document.write( "Question 259755: solve by the elimination method 7r-4s=33 4r+7s=56\r
\n" ); document.write( "\n" ); document.write( "I have tried setting it up like this \r
\n" ); document.write( "\n" ); document.write( "7r-4s=33
\n" ); document.write( "4r+7s=56
\n" ); document.write( "--------
\n" ); document.write( "3r 3s=89\r
\n" ); document.write( "\n" ); document.write( "I don't know if I am supposed to put a + or - between the 2 and to eliminate the variables I divide by 3 and get r,s = 29.6\r
\n" ); document.write( "\n" ); document.write( "this does not look right. Any help on how to solve would be appreciated. Thanks
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Algebra.Com's Answer #191205 by richwmiller(17219)\"\" \"About 
You can put this solution on YOUR website!
You are right it doesn't look right!
\n" ); document.write( "r=7 s=4\r
\n" ); document.write( "\n" ); document.write( "7r-4s=33
\n" ); document.write( "4r+7s=56
\n" ); document.write( "--------
\n" ); document.write( "3r 3s=89
\n" ); document.write( "your addition needs work.
\n" ); document.write( "7+4=11 not 3\r
\n" ); document.write( "\n" ); document.write( "but worse your approach is wrong
\n" ); document.write( "It is too early to add or subtract.
\n" ); document.write( "you have to get the same coefficient for x or y so that when you add or subtract either the x or the y disappear or are eliminated .\r
\n" ); document.write( "\n" ); document.write( "multiply the first by 7 and the second by 4
\n" ); document.write( "then we will add them to get rid of y's\r
\n" ); document.write( "\n" ); document.write( "7*(7r-4s=33)
\n" ); document.write( "4*(4r+7s=56)
\n" ); document.write( "49r-28s=7*33
\n" ); document.write( "16r+28s=4*56
\n" ); document.write( "add
\n" ); document.write( "65r=7*33+4*56
\n" ); document.write( "65r=455
\n" ); document.write( "r=7\r
\n" ); document.write( "\n" ); document.write( "
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