Question 235069
Given:
Equation 1: {{{B = A + 1}}}  This is because they are consecutive integers
Equation 2: {{{A*B = 8 + 2*(A+B)}}}

Solution:
Start by plugging A + 1 in equation 2 for B
{{{A*B = 8 + 2*(A+B)}}}
{{{A*(A+1) = 8 + 2*(A+(A+1))}}} Then simplify the equation
{{{A^2 + A = 8 + 2*(2A + 1)}}} Simplify again
{{{A^2 + A = 8 + 4A + 2)}}} Combine like terms
{{{A^2 + A = 10 + 4A)}}} Subtract 4A from both sides
{{{A^2 - 3A = 10)}}} Subtract 10 from both sides
{{{A^2 - 3A -10 = 0)}}}  This equation can be foiled
{{{A^2 - 3A -10 = (A+2)*(A-5) = 0)}}}
That gives you 2 possible answers for A.
A + 2 = 0 or rewritten as A = -2
& A - 5 = 0 or rewritten as A = 5

Since 5 is the only positive number it is most likely the answer you would want in this case.  But you should always check both.

Now plug 5 in equation 1 for A
{{{B = A + 1}}}
{{{B = 5 + 1}}}
{{{B = 6}}}

Now check the answers
{{{A*B = 8 + 2*(A+B)}}}
{{{5*6 = 8 + 2*(5+6)}}}
{{{30 = 8 + 2*(11)}}}
{{{30 = 8 + 22}}}
{{{30 = 30}}}