document.write( "Question 433708: 2r - 3s =-3\r
\n" ); document.write( "\n" ); document.write( "3r+2s=28\r
\n" ); document.write( "\n" ); document.write( "solve by the elimination method
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Algebra.Com's Answer #300666 by blwinbbbles(106)\"\" \"About 
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In order to use the elimination method we will have to manipulate the equations.\r
\n" ); document.write( "\n" ); document.write( "We have\r
\n" ); document.write( "\n" ); document.write( "1)2r-3s= -3 and
\n" ); document.write( "2)3r+2s= 28\r
\n" ); document.write( "\n" ); document.write( "We must pick one of the variable to eliminate so let's pick r..we have to make the r's in each equation the opposite value of each other in order to eliminate them. So the lowest common number they (the r values)have is 6..mulitply equation 1 by (-3) and equation 2 by (2) resulting in the following:\r
\n" ); document.write( "\n" ); document.write( "1) -6r + 9s = 9
\n" ); document.write( "2) 6r + 4s = 56 \r
\n" ); document.write( "\n" ); document.write( "Now add them together and you get\r
\n" ); document.write( "\n" ); document.write( "13s = 65 divide both sides by 13
\n" ); document.write( "s = 5\r
\n" ); document.write( "\n" ); document.write( "Now we do the same for s..again the lowest common number is 6 so multiply equation 1 by (2) and equation 2 b (3) resulting in the following:\r
\n" ); document.write( "\n" ); document.write( "1) 4r - 6s = -6
\n" ); document.write( "2) 9r + 6s = 84\r
\n" ); document.write( "\n" ); document.write( "Add them together and you get
\n" ); document.write( "13r = 78 divide both sides by 13
\n" ); document.write( "r = 6\r
\n" ); document.write( "\n" ); document.write( "You now know s=5 and r=6 and you have solved the problem by elimination\r
\n" ); document.write( "\n" ); document.write( "Hope this helps\r
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