Question 973229
The smallest number is equal to 7.

Explanation:

For this problem, I'll be using variables to represent the numbers: x, y, and z. But since they are consecutive, y=x+1, and z=x+2, so the numbers can be written as x, x+1, and x+2. 

According to the problem, the product of the first and third number ({{{x(x+2)}}} or {{{x^2+2x}}}) is equal to one less than eight times the second number ({{{8(x+1)-1)}}} or {{{8x+7}}})
We can set the problem up like this:
{{{x^2+2x=8x+7}}}
Now we can solve it like an algebraic equation instead of a word problem. Start by moving the variable to one side of the equation by subtracting 8x from both sides, resulting in:
{{{x^2-6x=7}}}
Since there is a variable squared, we should use the quadratic formula. So we will have to subtract 7 as well, so that 0 will be on one side of the equal sign.
{{{x^2-6x-7=0}}}
According to the quadratic equation, if {{{Ax^2+Bx+C=0,}}} then {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
Now, we plug in the numbers.
{{{x = (6 +- sqrt( (-6)^2-4*1*(-7) ))/(2*1) }}}
in this case, a equals one because there is no number before {{{x^2}}}, so the coefficient must be 1. b is -6 since that is the coefficient for x, and c is -7.
Finally, after using a calculator to solve the equation, we get
{{{x=7}}}