NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2

Class 7 Maths NCERT Solutions Chapter 12 Algebraic Expressions Ex 12.2

Question 1.
Simplify combining like terms:

(i) 21b – 32 + 7b – 20b
Solution:
21b – 32 + 7b – 20b
= 21b + 7b – 20b – 32
= (21 + 7 – 20 )b – 32 = 8b – 32

(ii) – z² + 13 z² -5z + 7z³ – 15z
Solution:
– z² + 13 z² – 5 z +7z³ – 15z
= 7z³ – z² + 13z² – 5z – 15z
= 7z³ + (-1 + 13)z² + (- 5 – 15)z
= 7z³ +12z² – 20z

(iii) P – (p – q) – q – (q – p)
Solution:
p – (p – q) – q – (q – p)
= p – p + q – q – q + p
= (p – p + p) + (q – q – q) = p – q

(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
Solution:
3a – 2b – ab – (a – b + ab) + 3ab + b – a
= 3a – 2b – ab – a + b – ab + 3ab + b – a
= 3a – a – a – 2b + b + b – ab – ab + 3ab
= (3 -1 – 1)a + (-2 +1 +1)b + (-1 -1 + 3)ab
= a + (0)b + ab = a + ab

(v) 5x² y – 5x² + 3yx² – 3y²+ x² – y² + 8xy² – 3y²
Solution:
5x² y – 5x² + 3yx² – 3y² + x² – y² + 8xy² – 3y²
= 5x² y + 3yx² – 5x² + x² – 3y² – y² -3y² – 3y² + 8xy²
= (5 +3)x² y + (- 5 + 1)x² + (- 3 -1 – 3)y² + 8xy²
= 8x² y – 4x² – 7y² + 8xy²

(vi) (3y² + 5y – 4) – (8y – y² – 4)
Solution:
(3y² + 5y – 4) – (8y – y² – 4)
= 3y² + 5y – 4 – 8y – y² + 4
= 3y² + y² + 5y – 8y – 4 + 4
= (3 + 1)y² + (5 – 8)y + (- 4 + 4)
= 4y² – 3y

Question 2.
Add:
(i) 3mn, -5mn, 8mn, -4mn
Solution:
Required sum:
= 3mn + (-5mn) + 8mn + (-4mn)
= (3 – 5 + 8 – 4) mn
= (11 – 9) mn
= 2 mn

(ii) t – 8tz, 3tz – z, z – t
Solution:
Required sum:
= (t – 8tz) + (3tz – z) + (z – t)
= t – 8tz + 3tz – z + z – t
= t – t – 8tz + 3tz – z + z
= (1 – 1) t + (-8 + 3)tz + (-1 + 1)z
= (0) t + (-5)tz + (0)z
= 0 – 5tz + 0
= -57z

(iii) -7mn + 5, 12mn + 2, 9mn – 8, -2mn – 3
Solution:
Required sum:
= (-7mn + 5) + (12mn + 2) + (9mn – 8) + (2mn – 3)
= -7mn + 5 + 12mn + 2 + 9mn – 8 – 2mn – 3
= -7mn + 12mn + 9mn – 2mn + 5 + 2 – 8 – 3
= (-7  + 12 + 9 – 2)mn +(5 + 2 – 8 – 3)
= 12mn – 4

(iv) a + 6 – 3 ,6 – a + 3, a – 6 + 3
Solution:
Required sum:
= (a + b – 3) + (b – a + 3) + (a – b + 3)
= a + b – 3 + b – a + 3 + a – b + 3
= (a – a + a) + (b + b – b) + (- 3 + 3 + 3)
= (1 – 1 – 1)a + (1 + 1 – 1) b + 3
= a + b + 3

(v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
Solution:
Required sum:
= (14x + 10y – 12xy – 13) + (18 – 7x – 10y + 8xy) + 4xy
= 14x + 10y – 12xy – 13 + 18 – 7x -10y + 8xy + 4xy
= 14x – 7x + 10y – 10y – 12xy + 8xy + 4xy – 13  + 18
= (14 – 7)x + (10 – 10)y + (-12 +  8 + 4)xy + (- 13 + 18)
= 7x + (0)y + (0)xy + 5
= 7x + 5

(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
Solution:
Required sum:
= (5m – 7n) +(3n – 4m + 2) + (2m – 3mn – 5)
= 5m – 7n +3n – 4m + 2 + 2m – 3mn – 5
= 5m – 4m + 2m – 7n + 3n – 3mn + 2 – 5
= (5 – 4 + 2)m + (- 7 + 3)n – 3mn + (2 – 5)
= 3m – 4n – 3mn – 3

(vii) 4x² y, – 3xy², – 5xy², 5x²y
Solution:
Required sum:
= 4x² y (-3xy²) + (- 5xy²) + 5x² y
= 4x² y – 3xy² – 5xy² + 5x² y
= 4x² y + 5x² y – 3xy² – 5xy²
= (4 +5)x²y + (- 3 – 5)xy²
= 9x² y – 8xy²

(viii) 3p² q² – 4pq + 5, – 10p²q², 15 + 9 pq + 7p²q²
Solution:
Required sum:
= (3p²q² – 4pq +5)+ (- 10p²q²) + (15 +9 pq +7p²q²)
= 3p²q² – 4 pq + 5 – 10p²q² +15 – 9 pq + 7p²q²
= 3p²q² -10p²q² +7 p²q² – 4 pq + 9 pq +5 +15
= (3 – 10 + 7)p²q² + (- 4 + 9) pq + (5 +15)
= (0)p²q² + 5pq + 20
= 0 + 5pq + 20 = 5pq + 20

(ix) ab – 4a, 4b – ab, 4a – 4b
Solution:
Required sum:
=(ab – 4a) + (4b – ab) + (4a – 4b)
= ab – 4a + 4b – ab + 4a – 4b
= ab – ab – 4a + 4a + 4b – 4b
= (1 – 1)ab + (- 4 + 4)a + (4 – 4)b
= (0)ab + (0)a + (0)b = 0 + 0 + 0 = 0

(x) x² – y² – 1, y² – 1 – x²,1 – x² – y²
Solution:
Required sum:
= (x² – y² – 1) + (y² -1 – x²) + (1 – x² – y²)
= x² – y² – 1 + y² – 1 – x² + 1 – x² – y²
= x² – x² – x² – y² + y² – y² -1 -1 +1
= (1 – 1 – 1)x² + (- 1 + 1 – 1)y² +(-1-1+1)
= x² – y² – 1

Question 3.
Subtract:

(i) – 5y² from y²
Solution:
The required difference is given by
y² – (- 5y²) = y² + 5y² = (1 + 5)y² = 6y²

(ii) 6xy from – 12xy
Solution:
The required difference is given by
(- 12xy) – 6xy = (-12 – 6)xy
= – 18xy

(iii) (a – b) from (a + b)
Answer:
The required difference is given by
(a + b) – (a – b) = a + b- a + b
= a – a + b + b
= (1 – 1 )a + (1 + 1)b = 2b

(iv) a (b – 5) from b (5 – a)
Solution:
The required difference is given by
b (5 – a) – a (b – 5) = 5b – ab – ab + 5a
= 5a + 5b + (-1 – 1)ab
= 5a + 5b – 2 ab

(v) – m² 5mn from 4m² – 3mn + 8
Solution:
The required difference is given by
(4m² – 3mn + 8) – (-m² + 5mn)
= 4m² – 3mn + 8 + m² – 5mn
= 4m² + m² – 3mn – 5mn + 8
= (4 + 1)m² + (- 3 – 5)mn + 8
= 5m² – 8 mn + 8

(vi) – x² + 10x – 5 from 5x – 10
Solution:
The required difference is given by
(5x -10) – (- x² + 10x – 5)
= 5x -10 + x² -10x + 5
= x² + (5 – 10)x + (- 10 + 5)
= x² – 5x – 5

(vii) 5a² – 7ab + 5b² from 3ab – 2a² – 2b²
Solution:
The required difference is given by
(3ab – 2a² – 2b²) – (5a² – 7ab + 5b²)
= 3ab – 2a² – 2b² 5a² +7ab – 5b²
= – 2a² – 5a² – 2b² – 5b² + 3ab + 7ab
= (- 2 – 5)a² + ( -2 – 5)b² + (3 + 7)ab
= -7a² – 7b² +10ab

(viii) 4pq – 5q² – 3p² from 5p² + 3q² – pq
Solution:
The required difference is given by
(5p²+3q² – pq) – (4pq – 5q² – 3p²)
= 5p² + 3q² – pq – 4pq , 5q² + 3p²
= 5p² + 3q² + 5q² – pq – 4pq
= (5 + 3)p² + (3 + 5)q² +(- 1 – 1 – 4)pq
= 8p² + 8q² – 5pq

Question 4.
(a) What should be added to x² + xy + y² to obtain 2x 2 + 3xy?
Solution:
Required expression is equal to the subtraction of x² + xy + y² from 2x² + 3xy.
Hence, required expression
= (2x² + 3xy) – (x² + xy + y²)
= 2x² + 3xy – x² – xy – y²
= 2x² – x² +3xy – xy – y²
= (2 – 1)x² + (3-1 ) xy – y²
= x² + 2xy – y²

(b) What should be subtracted from 2a + 8b +10 to get – 3a + 7b + 16?
Solution:
Let P denote the required expression, then (2a + 8b +10) – P = – 3a + 7b +16
Hence, required expression P
= (2a + 8b + 10) – (-3a + 7b + 16)
= 2a + 8b +10 + 3a – 7b -16
= 2a + 3a + 8b-7b+ 10 – 16
= (2 + 3)a + (8 – 7)b + (10 -16)
= 5a + b – 6

Question 5.
What should be taken away from 3x² – 4y² + 5xy + 20 to obtain – x² – y² + 6xy + 20 ?
Solution:
Let P denote the required expression, then
(3x² – 4y² + 5xy + 20) – P = (- x² – y² + 6xy + 20)
Hence, required expression P
= (3x² – 4y² + 5xy + 20) – (-x² – y² + 6xy + 20)
= 3x² – 4y² + 5xy + 20 + x² + y² – 6xy -20
=3x² + x² – 4y² +y² + 5xyy – 6xy + 20 – 20
= (3 +1)x² + (- 4 +1)y2 +(5 -6)xy + (20 – 20)
= 4x² – 3y² – xy

Question 6.
(a) From the sum of 3x – y +11 and -y -11, subtract 3 x – y – 11.
Solution:
The sum of 3x – y +11 and -y -11 is given by
(3x – y + 11) + (- y – 11) = 3x – y + 11 -y – 11 = 3x – 2y
Now, we have to subtract 3x – y -11 from 3x – 2y
∴ Required expression = (3x – 2y) – (3x – y -11)
= 3x – 2y- 3x + y + 11 = -y + 11

(b) From the sum of 4 + 3x and 5 – 4x + 2x², subtract the sum of 3x² – 5x and – x² + 2x + 5.
Solution:
The sum of 4 + 3x and 5 – 4x + 2x² is given by
(4 + 3x) + (5 – 4x + 2x²) = 4 + 3x + 5 – 4x + 2x²
= 9 – x + 2x²
The sum of 3xThe sum of 4 + 3x and 5 – 4x + 2x² is given by
(4 + 3x) + (5 – 4x + 2x²) = 4 + 3x + 5 – 4x + 2x²
= 9 – x + 2x² – 5x and – x² + 2x + 5 is given by
(3x² – 5x) + (-x² + 2x + 5)
= 2x² – 3x + 5
Now, we have to subtract 2x² – 3x + 5 from 9 – x + 2x²
∴ Required expression = (9 – x + 2x²) – (2x² – 3x + 5)
= 9 – x + 2x² – 2x² + 3x – 5
= 2x + 4