# RBSE Solutions for Class 7 Maths Chapter 13 बीजीय व्यंजक Ex 13.2

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

 Board RBSE Textbook SIERT, Rajasthan Class Class 7 Subject Maths Chapter Chapter 13 Chapter Name बीजीय व्यंजक Exercise Ex 13.2 Number of Questions 4 Category RBSE Solutions

## Rajasthan Board RBSE Class 7 Maths Chapter 13 बीजीय व्यंजक Ex 13.2

प्रश्न 1
निम्नलिखित बीजीय व्यंजकों को जोड़िए।
(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
हल:
(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

प्रश्न 2
निम्नलिखित बीजीय व्यंजकों(RBSESolutions.com)को घटाइए।
(i) x2 में से -5x2
(ii) (a + b) में से (a – b)
(iii) 4x2 – 3xy +8 में से x2 + 5x + 4
(iv) 3xy – 2x2 – 2y2 2 5x2 – 7xy + 5y2
(v) 5p2+ 2q2 – pq2 में से 4pq – 5q2 – 3p2
(vi) 5x – 10 में से x2 + 10x – 5
हल:
(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

प्रश्न 3
x + y + 2 प्राप्त करने(RBSESolutions.com)के लिए 7x – 8y में से क्या घटाना चाहिए?
हल:
(x + y + z) प्राप्त करने के लिए 7x – 8y में से माना A घटाना चाहिए।
(7x – 8y) – A = x + y + z
A = (7x – 8y) – (x + y + z)
= 7x – 8y – x – y – 2
= 6x – 9y – 2

प्रश्न 4
2p + 6 में क्या जोड़े कि 3p – q + 6 प्राप्त हो जाए?
हल:
(2p + 6) में माना R जोड़ा जाए।
(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
आवश्यक व्यंजक = p – q

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