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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)

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  1. anonymous
    • 5 years ago
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    hmmmm. well, we can set up a series of equations to model this

  2. anonymous
    • 5 years ago
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    let's call the the three numbers A, B and C

  3. anonymous
    • 5 years ago
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    i need to Define the variable, write an equation, and solve the equation

  4. anonymous
    • 5 years ago
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    since they are consecutive integers, we know that: 1. B = A+1 2. C = B +1

  5. anonymous
    • 5 years ago
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    now we also know that: 1. B x C = 10A + 20

  6. anonymous
    • 5 years ago
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    so to start solving, how about we put everything in terms of A

  7. anonymous
    • 5 years ago
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    Since they are three consecutive integers, we know B=A+1, C=B+1, and hence C=A+2

  8. anonymous
    • 5 years ago
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    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

  9. anonymous
    • 5 years ago
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    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

  10. anonymous
    • 5 years ago
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    Since they asked for positive integers, we know that A=9 is the correct answer, since -2 is negative

  11. anonymous
    • 5 years ago
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    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

  12. anonymous
    • 5 years ago
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    thank you!

  13. anonymous
    • 5 years ago
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    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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