<PRE><B>Question 1164026</B>
&lt;br&gt;
first problem...&lt;br&gt;
In my experience, the algebra is often easier if, instead of calling the three consecutive integers x, x+2, and x+4, we call then x-2, x, and x+2.  So that&#39;s how I will set up the problem.&lt;br&gt;
Twice the smallest (x-2), plus 1 less then the middle (x), plus 4 more than the largest (x+2), equals -15:&lt;br&gt;
{{{2(x-2)+((x)-1)+((x+2)+4)=-15}}}&lt;br&gt;
Solve using basic algebra....&lt;br&gt;
Second problem...&lt;br&gt;
Let the two integers be x and x+1.&lt;br&gt;
The sum of 4 times the smaller (x) and 3 less then twice the larger (x+1) is 71:&lt;br&gt;
{{{4(x)+(2(x+1)-3) = 71}}}&lt;br&gt;
Again solve using basic algebra....&lt;br&gt;
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