Can I use multiple versions of jQuery on the same page?

Yes, it's doable due to jQuery's noconflict mode. http://blog.nemikor.com/2009/10/03/using-multiple-versions-of-jquery/

<!-- load jQuery 1.1.3 -->
<script type="text/javascript" src="http://example.com/jquery-1.1.3.js"></script>
<script type="text/javascript">
var jQuery_1_1_3 = $.noConflict(true);
</script>

<!-- load jQuery 1.3.2 -->
<script type="text/javascript" src="http://example.com/jquery-1.3.2.js"></script>
<script type="text/javascript">
var jQuery_1_3_2 = $.noConflict(true);
</script>

Then, instead of $('#selector').function();, you'd do jQuery_1_3_2('#selector').function(); or jQuery_1_1_3('#selector').function();.

After looking at this and trying it out I found it actually didn't allow more than one instance of jquery to run at a time. After searching around I found that this did just the trick and was a whole lot less code.

    <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js" type="text/javascript"></script>
    <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js" type="text/javascript"></script>
    <script>var $j = jQuery.noConflict(true);</script>
    <script>
      $(document).ready(function(){
       console.log($().jquery); // This prints v1.4.2
       console.log($j().jquery); // This prints v1.9.1
      });
   </script>

So then adding the "j" after the "$" was all I needed to do.

$j(function () {
        $j('.button-pro').on('click', function () {
            var el = $('#cnt' + this.id.replace('btn', ''));
            $j('#contentnew > div').not(el).animate({
                height: "toggle",
                opacity: "toggle"
            }, 100).hide();
            el.toggle();
        });
    });