Question

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)

Answer 1

hmmmm. well, we can set up a series of equations to model this

Answer 2

let's call the the three numbers A, B and C

Answer 3

i need to Define the variable, write an equation, and solve the equation

Answer 4

since they are consecutive integers, we know that:
1. B = A+1
2. C = B +1

Answer 5

now we also know that:
1. B x C = 10A + 20

Answer 6

so to start solving, how about we put everything in terms of A

Answer 7

Since they are three consecutive integers, we know B=A+1, C=B+1, and hence C=A+2

Answer 8

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

Answer 9

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

Answer 10

Since they asked for positive integers, we know that A=9 is the correct answer, since -2 is negative

Answer 11

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

Answer 12

thank you!

Answer 13

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!