<PRE><B>Question 191689</B>
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The reason that it doesn&#39;t &quot;look right&quot; is because it isn&#39;t.  However, you started out on the right track.


Take it in logical pieces:


Let &lt;i&gt;&lt;b&gt;x&lt;/b&gt;&lt;/i&gt; represent the larger of the numbers, just like the instructions say.


You got the first bit of it right when you translated: &quot;six more than twice the larger number&quot; into *[tex \LARGE 6 + 2x]


But then you tried to represent &quot;the sum of the small number and three&quot; as *[tex \LARGE x + 3], when in fact that expression represents the sum of the &lt;i&gt;&lt;b&gt;large&lt;/b&gt;&lt;/i&gt; number and three.


The key here is the first statement in the problem. &quot;The sum of two numbers is thirty-three.&quot;   If one of the numbers is &lt;i&gt;&lt;b&gt;x&lt;/b&gt;&lt;/i&gt;, then the other one has to be *[tex \LARGE 33 - x].  (Let one number be &lt;i&gt;&lt;b&gt;x&lt;/b&gt;&lt;/i&gt; and the other &lt;i&gt;&lt;b&gt;y&lt;/b&gt;&lt;/i&gt;.  *[tex \LARGE x + y = 33 \ \ \Rightarrow\ \ y = 33 - x])


Now you can say the sum of the small number (*[tex \LARGE 33 - x]) and three is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 33 - x + 3 = 36 - x]


The second thing you did incorrectly was try to make this into an equation.  The problem didn&#39;t ask for an equation, and it couldn&#39;t.  The problem didn&#39;t give you anything to which the expression might equate.  So, just an expression -- no equals sign.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (6 + 2x) - (36 - x)]


Finally, you need to collect like terms to simplify the expression.  Don&#39;t forget to use the distributive property across the second set of parentheses:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6 + 2x - 36 + x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x - 30]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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