class 7th exercise 12.2 first
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i am 10th standard send the question I don't have textbook
Answer:
here is your 12.2 exercise with solutions!!
Step-by-step explanation:
1. Simplify combining like terms:
(i) 21b – 32 + 7b – 20b
Solution:-
When term have the same algebraic factors, they are like terms.
Then,
= (21b + 7b – 20b) – 32
= b (21 + 7 – 20) – 32
= b (28 – 20) – 32
= b (8) – 32
= 8b – 32
(ii) – z2 + 13z2 – 5z + 7z3 – 15z
Solution:-
When term have the same algebraic factors, they are like terms.
Then,
= 7z3 + (-z2 + 13z2) + (-5z – 15z)
= 7z3 + z2 (-1 + 13) + z (-5 – 15)
= 7z3 + z2 (12) + z (-20)
= 7z3 + 12z2 – 20z
(iii) p – (p – q) – q – (q – p)
Solution:-
When term have the same algebraic factors, they are like terms.
Then,
= p – p + q – q – q + p
= p – q
(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
Solution:-
When term have the same algebraic factors, they are like terms.
Then,
= 3a – 2b – ab – a + b – ab + 3ab + b – a
= 3a – a – a – 2b + b + b – ab – ab + 3ab
= a (1 – 1- 1) + b (-2 + 1 + 1) + ab (-1 -1 + 3)
= a (1 – 2) + b (-2 + 2) + ab (-2 + 3)
= a (1) + b (0) + ab (1)
= a + ab
(v) 5x2y – 5x2 + 3yx2 – 3y2 + x2 – y2 + 8xy2 – 3y2
Solution:-
When term have the same algebraic factors, they are like terms.
Then,
= 5x2y + 3yx2 – 5x2 + x2 – 3y2 – y2 – 3y2
= x2y (5 + 3) + x2 (- 5 + 1) + y2 (-3 – 1 -3) + 8xy2
= x2y (8) + x2 (-4) + y2 (-7) + 8xy2
= 8x2y – 4x2 – 7y2 + 8xy2
(vi) (3y2 + 5y – 4) – (8y – y2 – 4)
Solution:-
When term have the same algebraic factors, they are like terms.
Then,
= 3y2 + 5y – 4 – 8y + y2 + 4
= 3y2 + y2 + 5y – 8y – 4 + 4
= y2 (3 + 1) + y (5 – 8) + (-4 + 4)
= y2 (4) + y (-3) + (0)
= 4y2 – 3y