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 \n" );
document.write( "
Algebra.Com's Answer #191205 by richwmiller(17219)![]() ![]() 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( " \n" ); document.write( " |