find three consecutive positive numbers such that the product of the first and third minus the second, is 1 more than 6 times the third

" find three consecutive positive numbers"

Let the middle integer be M.

"such that the product of the first and third minus the second,"
(M-1)*(M+1) - M

"is 1 more than 6 times the third"
= 6(M+1)+1

The resulting formula is therefore
(M-1)(M+1)-M = 6(M+1)+1

This equation has rational factors, or you could use the quadratic formula to find the solution. The answer for M is less than 10.

I still don't understand it

This is worked exactly the same way as mathmate did it but uses a little different terminology. Perhaps this will help.

These are consecutive numbers; therefore, if we let x = first number, x + 1 = y is the second number and y+1 = z = the third number.
Then product of 1st and 3rd is
xz and that minus the second makes it
xz-y and that = 6(3rd)+1 but the 3rd is 6z so
xz-y=6z+1

Now just substitute for x = y-1 and for z = y+1 and solve for y (which is the middle number).
(y-1)(y-1) - y = 6(y+1)+1
y=??
y-1 = x
y+1 = z

To solve this problem, let's represent the three consecutive positive numbers as n, n+1, and n+2.

According to the given information, "the product of the first and third minus the second is 1 more than 6 times the third." We can write this as an equation:

(n)(n+2) - (n+1) = 6(n+2) + 1

To solve the equation, we will simplify and solve for n:

n^2 + 2n - (n+1) = 6n + 12 + 1
n^2 + 2n - n - 1 = 6n + 13
n^2 + n - 1 = 6n + 13
n^2 - 5n - 14 = 0

Next, we can factor or use the quadratic formula to solve for n. Since we are looking for positive consecutive numbers, we can discard the negative solution.

Factoring:
(n - 7)(n + 2) = 0

Setting each factor to zero:
n - 7 = 0 or n + 2 = 0

Solving each equation:
n = 7 or n = -2

Since n represents a positive number, we can discard n = -2.

Therefore, the three consecutive positive numbers are 7, 8, and 9.