Question 973166

2.	Twice the sum of two consecutive odd integers is three less than the product of two less than the smaller odd integer and two more than the larger odd integer
<pre>Let the smaller integer be S
Then larger is: S + 2
Therefore, we get: 2(S + S + 2) = (S – 2) * (S + 2 + 2) – 3
2(2S + 2) = (S – 2)(S + 4) – 3
{{{4S + 4 = S^2 + 2S - 8 - 3}}}
{{{4S + 4 = S^2 + 2S - 11}}}
{{{S^2 + 2S - 4S - 11 - 4 = 0}}}
{{{S^2 - 2S - 15 = 0}}}
(S - 5)(S + 3) = 0
S, or smaller integer = 5		OR	 	  S = - 3

If the smaller integer is 5, then integers are: {{{highlight_green(system(5_and,7))}}}
However, if the smaller integer is - 3, then integers are: {{{highlight_green(system(- 3_and,- 1))}}}