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anonymous
 5 years ago
help me with this problem: Find three consecutive positive integers such that the product of the second and third integers is twenty more than ten time the first integer. (HEEEEELP)
anonymous
 5 years ago
help me with this problem: Find three consecutive positive integers such that the product of the second and third integers is twenty more than ten time the first integer. (HEEEEELP)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmmmm. well, we can set up a series of equations to model this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let's call the the three numbers A, B and C

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i need to Define the variable, write an equation, and solve the equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since they are consecutive integers, we know that: 1. B = A+1 2. C = B +1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now we also know that: 1. B x C = 10A + 20

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so to start solving, how about we put everything in terms of A

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Since they are three consecutive integers, we know B=A+1, C=B+1, and hence C=A+2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so we now we need to substitute to make everything in that second part: B x C = 10A + 20, be interms of A. Namely, let's replace B with A+1, and C with A+2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so: 1. (A+1)(A+2) = 10A + 20 2. A^2 + A +2A +2 = 10A + 20 3. A^2 + 3A 10A +2 = 20 4. A^2  7A + 2 = 20 5. A^2 7A 18 = 0 6. (A+2)(A9) = 0 A = 2 or A = 9

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Since they asked for positive integers, we know that A=9 is the correct answer, since 2 is negative

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so if A=9, then we can plug it back into our other equations. A=9, B = A+1, so B=10 C = B+1, so c+11 So the answer is 9,10,11

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let's also check it. So they said that "the product of the second and third integers is twenty more than ten time the first integer" So 11x10 = 10x9 + 20 110 = 90 +20 110 = 110 . yep, np!
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