document.write( "Question 1175442: One number is twice another number. Twice the lesser number increased by the greater number is at least 45. What is the least possible integral value for the lesser number?\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
Algebra.Com's Answer #801088 by Edwin McCravy(20064)\"\" \"About 
You can put this solution on YOUR website!
\r\n" );
document.write( "It is not necessary to state that the lesser number is positive, since\r\n" );
document.write( "if it were negative, both the lesser and greater would be negative, and\r\n" );
document.write( "you cannot increase a negative with a negative.  And 0 is out because\r\n" );
document.write( "twice zero is zero, and there could be no lesser and greater.\r\n" );
document.write( "\r\n" );
document.write( "x = lesser (positive) integer\r\n" );
document.write( "2x = greater integer\r\n" );
document.write( "\r\n" );
document.write( "2x + 2x ≥ 45\r\n" );
document.write( "     4x ≥ 45\r\n" );
document.write( "      x ≥ 11.25\r\n" );
document.write( "\r\n" );
document.write( "Since x is an integer, x=12.  \r\n" );
document.write( "\r\n" );
document.write( "Edwin

\n" ); document.write( " \n" ); document.write( "
\n" );