document.write( "Question 1123520: Find four consecutive integers such that 4 times the sum of the first and the fourth is greater than 6 times the third. \n" ); document.write( "

Algebra.Com's Answer #740068 by ikleyn(52832) $\"\"$ $\"About$

You can put this solution on YOUR website!
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document.write( "Let the numbers are n, n+1, n+2 and n+3.\r\n" );
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document.write( "Then the condition says \r\n" );
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document.write( "4*(n + (n+3)) > 6*(n+2)\r\n" );
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document.write( "4n + 4n + 12 > 6n + 12\r\n" );
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document.write( "8n + 12 > 6n + 12\r\n" );
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document.write( "8n - 6n > 12 - 12\r\n" );
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document.write( "2n > 0\r\n" );
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document.write( "n > 0.\r\n" );
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\n" ); document.write( "\n" ); document.write( "Answer. Any four consecutive integers n, n+1, n+2 and n+3 with n > 0 satisfy the problem's condition.\r
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\n" ); document.write( "\n" ); document.write( "Be aware. The solution by @JessLynnLancaster24 is totally wrong and contains a lot of errors,
\n" ); document.write( "demonstrating extremely low level in Math.\r
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\n" ); document.write( "\n" ); document.write( "This forum is not the place for such solutions and such \"tutors\".\r
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