document.write( "Question 390483: x+1/3y=-11
\n" ); document.write( "3/4x-1/2y=-15 \n" ); document.write( "
Algebra.Com's Answer #276985 by emargo19(101)\"\" \"About 
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Using the substitution method:-
\n" ); document.write( "x + 1/3y = -11 ----- equation 1
\n" ); document.write( "3/4x - 1/2y = -15 ----- equation 2\r
\n" ); document.write( "\n" ); document.write( "x = - 11 - 1/3y
\n" ); document.write( "3/4 ( -11 - y/3 ) - y/2 = - 15
\n" ); document.write( "- 33/4 - 3y/12 -y/2 = - 15
\n" ); document.write( "- 3y/12 - y/2 = -15 + 33/4
\n" ); document.write( "(-3y-6y)/12 = ( -60 + 33) / 4
\n" ); document.write( "-9y/12 = - 6.75
\n" ); document.write( "y = (-6.75 * 12)/-9
\n" ); document.write( "y = 9
\n" ); document.write( "Next lets substitute y=9
\n" ); document.write( "x + 1/3y = - 11 ----- equation 1
\n" ); document.write( "x + 3 = - 11
\n" ); document.write( "x = - 14\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
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