<PRE><B>Question 1092272</B>
&lt;br&gt;I nearly always find it easier to solve problems like this using a single variable instead of two.  Given that the sum of the two numbers is 15, I would do&lt;br&gt;
let x = large number
then (15-x) = smaller number&lt;br&gt;
The sum of the squares of the two numbers is 9 more than 13 times the larger number:&lt;br&gt;
{{{x^2 + (15-x)^2 = 13x+9}}}
{{{x^2 + 225 - 30x + x^2 = 13x+9}}}
{{{2x^2-43x+216 = 0}}}
{{{(2x-27)(x-8) = 0}}}&lt;br&gt;
This gives us two potential solutions:&lt;br&gt;
(1) x = 8; 15-x = 7  OR
(2) x = 27/2; 15-x = 3/2&lt;br&gt;
Since our solution method involved squaring expressions, we need to check to see which of the solutions actually satisfy the original problem.&lt;br&gt;
For the first...&lt;br&gt;
{{{8^2+7^2 = 64+49 = 113}}}  and
{{{13(8)+9 = 113}}}&lt;br&gt;
So that solution is valid.&lt;br&gt;
For the second...&lt;br&gt;
{{{(27/2)^2 + (3/2)^2 = 729/4+9/4 = 738/4 = 369/2}}}
{{{13(27/2)+9 = 351/2+18/2 = 369/2}}}&lt;br&gt;
This solution is also valid.&lt;br&gt;
So there are two solutions to the problem: 8 and 7; or 27/2 and 3/2.</PRE>