MP Board Class 7th Maths Solutions Chapter 12 बीजीय व्यंजक Ex 12.2

प्रश्न 1.
समान पदों को संयोजित (मिलान) करके सरल कीजिए:
(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 – 8xy2 – 3y2
(vi) (3y2 + 5y – 4) – (8y – y2 – 4)
हल:
(i) 21b – 32 + 7b – 20b
= 21b + 7b – 20b – 32
= (21 + 7 – 20) b – 32
= 8b – 32

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

(iii) p – (p – q) – q – (q – p)
= p – p + q – q – q + p
= p – p + p + q – q – q
= (1 – 1 + 1)p + (1 – 1 – 1)q
= p – q

(iv) 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
= (1)a + (0) b + (1) ab = a + ab

(v) 5x2 y – 5x2 + 3yx2 – 3y2 + x2 – y2 + 8xy2
= (5x2y + 3yx2) + (8xy2) + (-5x2 + x2) + (-3y2 – y2 – 3y2)
= (5 + 3) x2y + 8xy2 + (- 5 + 1)x2 + (- 3 – 1 – 3) y2
= 8x2y + 8xy2 – 4x2 – 7y2

(vi) (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 + 0 = 4y2 – 3y

MP Board Solutions

प्रश्न 2.
जोड़िए:
(i) 3mn, -5mn, 8mn, -4mn
(ii) t – 8tz, 3tz – z, z – 1
(iii) -7mn + 5,12mn + 2, 9mn – 8, -2 mn – 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) 3p2q2 – 4pq + 5, – 10p2q2, 15 + 9pq + 7p2q2
(ix) ab – 4a, 4b – ab, 4a – 4b
(x) x2 – y2 – 1 – y2 – 1 – x2, 1 – x2 – y2
हल:
(i) अभीष्ट योग
= 3mn + (-5mn) + 8mn + (-4mn)
= [3+ (-5) + 8 + (-4)] mn
= [11 – 9] mn = 2mn

(ii) अभीष्ट योग
= (1 – 8tz) + (3tz – z) + (z – t)
= t – 8tz + 3tz – z + z – t
= (t – t) + (-z + z) + (-8tz + 3tz)
= (1 – 1) t + (- 1 + 1) z + (- 8 + 3) tz
= (0) t + (0) z + (-5)tz = – 5tz

(iii) अभीष्ट योग = (-7mn + 5) + (12mn + 2) + (9mn – 8) + (-2mn – 3)
= -7mn + 5 + 12mn + 2 + 9mn – 8 – 2mm – 3
= (-7mm + 12mn + 9mn – 2mn) + (5 + 2 – 8 – 3)
= (-7 + 12 + 9 – 2) mn + (7 – 11)
= (21 – 9) mn + (-4) = 12mn – 4

(iv) अभीष्ट योग = (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)
= (1) a + (1) b + (3)
= a + b + 3

(v) अभीष्ट योग
= (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 + (0) y + (0) xy + 5
= 7x + 5.

(vi) अभीष्ट योग
= (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 – 3 mn – 3
= 3m – 4n – 3mn – 3

(vii) अभीष्ट योग
= 4x2y + (-3xy2) + (-5xy2) + 5x2y
= 4x2y – 3xy – 5xy2 + 5x2y
= (4 + 5) x2y + (- 3 – 5) xy2
= 9x2y – 8xy2

(viii) अभीष्ट योग
= (3p2q2 – 4pq + 5) + (-10 p2q2) + (15 + 9pq + 7p2q2)
= 3p2q2 – 4pq + 5 – 10p2q2 + 15 + 9pq + 7p2q2
= 3p2q2 – 10p2q2 + 7p2q2 – 4pq + 9pq + 5 + 15
= (3 – 10 + 7)p2q2 + (- 4 + 9) pq + (5 + 15)
= (0)p2q2 + 5pq + 20
= 5pq + 20

(ix) अभीष्ट योग
= (ab – 4a) + (4b – ab) + (4a – 4b)
= ab – 4a + 4b – ab + 4a – 4b
= ab – ab – 4a + 4a + 4b – 4b
= (0) ab + (0) a + (0) b
= 0 + 0 + 0 = 0

(x) अभीष्ट योग
= (x2 – y2 – 1) + (y2 – 1 – x2) + (1 – x2 – y2)
= x2 – y2 – 1 + y2 – 1 – x + 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

MP Board Solutions

प्रश्न 3.
घटाइए:
(i) y2 में से – 5y2
(ii) – 12xy में से 6xy
(iii) (a + b) में से (a – b)
(iv) b (5 – a) में से a (b – 5)
(v) 4m2 – 3mn + 8 में से – m2 + 5mn
(vi) 5x – 10 में से – x2 + 10x – 5
(vii) 3ab – 2a2 – 2b2 में से 5a2 – 7ab + 5b2
(viii) 5p2 + 3q2 – Pq में से 4pq – 5q2 – 3p2
हल:
(i) अभीष्ट अन्तर
y2 – (-5y2)
= y2 + 5 = 6y2

(ii) अभीष्ट अन्तर = – 12xy – 6xy = -18xy

(iii) अभीष्ट अन्तर
= (a + b) – (a – b) = a + b – a + b
= (1 – 1)a + (1 + 1) b = 2b

(iv) अभीष्ट अन्तर
= b (5 – a) – a (b – 5)
= 5b – ab – ab + 5a
= 5a + 5b + ( – 1 – 1) ab
= 5a + 5b – 2ab

(v) अभीष्ट अन्तर
= (4m2 – 3mn + 8) – (- m2 + 5mn)
= 4m2 – 3mn + 8 + m2 – 5mn
= 4m2 + m2 – 3mn – 5mn + 8
= (4 + 1)m + (- 3 – 5)mn + 8
= 5m2 – 8mn + 8

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

(vii) अभीष्ट अन्तर
= (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
= -7a – 7b2 + 10ab

(viii) अभीष्ट अन्तर
= (5p2 + 3q2 – pq) – (4pq – 5q2 – 3p2)
= 5p2 + 3q2 – pq – 4pq + 5q2 + 3p2
= 5p2 + 3p2 + 3q2 + 5q2 – pq – 4pq
= (5 + 3)p2 + (3 + 5) q2 + (- 1 – 4)pq
= 8p2 + 8q2 – 5pq

प्रश्न 4.
(a) 2x2 + 3xy प्राप्त करने के लिए x2 + xy + ya में क्या जोड़ना चाहिए?
(b) -3a + 7b + 16 प्राप्त करने के लिए 2a + 8b + 10 में से क्या घटाना चाहिए ?
हल:
(a) अभीष्ट व्यंजक
= (2x2 + 3xy) – (x2 + xy + y2)
= 2x2 + 3xy – x2 – xy – y2
= 2x2 – x2 – y2 + 3xy – xy
= (2 – 1) x2 – y2 + (3 – 1)xy
= x2 – y2 + 2xy

(b) अभीष्ट व्यंजक
= (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

MP Board Solutions

प्रश्न 5.
-x2 – y2 + 6xy + 20 प्राप्त करने के लिए, 3x2 – 4y2 + 5xy + 20 में से क्या निकाल लेना चाहिए?
हल:
अभीष्ट व्यंजक = (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 + (20 – 20)
= 4x2 – 3y2 – xy + 0
= 4x2 – 3y2 – xy

MP Board Solutions

प्रश्न 6.
(a) 3x – y + 11 और – y – 11 के योग में से 3x – y – 11 को घटाइए।
(b) 4 + 3x और 5 – 4x + 2x2 के योग में से 3x2 – 5x और – x2 + 2x + 5 के योग को घटाइए।
हल:
(a) 3x – y + 11 और – y – 11 का योग
= (3x – y + 11) + (- y – 11)
= 3x – y + 11 – y – 11
= 3x – y – y + 11 – 11
= 3x – 2y
अब 3x – 2y में से 3x – y – 11 को घटाने पर,
अभीष्ट अन्तर = (3x – 2y) – (3x – y – 11)
= 3x – 2y – 3x + y + 11
= (3x – 3x) + (-2y + y) + 11
= 0 – y + 11
= – y + 11

(b) 4 + 3x और 5 – 4x + 2x2 का योग
= 4 + 3x + 5 – 4 x + 2x2
= (4 + 5) + (3x – 4x) + 2x2
= 9 – x + 2x2
3x2 – 5x और = x2 + 2x + 5 का योग
= 3x2 – 5x – x2 + 2x + 5
= 3x2 – x2 – 5x + 2x + 5
= (3 – 1)x2 + (- 5 + 2)x + 5
= 2x2 – 3x + 5
अब, प्रश्नानुसार
अभीष्ट अन्तर = (9 – x + 2x2) – (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 ⇒ 2x +4
पाठ्य-पुस्तक पृष्ठ संख्या # 258

MP Board Class 7th Maths Solutions

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