Question 123341
{{{9x^2+25y^2=46}}}, {{{xy=-1}}}. Find {{{3x+5y}}}
<pre><font size = 4><b>
Notice that 

{{{(3x+5y)^2=9x^2+30xy+25y^2}}}, so

{{{(3x+5y)^2=(9x^2+25y^2)+30xy}}}, and since {{{9x^2+25y^2=46}}},

{{{(3x+5y)^2=46+30xy}}}, and since {{{xy=-1}}},

{{{(3x+5y)^2=46+30(-1)}}}, or

{{{(3x+5y)^2=46-30}}}, or

{{{(3x+5y)^2=16}}}. Taking square roots of both sides:

{{{3x+5y}}}= ±{{{4}}}.

Edwin</pre>