
Solution :
As per the question,
= x2 – xy + y2 – x + y + 3 + x2 – 3y2 + 4x – 1
= 2x2 + 3xy – 2y2 – x + y + 2
Hence, should be subtracted 2x2 + 3xy – 2y2 – x + y + 2 to obtain – x2 + 3y2 – 4xy + 1.
Question no – (19)
Solution :
According to the question,
= x – 2y + 3z – 3x – 5y + 7
= – 2x – 7y + 3z + 7
Question no – (20)
Solution :
According to the question,
= 2x2 – 3y2 + xy – x2 + 2xy – 3y2
= x2 + 3xy – 6y2
Question no – (21)
Solution :
As per the question,
= a2 – 3ab + 2b2 – 2a2 + 7ab – 9b2
= – a2 + 4ab – 7b2
Question no – (22)
Solution :
As per the question,
= x3 + 2x2 – 3xz + 2 – 12x3 + 4x2 – 3x + 7
= 11x3 + 6x2 – 6x + 9
Hence, 11x3 + 6x2 – 6x + 9 must be added to make the sum x3 + 2x2 – 3x + 2.
Question no – (23)
Solution :
From the question we get,
P = 7x2 + 5xy – 9y2,
Q = 4y2 – 3x2 – 6xy
R = – 4x2 + xy 5y2
According to the question,
= P + Q + R = 0
∴ (7x2 + 5xy – 9y2) + (4y2 + -3x2 – 6xy) + (4x2 + xy + 5y2 )
= 7x2 + 5xy – 9y2 + 4y2 + – 3x2 – 6xy – 4x2 + xy + 5y2
= 0
Question no – (24)
Solution :
From the question we get,
P = a2 – b2 + 2ab,
Q = a2 + 4b2 – 6ab,
R = b2 + b,
S = a2 – 4ab and
T = – 2a2 + b2 – ab + a
According to question,
= P + Q + R + S – T
∴ [a2 – b2 + 2ab + a2 + 4b2 – 6ab + b2 + 3 a2 – 4ab] – [- 2a2 + b2 – ab + a]
= 3a2 + 4b2 – 8ab + b + 2a2 – b2 – a
= 5a2 + 3b2 – 7ab – a + b
Algebraic Expressions Exercise 7.3 Solution
Question no – (1)
Solution :
(i) x + y – 3z + y
= x + y – (3z – y)
(ii) 3x – 2y – 5z – 4
= 3x – 2y – (5z + 4)
(iii) 3a – 2b + 4c – 5
= 3a – 2b (5 – 4c)
(iv) 7a + 3b + 2c + 4
= 7a + 3b – (- 4 – 2c)
(v) 2a2 – b2 – 3ab + 6
= 2a2 – b2 – 3 (ab + 6)
(vi) a2 + b2 – c2 + ab – 3ac
= a2 + b2 + c2 – (3ac – ab)
Question no – (2)
Solution :
(i) The sum of a – b and 3a – 2b + 5 is subtracted from 4a + 3b – 7.
= [4a – 2b + 7] – [(a – b) + (3a – 2b + 5)]
(ii) Three times the sum of 2x + y – {5 – (x – 3y)} and 7x – 4y + 3 is subtracted from 3x – 4y + 7.
= (3x – 4y + 7) – 3 [{(2x + y) – {5 – 1x – 3y)} + 7x – 4y + 3}]
(iii) The subtraction of x2 – y2 + 4xy from 2x2 + y2 – 3xy is added to 9x2 – 3y2 – xy.
= [2x2 + y2 – 3x2 – (x2 – y2 + 4xy)] + (9x2 – 3y2 – xy]
Algebraic Expressions Exercise 7.4 Solution
Question no – (1)
Solution :
Given, 2x + (5x – 3y)
∴ 2x + 5x – 3y
= 7x – 3y
Question no – (2)
Solution :
Given, 3x – (y – 2x)
∴ 3x – y + 2x
= 5x – y
Question no – (3)
Solution :
Given, 5a – (3b – 2a + 4c)
∴ 5a – 3b + 2a – 4c
= 7a – 3b – 4c
Question no – (4)
Solution :
Given, -2(x2 – y2 + xy) – 3(x2 + y2 – xy)
∴ -2x2 + 2y2 – 2xy – 3x2 – y2 + 3xy
= -5x2 – y2 + xy
Question no – (5)
Solution :
Given, 3x + 2y – {x – (2y – 3)}
∴ 3x + 2y – x + 2y – 3
= 2x + 4y – 3
Question no – (6)
Solution :
Given, 5a – {3a – (2 – a) + 4}
∴ 5a – 3a + 2 – a – 4
= a – 2
Question no – (7)
Solution :
Given, a – [b – {a – (b – 1) + 3a}]
∴ a – [b – 4a + b – 1]
= a – 2b + 4a + 1
= 5a – 2b + 1
Question no – (8)
Solution :
In the given question,
= a – [2b – {3a – (2b – 3c)}]
∴ a – [2b – 3a + 2b – 3c]
= a – 4b + 3a + 3c
= 4a – 4b – 3c
Question no – (9)
Solution :
As per the question,
-x + [5y – {2x – (3y – 5x)}]
∴ – x + [5y – {2x – 3y + 5x}]
= – x + [5y – 7x + 3y]
= – x + 8y – 7x
= + 8y – 8x
Question no – (10)
Solution :
Given in the question,
= 2a – [4b {4a – 3 (2a – b)}]
∴ 2a – [4b – {4a – 6a + 3b}]
= 2a – [4b + 2a – 3b]
= 2a – b – 2a
= – b
Question no – (11)
Solution :
Given, – a – [a + {a + b – 2a – (a – 2b)} – b]
∴ – a -[a + {a + b – 2a – a 2b – b}]
= – a – [a + (- 2a + 2b)]
= – a – + 2a – 2b
= – 2b
Question no – (12)
Solution :
In the given question,
= 2x – 3y – [3x – 2y – {x – z – (x – 2y)}]
∴ 2x – 3y – [3x – 2y – x + z + x – 2y]
= 2x – 3y – [3x – 4y + z]
= 2x – 3y – 3x + 4y – z
= – x + y – z
Question no – (13)
Solution :
In the question,
= 5 + [x – {2y – (6x + y – 4) + 2x} – {x – (y – 2)}]
∴ 5 + [x – 2y + 6x + y – 4 – 2x – x + y – 2]
= 5 + 4x – 6
= 4x – 1
Question no – (14)
Solution :
Given, x2 – [3x + {2x – (x2 – 1)} + 2]
∴ x2 – [3x + 2x – x2 +1 + 2]
= x2 + x2 – 5x – 3
= 2x2 – 5x – 3
Question no – (15)
Solution :
In the given question,
= 20 – [5xy + 3 {x2 – (xy – y) – (x – y)}]
∴ 20 – [5xy + 3x2 – 3xy + 3y – 3x + 3y]
= 20 – [2xy + 3x2 + 6y – 3x]
= 20 – 2xy – 3×2 – 6y + 3x
Question no – (16)
Solution :
Given, 85 – [12x – 7 (8x – 3) – 2 {10x – 5(2 – 4x)}]
∴ 85 – [12x – 7 (8x – 3) – 2{10x – 10 + 20x}]
= 85 – 12x + 56x + 21 + 60x – 20
= 44 + 104x
Question no – (17)
Solution :
Given in the question,
= xy – [yz – zx – {yx – (3y – xz) – (xy – zy)}]
∴ xy – [yz – zx – yx + 3y – xz + xy – zy]
= xy – yz + zx + xy – 3y + xz – xy + zy
= xy + 2xz – 3y
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