document.write( "Question 1198539: The second of three numbers is one less than the first. The third number is 5 less than twice the second. If the third number exceeds the first number by 12, find the three numbers. \n" ); document.write( "
Algebra.Com's Answer #832149 by MathLover1(20850)\"\" \"About 
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\n" ); document.write( "let the first number be \"x\"\r
\n" ); document.write( "\n" ); document.write( "if the second number is one less than the first, the second number is\r
\n" ); document.write( "\n" ); document.write( "\"x-1\"\r
\n" ); document.write( "\n" ); document.write( "if the third number is \"5\" less than \"twice\" the second, the third number is \"2%28x-1%29-5=2x+-+7\"\r
\n" ); document.write( "\n" ); document.write( "if the third number exceeds the first number by \"12\", the third number is \"x%2B12\"\r
\n" ); document.write( "\n" ); document.write( "then\r
\n" ); document.write( "\n" ); document.write( "\"2x+-+7=x%2B12\"\r
\n" ); document.write( "\n" ); document.write( "\"2x+-+x=7%2B12\"\r
\n" ); document.write( "\n" ); document.write( "\"x=19\"\r
\n" ); document.write( "\n" ); document.write( "\"x-1=19-1=18\"\r
\n" ); document.write( "\n" ); document.write( "\"x%2B12=19%2B12=31\"\r
\n" ); document.write( "\n" ); document.write( " the three numbers are:
\n" ); document.write( "\"19\"
\n" ); document.write( "\"18\"
\n" ); document.write( "\"31\"\r
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