document.write( "Question 107332This question is from textbook Introductory and Intermediate Algebra
\n" ); document.write( ": I'm not really clear on how to set up this type of problem. Any insight would be greatly appreciated. x+y=12
\n" ); document.write( " y=3x
\n" ); document.write( "

Algebra.Com's Answer #78125 by ptaylor(2198)\"\" \"About 
You can put this solution on YOUR website!
x+y=12-------------------eq1
\n" ); document.write( "y=3x----------------------eq2\r
\n" ); document.write( "\n" ); document.write( "I think that substituting \"y=3x\" from eq2 into eq1 is the easiest way to work this problem. If we do that, we get:\r
\n" ); document.write( "\n" ); document.write( "x+3x=12 or
\n" ); document.write( "4x=12 divide both sides by 4
\n" ); document.write( "x=3
\n" ); document.write( "From eq2:
\n" ); document.write( "y=3x=3*3=9\r
\n" ); document.write( "\n" ); document.write( "So:
\n" ); document.write( "x=3
\n" ); document.write( "and
\n" ); document.write( "y=9\r
\n" ); document.write( "\n" ); document.write( "We could also subtract eq2 from eq1 and we would get:\r
\n" ); document.write( "\n" ); document.write( "x=12-3x add 3x to both sides\r
\n" ); document.write( "\n" ); document.write( "x+3x=12-3x+3x or
\n" ); document.write( "4x=12 same as before\r
\n" ); document.write( "\n" ); document.write( "CK\r
\n" ); document.write( "\n" ); document.write( "from eq1
\n" ); document.write( "x+y=12
\n" ); document.write( "3+9=12
\n" ); document.write( "12=12
\n" ); document.write( "from eq2
\n" ); document.write( "y=3x
\n" ); document.write( "9=3*3
\n" ); document.write( "9=9\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Hope this helps----ptaylor
\n" ); document.write( "
\n" );