RBSE Solutions for Class 7 Maths Chapter 13 Algebraic Expression Ex 13.2

RBSE Solutions for Class 7 Maths Chapter 13 Algebraic Expression Ex 13.2 is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 13 Algebraic Expression Exercise 13.2.

Rajasthan Board RBSE Class 7 Maths Chapter 13 Algebraic Expression Ex 13.2

Question 1
Add the following (RBSESolutions.com) algebaric expressions :
(i) t – 4tz, 2t + 6tz
(ii) 7xy, 5xy, 3xy, – 2xy
(iii) 5x – 7y, 3y – 4x + 2, 2x – 3xy – 5
(iv) m² – n² – 1, n² -1 – m², 1 – m² – n²
(v) 3x + 11 + 8z, 5x – 7
(vi) a²b + ab + ab², -a²b + 2ba + 2a²b²
(vii) x – y, y – z, z – x
Solution:
(i) (t – 4tz) + (2t + 6tz) = t – 4tz + 2t + 6tz
= t + 2t – 4tz + 6tz
= (1 + 2)t + (-4 + 6)tz
= 31 + 212

(ii) 7xy + 5xy + 3xy – 2xy = (7 + 5 + 3 – 2)xy
= (15 – 2) xy
= (13)xy
= 13xy

(iii) (5x – 7y) + (3y – 4x + 2) + (2x – 3xy – 5)
= 5x – 7y + 3y – 4x + 2 + 2x – 3xy – 5
= 5x – 4x + 2x – 7y + 3y – 3xy + 2 – 5
=(5 – 4+ 2)x + (-7 + 3)y – 3xy – 3
= (7 – 4)x + (-4)y – 3xy – 3
= 3x – 4y – 3xy – 3

(iv) (m² – n² – 1) + (n² – 1 – m²) + (1 – m² – n²)
= m² – n² – 1 + n² – 1 – m² + 1 – m² – n²
= m² – m² – m² – n² + n² – n² – 1 – 1 + 1
= (1 – 1 – 1)m² + (- 1 + 1 – 1)n² – 2 + 1
= (1 – 2)m² + (- 2 + 1)n² – 1
= (-1)m² +(-1) n² – 1
= -m² – n² – 1

(v) (3x + 11 + 8z) + (5x – 7)
= 3x + 11 8z + 5x – 7
= 3x + 5x + 8z + 11 – 7
= (3 + 5)x + 8z + 4
= 8x + 8z + 4

(vi) (a²b + ab + ab²) + (-a²b + 2ba + 2a²b²)
= a²b + ab + ab² – a²b + 2ba + 2a²b²
= a²b – a²b + 2a²b² + ab² + ab + 2ba
= (1 – 1)a²b + 2a²b² + ab + ab + 2ba (ab = ba)
= (0)a²b + 2a²b² + ab² + (1 + 2)ab
= 0 + 2a²b² + ab² + 3ab
= 2a²b² + ab² + 3ab

(vii) (x – y) + (y – z) + ( Z – x)
x – y + y – z + z – x
= x – x – y + y – z + z
= (1 – 1)x + (- 1 + 1)y + 1(-1 + 1)z
= 0x + 0y + 0z
= 0 + 0 + 0
= 0

Question 2
Subtract the following (RBSESolutions.com) algebraic expression.
(i) subtract – 5x² from x²
(ii) (a – b) from (a + b)
(iii) x² + 5x + 4 from 4x² – 3xy + 8
(iv) 5x² – 7xy + 5y² from 3xy – 2x² – 2y²
(v) 4pq – 5q² – 3p² from 5p² + 2q² – pq²
(vi) x²+ 10x – 5 from 5x – 10
Solution:
(i) x² – (-5x²) = x² + 5x²
= (1 + 5) x²
= (6)x² = 6x²

(ii) (a + b) (a – b)= a + b – a + b
= a – a + b + b
= (1 – 1)a + (1 + 1)b
= 0a + 2b = 2b

(iii) (4x² – 3xy + 8) – (x² + 5x + 4)
= 4x² – 3xy + 8 – x² – 5x – 4
= 4x² – x² – 3xy – 5x + 8 – 4
= (4 – 1)x² – 3xy – 5x + 4
= 3x² – 3xy – 5x + 4

(iv) (3xy – 2x² – 2y²) – (5x² – 7xy + 5y²)
= 3xy – 2x² – 2y² – 5x² + 7xy – 5y²
= 3xy + 7xy – 2x² – 5x² – 2y² – 5y²
= (3 + 7)xy + (-2 – 5)x² + (-2 – 5)y²
= 10xy – 7x² – 7y²
= -7x² – 7y² + 10xy

(v) (5p² + 2q² – pq²) – (4pq – 5q² – 3p²)
= 5p² + 2q² – pq² – 4pq +5q² + 3p²
= 5p² + 3p² + 2q² +5q² – pq² – 4pq
= (5 + 3)p² + (2 + 5)q² – pq² – 4pq
= 8p² + 7q² – pq² – 4pq

(vi) (5x – 10) – (x² + 10x – 5)
= 5x -10 – x² – 10x +5
= -x² + 5x – 10x – 10 + 5
= -x² + (5-10)x – 5
= -x² – 5x – 5

Question 3
What should be (RBSESolutions.com) subtracted from 7x – 8y to get x + y + z?
Solution:
Let A should be subtracted from 7x – 8y to get( x+y + z)
(7x – 8y) – A = x + y + z
A = (7x – 8y) – (x + y + z)
= 7x – 8y – x – y – 2
= 6x – 9y – 2

Question 4
What should (RBSESolutions.com) be added to 2p + 6 to get 3p – q + 6 ?
Solution:
Let should be added to (2p + 6) to get 3p – q + 6.
(2p + 6) + R = (3p – 9 + 6)
R = (3p – q + 6) – (2p + 6)
= 3p – q + 6 – 2p – 6
= 3p – 2p – q + 6 – 6
= p – q + 0
Required expression = p – q

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