BSE Odisha 7th Class Maths Solutions Chapter 6 ବୀଜଗଣିତ Ex 6.2

Odisha State Board BSE Odisha 7th Class Maths Solutions Chapter 6 ବୀଜଗଣିତ Ex 6.2 Textbook Exercise Questions and Answers.

BSE Odisha Class 7 Maths Solutions Chapter 6 ବୀଜଗଣିତ Ex 6.2

Question 1.
ସଦୃଶ ପଦଗୁଡ଼ିକୁ ଏକାଠି କରି ସରଳ କର ।

(କ) 21b – 74 + 3b – 2a
ସମାଧାନ:
21b – 7a – 3b – 2a
= 21b – 3b – (-7a) – (-2a)
= (21 – 3)b – {(-7) – (-2)} a
= 24b – (-9a) = 24b – 9a

(ଖ) -z² +  13z² – 5z +  7z³ – 15z
ସମାଧାନ:
-z² +  13z² – 5z +  7z³ – 15z
= 7z³ – z² + 13z² – 5z – 15z
= 7z³ – (-1 + 13)z² + {(-5) + (-15)}z
= 7z³ + 112z² – 20z

(ଗ) 3a – 2b – c – 5b – 6c – 2a
ସମାଧାନ:
3a – 2b – c – 5b – 6c – 2a
= 3a – 2a- 2b – 5b – c – 6c
= (3 + 2)a + (-2 – 5)b + (-1 + 6)c
= 5a – 7b + 5c

(ଘ) 6ab + 2a – 3ab – ab + 5a
ସମାଧାନ:
6ab + 2a – 3ab – ab + 5a
= 6ab – 3ab – ab + 2a + 5a
= (6 – 3 – 1)ab + (2 + 5)a
= 2ab + 7a

(ଙ) 5x²y – 5x² + 3yx² – 3y² +  x² + y² + 4xy² – 2y²
ସମାଧାନ:
5x²y – 5x² + 3yx² – 3y² +  x² + y² + 4xy² – 2y²
= 5x²y + 3yx² – 5x² + x² – 3y² + y² – 2y² + 4xy²
= (5 + 3)x²y + (-5 + 1)x² + (-3 + 1 – 2)y² + 4xy²
= 8x²y – 4x² – 4y² + 4xy²

Question 2.
ଯୋଗଫଳ ସ୍ଥିର କର ।

(କ) 3mn, -5mn, 8mn, -4mn
ସମାଧାନ:
ଯୋଗଫଳ = 3mn – 5mn + 8mn – 4mn
= (3 – 5 + 8 – 4)mn = (11 – 9)mn = 2mn

(ଖ) 5a, 8a, -9a, -2a
ସମାଧାନ:
ଯୋଗଫଳ = 5a + 8a – 9a – 2a = (5 + 8 – 9 – 2)a = 2a

(ଗ) a + b – 3, b – 2a + 3
ସମାଧାନ:
ଯୋଗଫଳ = a + b – 3 + b – 2a + 3 = a – 2a + b + b – 3 + 3
= {1 + (-2)}a + (1 + 1)b + {(-3) + 3} = -1 × a + 2b + 0 = 2b – a

(ଘ) -7mn + 5, 2mn + 2
ସମାଧାନ:
ଯୋଗଫଳ = -7mn + 5 + 2mn + 2 = -7mn + 2mn + 5 + 2
= (-7 + 2)mn + (5 + 2) = -5mn + 7

(ଙ) x² – 2y + 3, 3y² + 5y – 7
ସମାଧାନ:
ଯୋଗଫଳ = x² – 2y + 3 + 3y² + 5y – 7 = x² + 3y² + (-2y) + 5y + 3 – 7
= x² + 3y² + (-2 + 5)y + (3 – 7) = x² + 3y² + 3y – 4

(ଚ) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy
ସମାଧାନ:
ଯୋଗଫଳ = 14x + 10y – 12xy – 13 + 18 – 7x – 10y + 8xy
= 14x – 7x + 10y – 10y – 12xy + 8xy – 13 + 18
= (14 – 7)x + (10 – 10)y + (-12 + 8)xy + (18 – 13)
= 7x + 0 . y – 4xy + 5 = 7x – 4xy + 5

(ଛ) 5m – n + 5, 3m + 4n – 1
ସମାଧାନ:
ଯୋଗଫଳ = 5m – n + 5 + 3m + 4n – 1 = 5m + 3m – n + 4n + 5 – 1
= (5 + 3)m + (-1 + 4)n + (5 – 1) = 8m + 3n + 4

(ଜ) x² – y² – 1, y² – 1 – x², 1 – x²y²
ସମାଧାନ:
ଯୋଗଫଳ = x² – y² – 1 + y² – 1 – x² + 1 – x²y²
= x² – x² – y² + y² – 1 – 1 + 1 – x²y²
= (x – x²) + (-y² + y²) + (-1 + 1 )- 1 – x²y²
= 0 + 0 + 0 – 1 – x²y²
= -x²y² – 1