if b* a is 20 find all the possible pairs of integral values that a and b can take
Answers
Answer:
1×20
20×1
-1×-20
-20×-1
2×10
10×2
-2×-10
-10×-2
4×5
5×4
-4×-5
-5×-4
Hope it's clear.
Step-by-step explanation:
ab = a + b + 20
ab = a + b + 20→ ab – b = a + 20 → b(a – 1) = a + 20
ab = a + b + 20→ ab – b = a + 20 → b(a – 1) = a + 20→ b = (a + 20) / (a – 1) ………. (1)
ab = a + b + 20→ ab – b = a + 20 → b(a – 1) = a + 20→ b = (a + 20) / (a – 1) ………. (1)Further, ab = a + b + 20
ab = a + b + 20→ ab – b = a + 20 → b(a – 1) = a + 20→ b = (a + 20) / (a – 1) ………. (1)Further, ab = a + b + 20→ ab – a = b + 20 → a(b – 1) = b + 20
ab = a + b + 20→ ab – b = a + 20 → b(a – 1) = a + 20→ b = (a + 20) / (a – 1) ………. (1)Further, ab = a + b + 20→ ab – a = b + 20 → a(b – 1) = b + 20→ a = (b + 20) / (b – 1) ………. (2)
ab = a + b + 20→ ab – b = a + 20 → b(a – 1) = a + 20→ b = (a + 20) / (a – 1) ………. (1)Further, ab = a + b + 20→ ab – a = b + 20 → a(b – 1) = b + 20→ a = (b + 20) / (b – 1) ………. (2)Note the similarity in equations (1) and (2).
ab = a + b + 20→ ab – b = a + 20 → b(a – 1) = a + 20→ b = (a + 20) / (a – 1) ………. (1)Further, ab = a + b + 20→ ab – a = b + 20 → a(b – 1) = b + 20→ a = (b + 20) / (b – 1) ………. (2)Note the similarity in equations (1) and (2).For “b” to be a positive integer, the various values that the integer “a” can take are 2, 4, 8 and 22, for which the values of integer “b” are 22, 8, 4 and 2 based on equation (1).
ab = a + b + 20→ ab – b = a + 20 → b(a – 1) = a + 20→ b = (a + 20) / (a – 1) ………. (1)Further, ab = a + b + 20→ ab – a = b + 20 → a(b – 1) = b + 20→ a = (b + 20) / (b – 1) ………. (2)Note the similarity in equations (1) and (2).For “b” to be a positive integer, the various values that the integer “a” can take are 2, 4, 8 and 22, for which the values of integer “b” are 22, 8, 4 and 2 based on equation (1).As the value of integer “a” increases, the value of integer “b” reduces. Further, from equation (2), we see that the least value that integer “b” can take is 2.
ab = a + b + 20→ ab – b = a + 20 → b(a – 1) = a + 20→ b = (a + 20) / (a – 1) ………. (1)Further, ab = a + b + 20→ ab – a = b + 20 → a(b – 1) = b + 20→ a = (b + 20) / (b – 1) ………. (2)Note the similarity in equations (1) and (2).For “b” to be a positive integer, the various values that the integer “a” can take are 2, 4, 8 and 22, for which the values of integer “b” are 22, 8, 4 and 2 based on equation (1).As the value of integer “a” increases, the value of integer “b” reduces. Further, from equation (2), we see that the least value that integer “b” can take is 2.Hence, both integers “a” and “b” cannot take values greater than 22.
ab = a + b + 20→ ab – b = a + 20 → b(a – 1) = a + 20→ b = (a + 20) / (a – 1) ………. (1)Further, ab = a + b + 20→ ab – a = b + 20 → a(b – 1) = b + 20→ a = (b + 20) / (b – 1) ………. (2)Note the similarity in equations (1) and (2).For “b” to be a positive integer, the various values that the integer “a” can take are 2, 4, 8 and 22, for which the values of integer “b” are 22, 8, 4 and 2 based on equation (1).As the value of integer “a” increases, the value of integer “b” reduces. Further, from equation (2), we see that the least value that integer “b” can take is 2.Hence, both integers “a” and “b” cannot take values greater than 22.Thus, we see that there are only four pairs of positive integers (a, b) satisfying the given requirement and these ordered pairs are (2, 22), (4, 8), (8, 4) and (22, 2).