the product of the second and third of three consecutive integers is 2 more than 12 time the first integer

Answer 1

Hi Aura, thanks for the question, these always used to give me a hard time.

So we have three numbers, I'll call them f, s, and t (for first, second, and third)

They tell us that s x t = 2 + 12f

"The product of the second and third (s x t) is two more than 12 times the first (f)"

We also know that these are three consecutive integers, in other words, each number is 1 greater than the last,

so we can say that s = f + 1,

and that t = f + 2, since it's one more than s

Now the reason why we're turning everything into an equation that has only f in it is so that we can solve for it, since it would be a little (just a liiiiiittle bit) difficult to solve if we had all three variables in the same equation.

Now we put them all together:

s x t = 2 + 12f

(f + 1) x t = 2 + 12f (substituting s = ...)

(f + 1) x (f + 2) = 2 + 12f (t = ...)

f² + 3f + 2 = 2 + 12f (multiply them together (FOIL))

f² + 3f + 2 - 2 - 12f = 0 (switching everything to one side so we can get it more tidy)

f² - 9f= 0

f(f - 9) = 0 (factor out an f)

and now f can be one of two numbers, 0, or 9. It's usually a good idea to check the answers you get to make sure that they fit what we're looking for.

We know we just found out f, the first number, so now we can find s and t by adding 1 and 2 to f, respectively.

s = (0) + 1 = 1 to check 0

s = (9) + 1 = 10 to check 9

t = (0) + 2 = 2

t = (9) + 2 = 11

Now let's see if the original equation holds true for both or not:

s x t = 2 + 12f

(1) x (2) = 2 + 12(0) √ so 0 works as the first number. (which means 0, 1, 2 do too)

2 = 2 + 0

(10) x (11) = 2 + 12(9) √ and 9 works as well (9, 10, 11)

110 = 2 + 108

So after all of this, we know that those three consecutive numbers could be either 0, 1, and 2; or 9, 10, and 11

Answer 2

This is an interesting way of asking an algebra problem through words.

There are two major things we know so we can start assigning values to our main variable.

"The product of the second and third of three consecutive numbers is 2 more than some value"

AND

"That value is 12 times the first integer."

So, we have 3 consecutive integers (ex: 1, 2, 3 or 5, 6, 7 etc.) we just don't know them yet. And the second half of our problem refers back to the first of those three numbers again. So let's call that first number X.

For those 3 consecutive integers we can now call them: x, x+1, and x+2.

The second half of our question says that the value is 12 times the first integer, so, 12x, great.

The first half says the product (multiplication) of the second and third integers (x+1 and x+2) equals 2 more than that. So... (x+1)(x+2) = 12x+2

We could stop there, or continue to simplify and now solve for x (and the rest of the 3 consecutive numbers) if that is what the question is looking for.

SIMPLIFYING:

(x+1)(x+2) => x^2 + 3x + 2

x^2 + 3x + 2 = 12x + 2

-12x -12x

x^2 - 9x + 2 = 2

-2 -2

x^2 - 9x = 0

x (x - 9) = 0

x = 0 AND x - 9 = 0

x = 0 AND x = 9

CHECK

x = 0

(0+1)(0+2) = 12(0)+2

1*2 = 0+2

2 = 2 (checks out)

x = 9

(9+1)(9+2) = 12(9) +2

10*11 = 108+2

110 = 110 (checks out)

if x = 0 and x = 9, our consecutive numbers are 0, 1, 2 AND 9, 10, 11