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Question Number 45291 by naka3546 last updated on 11/Oct/18

How many possible triple of (a,b,c) ∈ integers so that : ∣ a + b ∣ + c = 19 ab + ∣ c ∣ = 97

Howmanypossibletripleof(a,b,c)∈integerssothat:∣a+b∣+c=19ab+∣c∣=97

Answered by MJS last updated on 12/Oct/18

c=19−∣a+b∣ c=97−ab ∨ c=ab−97 19−∣a+b∣=97−ab ∨ 19−∣a+b∣=ab−97 ∣a+b∣=ab−78 ∨ ∣a+b∣=116−ab a+b=ab−78 ∨ a+b=78−ab ∨ a+b=116−ab ∨ a+b=ab−116 b=((a+78)/(a−1)) ∨ b=((78−a)/(a+1)) ∨ b=((116−a)/(a+1)) ∨ b=((a+116)/(a−1)) case 1 a=0 b=−78 no solution for c a=2 b=80 no solution for c case 2 a=−80 b=−2 no solution for c a=0 b=78 no solution for c case 3 a=−118 b=−2 no solution for c a=−40 b=−4 no solution for c a=−14 b=−10 no solution for c a=0 b=116 c=−97 a=2 b=38 c=−21 a=8 b=12 c=−1 case 4 a=−116 b=0 c=−97 a=−38 b=−2 c=−21 a=−12 b=−8 c=−1 a=2 b=118 no solution for c a=4 b=40 no solution for c a=10 b=14 no solution for c ⇒ because a and b are interchangeable we have 12 solutions

c=19−∣a+b∣c=97−ab∨c=ab−9719−∣a+b∣=97−ab∨19−∣a+b∣=ab−97∣a+b∣=ab−78∨∣a+b∣=116−aba+b=ab−78∨a+b=78−ab∨a+b=116−ab∨a+b=ab−116b=a+78a−1∨b=78−aa+1∨b=116−aa+1∨b=a+116a−1case1a=0b=−78nosolutionforca=2b=80nosolutionforccase2a=−80b=−2nosolutionforca=0b=78nosolutionforccase3a=−118b=−2nosolutionforca=−40b=−4nosolutionforca=−14b=−10nosolutionforca=0b=116c=−97a=2b=38c=−21a=8b=12c=−1case4a=−116b=0c=−97a=−38b=−2c=−21a=−12b=−8c=−1a=2b=118nosolutionforca=4b=40nosolutionforca=10b=14nosolutionforc⇒becauseaandbareinterchangeablewehave12solutions

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