document.write( "Question 1066867: I need help to solve the following for x: \"+p%5E2%281-x%29-2pqx=q%5E2%281%2Bx%29\"
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Algebra.Com's Answer #682088 by math_helper(2461)\"\" \"About 
You can put this solution on YOUR website!
Since we want to solve for 'x', its best to get all the terms involving x onto one side:\r
\n" ); document.write( "\n" ); document.write( " \"+p%5E2%281-x%29-2pqx+=+q%5E2%281%2Bx%29+\"
\n" ); document.write( " \"++p%5E2+-+p%5E2x+-+2pqx+=+q%5E2%2Bq%5E2x+\"
\n" ); document.write( "Now bring all 'x' related terms to the left, and everything else to the right:
\n" ); document.write( " \"++-p%5E2x+-+2pqx+-+q%5E2x+=+q%5E2+-+p%5E2+\"
\n" ); document.write( "Multiply both sides by -1, this has the effect of changing the left to all +, and swaps the signs on the right:
\n" ); document.write( " \"++p%5E2x+%2B+2pqx+%2B+q%5E2x+=+p%5E2+-+q%5E2+\"
\n" ); document.write( " \"+++x+%28p%5E2+%2B+2pq+%2B+q%5E2%29+=+p%5E2+-+q%5E2+\"
\n" ); document.write( " \"++++x+=++%28p%5E2+-+q%5E2%29+%2F+%28p%5E2+%2B+2pq+%2B+q%5E2+%29+\"\r
\n" ); document.write( "\n" ); document.write( "Not done yet, there is further simplification:\r
\n" ); document.write( "\n" ); document.write( " Numerator: \"+p%5E2+-+q%5E2+=+%28p-q%29%28p%2Bq%29+\"
\n" ); document.write( " Denominator: \"+p%5E2+%2B+2pq+%2B+q%5E2+=+%28p%2Bq%29%5E2+\"\r
\n" ); document.write( "\n" ); document.write( " So \"++++x+=++%28%28p-q%29%28p%2Bq%29%29+%2F+%28%28p%2Bq%29%28p%2Bq%29%29+\" \r
\n" ); document.write( "\n" ); document.write( "Canceling a \"p+q\" from numerator and denominator:
\n" ); document.write( " \"++highlight%28x+=+%28p-q%29%2F%28p%2Bq%29+%29\"\r
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