document.write( "Question 485788: if i have 8 creatures and some have 4 legs and others have 6 legs and all the legs add up to 36. how many of each kind of creature can i have?
\n" ); document.write( "solving two ways \n" ); document.write( "

Algebra.Com's Answer #332180 by Edwin McCravy(20055) $\"\"$ $\"About$

You can put this solution on YOUR website!
if i have 8 creatures and some have 4 legs and others have 6 legs and all the legs add up to 36. how many of each kind of creature can i have?
\n" ); document.write( "solving two ways
\n" ); document.write( "

\r\n" );
document.write( "Solving it in your head without algebra:\r\n" );
document.write( "\r\n" );
document.write( "Consider all 8 creatures as having 4-legs, and some creatures have \r\n" );
document.write( "1 extra pair of legs.  So 8×4 = 32 legs are accounted for.  The \r\n" );
document.write( "extra 4 legs means that two of the 8 creatures have an extra pair \r\n" );
document.write( "of legs. So, 6 of the 8 have only 4 legs and the other two have an \r\n" );
document.write( "extra pair, or six legs. \r\n" );
document.write( "\r\n" );
document.write( "Solving by algebra:\r\n" );
document.write( "\r\n" );
document.write( "Let x = the number of 4-legged creatures\r\n" );
document.write( "Let y = the number of 6 legges creatures\r\n" );
document.write( "\r\n" );
document.write( " x +  y =  8\r\n" );
document.write( "4x + 6y = 36\r\n" );
document.write( "\r\n" );
document.write( "Solve the system and get x = 6 and y = 2\r\n" );
document.write( "\r\n" );
document.write( "Edwin

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

\n" );