GSEB Solutions Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2

   

Gujarat Board GSEB Textbook Solutions Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2

GSEB Solutions Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2

Question 1.
Simplify by combining like terms:
(i) 21b – 32 + 7b – 20b
(ii) -z² + 13z² – 5z + 7z3 – 15z
(iii) p – (p – q) – q – (q – p)
(iv) 3a – 2b – ab – (a – b + ab ) + 3ab + b -a
(v) 5x²y – 5x² + 3yx² – 3y² + x² – y² + 8xy² – 3y²
(vi) (3y² + 5y – 4) – (8y – y² – 4)
Solution:
(i) 21b – 32 + lb – 20b
Combining the like terms, we have
= (21b + 7b – 20b) + (-32)
= (21 + 7 – 20)b + (-32)
= (8)b + (-32) = 8b – 32

(ii) -z² + 13z² – 5z + 7z3 – 15z
Collecting the like terms, we have:
= (7z3) + (- z² + 13z²) + (- 5z – 15z)
= (7z3) + (-1 + 13)z² + (- 5 – 15)z
= (7)z3 + (12)z² + (- 20)z
= 7z3 + 12z² – 20z

(iii) p – (p – q) – q – (q – p)
We have: p – (p – q) – q – (q – p)
=p – p + q – q – q + p
Collecting the like terms, we have:
(p – p + p) + (q – q – q)
= (1 – 1 + 1)p + (1 – 1 – 1 )q
= (1)p + (- 1 )q
= p – q

(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
We have:
3a – 2b – ab – (a – b + ab) + 3ab + b – a
= 3a – 2b – ab – a + b – ab + 3ab + b – a
Collecting the like terms, we have:
(3a – a – a) + (-2b + b + b) + (- ab – ab + 3ab)
= (3 – 1 – 1)a + (- 2 + 1 + 1)b + (- 1 – 1 + 3)ab
= (1)a + (0)b + (+1)ab = a + ab

(v) 5x²y – 5x² + 3yx² – 3y² + x² – y² + 8xy² – 3y
Collecting the like terms, we have:
(5x²y + 3yx²) + (8xy²) + (- 5x² + x²) + (- 3y² – y² – 3y² )
= (5 + 3 )x² y + (8 )xy² + (- 5 + 1)x² + (- 3 1 – 3)y²
= 8x²y + 8xy² + (- 4)x² + (- 7)y²
= 8x²y + 8xy² – 4x² – 7y²

(vi) (3y² + 5y – 4) – (8y – y² – 4)
We have:
(3y² + 5y – 4) – (8y – y² – 4) = 3y² + 5y – 4 – 8y + y² + 4
Collecting the like terms, we have:
(3y² + y²) + (5y – 8y) + (- 4 + 4)
= (3 + 1)y² + (5 – 8)y + (- 4 + 4)
= (4)y² + (- 3 )y + (0)
= 4y² – 3y

GSEB Solutions Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2

Question 2.
Add:
(i) 3mn, -5mn, mn, – 4mn
(ii) t – 8tz, 3tz – z, z – t
(iii) -7mn + 5, 12mn + 2, 9mn – 8, -2mn – 3
(iv) a + b -3, b – a + 3, a – b + 3
(v) 14x+ 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
(vii) 4x²y, -3xy², -5xy², 5x²y
(viii) 3p² q² – 4pq + 5, -10p² q² , 15 + 9pq + 7p² q²
(ix) ab – 4a, 4b – ab, 4a – 4b
(x) x² – y² – 1, y² – 1 – x², 1 – x² – y²
Solution:
(i) 3mn, – 5mn, 8mn, – 4mn
We have:
3 mn + (-5 mn) + %mn + (-4 mn)
= [3 + (-5) + 8 + (-4)]mn
= [11 + (-9 )]mn
= [2]mn
= 2 mn

(ii) t – 8tz, 3tz – z, z – t
We have:
(t – 8tz) + (3tz – z) + (z – t)
= t – 8tz + 3tz – z + z – t
= (t – t) + (-z + z) + (-8 tz + 3 tz)
= (1 – 1 )t + (-1 + 1)z + (-8 + 3 )tz
= (0)t + (0)z + (-5 )tz
= -5tz

(iii) -7mn + 5, 12mn + 2, 9mn – 8, -2mn – 3
We have:
(-7mn + 5) + (12mn + 2) + (9mn – 8) + (-2mn – 3)
= -7 mn + 5 + 12mn + 2 + 9 mn – 8 – 2mn – 3
= (-7mn + 12 mn + 9 mn – 2 mn) + (5 + 2 – 8 – 3)
= (-7 + 12 + 9 – 2)mn + (5 + 2 – 8 – 3)
= (21 – 9)mn + (7 – 11)
= 12mn + (-4)
= 12mn – 4

(iv) a + b – 3, b – a + 3, a – b + 3
We have:
(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 + 6)
= (2 – 1)a + (2 – 1)b + (-3 + 6)
= (1)a + (1)b + (3)
= a + b + 3

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

(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
We have:
(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 + (-1 + 3)n – 3mn + (-3)
= (7 – 4)m + (- 4)n – 3mn – 3
= 3m – 4n – 3mn – 3

(vii) 4x²y, -3xy², -5xy², 5x²y
We have:
4x²y + (-3xy² ) + (-5xy²) + 5x²y
= 4x²y – 3xy² – 5xy² + 5x² y
= (4x²y + 5 x²y) + [(-3xy² ) + (-5xy²)]
= (4x²y + 5)x²y + [(-3) + (-5)]xy²
= (9)x²y + (-8)xy²
= 9x²y – 8xy²

(viii) 3p² q² – 4pq + 5, -10p²q², 15 + 9pq + 7p²q²
We have:
(3p²q² – 4pq + 5) + (-10p²q²) + (15 + 9pq + 7p²q² )
= 3p²q² – 4pq + 5 – 10p²q² + 15 + 9pq + 7p²q²
= (3p²q² – 10p²q² + 7p²q²) + (-4pq + 9pq) + (5 + 15)
= (3 – 10 + 7)p²q² + (-4 + 9)qp + 20
= (0)p²q² + 5pq + 20
= 5pq + 20

(ix) ab – 4a, 4b – ab, 4a – 4b
We have:
(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
We have:
(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 + (- 2 + 1)
= (1 – 2)x² + (- 2 + 1)y² + (- 1)
= (- 1)x² + (- 1)y² + (- 1)
= – x² – y² – 1

GSEB Solutions Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2

Question 3.
Subtract:
(i) -5y² from y²
(ii) 6xy from -12xy
(iii) (a – b) from (a + b)
(iv) a(b – 5) from b(5 – a)
(v) -m² + 5mn from 4m² – 3mn + 8
(vi) -x² + 10x – 5 from 5x – 10
(vii) 5a² – 7ab + 5b² from 3ab – 2a² – 2b²
(viii) 4pq – 5q² – 3p² from 5p² + 3q² – pq
Solution:
(i) Subtract -5y² from y²
We have: y² – (-5y²) = y² + 5 – y² = (1 + 5)y² = 6y²

(ii) Subtract 6xy from – 12xy We have:
– 12xy – 6xy = (- 12 – 6 )xy = – 18xy

(iii) Subtract (a – b) from (a + b)
We have:
(a + b) – (a – b) = a + b – a + b
= (a – a) + (b + b)
= (1 – 1 )a + (1 + 1 )b
= (0)a + (2)b
= 2b

(iv) Subtract a(b – 5) from 6(5 – a)
We have:
b(5 – a) – a(b – 5) = 56 – ab – ab + 5a
= 5b + 5a + (- ab – ab)
= (5b + 5a) + (-1 – 1)ab
= 5a + 5b + (- 2)ab
= 5a + 5b – 2ab

(v) Subtract – m² + 5mn from 4m² – 3mn + 8
We have:
(4m² – 3mn) + 8 – (-m² + 5mn)
= 4m² – 3mn + 8 + m² – 5mn
= (4m² + m²) + (- 3mn – 5mn) + 8
= (4 + 1)m² + (- 3 – 5)mn + 8
= (5)m² + (- 8)mn + 8
= 5m² – 8mn + 8

(vi) Subtract – x² + 10x – 5 from 5x – 10
We have:
(5x – 10) – [- x² + 10x – 5] = 5x – 10 + x² – 10x + 5
= x² + (5x – 10x) + (- 10 + 5)
= x² + (5 – 10)x + (- 5)
= x² – 5x – 5

(vii) Subtract 5a² – 7ab + 5b² from 3ab – 2a² – 2b²,
We have:
(3ab – 2a² – 2b²) – (5a² – 7ab + 5b²)
= 3ab – 2a² – 2b² – 5a² + 7ab – 5b²
= (3ab + 7ab) + (-2a² – 5a²) + (-2b² – 5b²)
= (3 + 7)ab + (- 2 – 5)a² + (-2 – 5)b²
= (10)ab + (-7)a² + (-7)b²
= 10ab – 7a² – 7b²

(viii) Subtract 4pq – 5q² – 3p² from 5p² + 3q² – pq,
We have:
(5p² + 3 q² – pq) – (4pq – 5q² – 3p²)
= 5p² + 3 q² – pq – 4pq + 5 q² + 3p²
= (5p² + 3p²) + (3q² + 5q²) + (- pq – 4pq)
= (5 + 3)p² + (3 + 5)q² + (- 1 – 4)pq
= (8)p² + (8)q² + (- 5 )pq
= 8p² + 8q² – 5pq

Question 4.
(a) What should be added to x² + xy + y² to obtain 2x² + 3xy?
(b) What should be subtracted from 2a + 86 + 10 to get – 3a + lb + 16?
Solution:
(a) The required expression = (2x² + 3xy) – (x² + xy + y²)
= 2x² + 3 xy – x² – xy – y²
= (2x² – x²) – y² + (3xy – xy)
= (2 – 1)x² – y² + (3 – 1 )xy
= (1)x² – y² + (2)xy
= x² – y² + 2xy

(b) The required expression
= (2a + 86 + 10) – (-3a + 7b + 16)
= 2a + 86 + 10 + 3a – 7b – 16
= (2a + 3a) + (8b – 7b) + (10 – 16)
= (2 + 3)a + (8 – 7)b + (-6)
= (5)a + (1)b + (-6)
= 5a + b – 6

Question 5.
What should be taken away from 3x² – 4y² + 5xy + 20 to obtain – x² – y² + 6xy + 20?
Solution:
The required expression
= (3x² – 4y² + 5xy + 20) – (-x² – y² + 6xy + 20)
= 3x² – 4y² + 5xy + 20 + x² + y² – 6xy – 20
= (3x² + x²) + (-4y² + y²) + (5xy – 6xy) + (20 – 20)
= (3 + 1)x² + (- 4 + 1)y² + (5 – 6)xy + (0)
= 4x² + (- 3)y² + (-1)xy + 0
= 4x² – 3y² – xy

GSEB Solutions Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2

Question 6.
(a) From the sum of 3x – y + 11 and – y – 11, subtract 3x – y – 11.
(b) From the sum of 4 + 3x and 5 – 4x + 2x², subtract the sum of 3x² – 5x and – x² + 2x + 5.
Solution:
(a) Sum of 3x – y + 11 and – y – 11
= (3x – y + 11) + (- y – 11)
= 3x – y + 11 – y – 11
= 3x – y – y + 11 – 11
= 3x – 2y + 0
= 3x – 2y
Now, subtracting 3x – y – 11 from 3x – 2y:
(3x – 2y) – (3x – y – 11) = 3x – 2y – 3x + y + 11
= (3x – 3x) + (- 2y + y) + 11
= 0 + (- y) + 11 = – y + 11

(b) Sum of 4 + 3x and 5 – 4x + 2x²
= 4 + 3x + 5 – 4x + 2x²
= (4 + 5) + (3x – 4x) + 2x²
= 9 + (- x) + 2x²
= 9 – x + 2x²

Sum of (3x² – 5x) and (- x² + 2x + 5)
= 3x² – 5x – x² + 2x + 5
= (3x² – x²) + (- 5x + 2x) + 5
= (3 – 1)x² + (- 5 + 2)x + 5
= 2x² + (- 3)x + 5
= 2x² – 3x + 5

Now, according to the question, we have:
(9 – x + x²) – (2x² – 3x + 5)
= 9 – x + 2x² – 2x² + 3x – 5
= (9 – 5) + (- x + 3x) + (2x² – 2x²)
= (4) + (- 1 + 3)x + (2 – 2)x²
= 4 + (2)x + (0)x²
= 4 + 2x or 2x + 4

GSEB Solutions Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2

Leave a Comment

Your email address will not be published. Required fields are marked *