document.write( "Question 6923: Longjonsilver had walked me through the following problem, but didn't actually solve it for me. I would like to see if the answer I came up with is correct. Thank you.\r
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\n" ); document.write( "\n" ); document.write( "x^2+y^2=13 = 2x^2+2y^2=26
\n" ); document.write( "y=2x^2-5 = 2x^2=y+5
\n" ); document.write( "y+5+2y^2=26
\n" ); document.write( "2y^2+y-21=0
\n" ); document.write( "(2y+7)(y-3)
\n" ); document.write( "2y+7=0 y-3=0
\n" ); document.write( "y=-7/2 or y=3\r
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\n" ); document.write( "\n" ); document.write( "x^2-7/2^2=13
\n" ); document.write( "x^2+49/4=13
\n" ); document.write( "x^2=52/4-49/4
\n" ); document.write( "x^2=3/4
\n" ); document.write( "x=+ or - sqrt3/4\r
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\n" ); document.write( "\n" ); document.write( "x^2+3^2=13
\n" ); document.write( "x^2+9=13
\n" ); document.write( "x^2=4
\n" ); document.write( "x=+ or - 2\r
\n" ); document.write( "\n" ); document.write( "ORDERED PAIRS ARE (-7/2, SQRT3/4) (-7/2,-SQRT3/4) (3,2) (3,-2)\r
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\n" ); document.write( "\n" ); document.write( "THANK YOU IN ADVANCE!!!
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Algebra.Com's Answer #3812 by longjonsilver(2297)\"\" \"About 
You can put this solution on YOUR website!
nearly perfect :-)\r
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\n" ); document.write( "\n" ); document.write( "the only \"wrong\" thing is that you have written the ordered pairs as (y,x)...we usually write them as (x, y), so correct the order within all 4 and then you have perfection.\r
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\n" ); document.write( "\n" ); document.write( "Well done
\n" ); document.write( "Jon.
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