document.write( "Question 285948: The problem is: One number is 8 more than the second number. Twice the second number is one more than the first. Find the numbers. \n" ); document.write( "
Algebra.Com's Answer #207349 by oberobic(2304)\"\" \"About 
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x = first number
\n" ); document.write( "y = second number
\n" ); document.write( "x = 8y :: the first number = 8 times the second number
\n" ); document.write( "2y = x+1 :: twice the second number is one more than the first number
\n" ); document.write( ".
\n" ); document.write( "substitute x = 8y...
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\n" ); document.write( "2y = 8y + 1
\n" ); document.write( "-6y = 1
\n" ); document.write( "y = 1/(-6) = -1/6
\n" ); document.write( ".
\n" ); document.write( "x = 8y
\n" ); document.write( "x = 8(-1/6) = -8/6 = -4/3
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\n" ); document.write( "Checking...is twice the second number = 1 more than first?
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\n" ); document.write( "2*y = 2(-1/6) = -2/6 = -1/3
\n" ); document.write( "x+1 = -4/3 + 1 = -1/3
\n" ); document.write( "Correct
\n" ); document.write( ".
\n" ); document.write( "Answer:
\n" ); document.write( "The numbers are x and y:
\n" ); document.write( "x = -4/3, and
\n" ); document.write( "y = -1/6.
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\n" ); document.write( "Done. \n" ); document.write( "
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