Question

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

Answer 1

" 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.

Answer 2

I still don't understand it

Answer 3

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