document.write( "Question 246286: Find three consecutive integers such that the sum of twice the first and 4 times the second is equal to 20 more than twice the third \n" ); document.write( "
Algebra.Com's Answer #179899 by richwmiller(17219)\"\" \"About 
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let n=first integer
\n" ); document.write( "n+1=second
\n" ); document.write( "n+2=third
\n" ); document.write( "2n+4(n+1)=20+2(n+2)
\n" ); document.write( "2n+4n+4=20+2n+4\r
\n" ); document.write( "\n" ); document.write( "combine and simplify
\n" ); document.write( "6n+4=24+2n
\n" ); document.write( "subtract 2n from both sides
\n" ); document.write( "4n+4=24
\n" ); document.write( "subtract 4 from both sides
\n" ); document.write( "4n=20
\n" ); document.write( "divide both sides by 4
\n" ); document.write( "n=5:n+1=6:n+2=7
\n" ); document.write( "integers are 5,6,7
\n" ); document.write( "check
\n" ); document.write( "2(5)+4(6)=20+2(7)
\n" ); document.write( "10+24=20+14
\n" ); document.write( "34=34
\n" ); document.write( "ok\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
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