anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
hmmmm. well, we can set up a series of equations to model this
anonymous
  • anonymous
let's call the the three numbers A, B and C
anonymous
  • anonymous
i need to Define the variable, write an equation, and solve the equation

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anonymous
  • anonymous
since they are consecutive integers, we know that: 1. B = A+1 2. C = B +1
anonymous
  • anonymous
now we also know that: 1. B x C = 10A + 20
anonymous
  • anonymous
so to start solving, how about we put everything in terms of A
anonymous
  • anonymous
Since they are three consecutive integers, we know B=A+1, C=B+1, and hence C=A+2
anonymous
  • anonymous
so 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
  • anonymous
so: 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)(A-9) = 0 A = -2 or A = 9
anonymous
  • anonymous
Since they asked for positive integers, we know that A=9 is the correct answer, since -2 is negative
anonymous
  • anonymous
so 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
  • anonymous
thank you!
anonymous
  • anonymous
Let'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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