document.write( "Question 262427: sqrt3y + sqrt1 = 4\r
\n" ); document.write( "\n" ); document.write( "the square root sign is above 3y + 1
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Algebra.Com's Answer #193341 by stanbon(75887)\"\" \"About 
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sqrt(3y + 1) = 4
\n" ); document.write( "the square root sign is above 3y + 1
\n" ); document.write( "Square both sides to get:
\n" ); document.write( "3y + 1 = 16
\n" ); document.write( "3y = 15
\n" ); document.write( "y = 3
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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