document.write( "Question 401355: I need to solve by using the elimination method but the fraction is throwing me off and I am not sure how to solve this one. Thank you \r
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document.write( "0.3x-0.2y=4\r
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document.write( "0.2x-0.3y=5/23 \n" );
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Algebra.Com's Answer #284028 by dnanos(83)![]() ![]() You can put this solution on YOUR website! First multiply by 10 both equations \n" ); document.write( "3x+2y=40 \n" ); document.write( "2x-3y=50/23 \n" ); document.write( "then multiply the first one by 2 and the second by -3 \n" ); document.write( "6x+4y=80 \n" ); document.write( "-6x+9y=-150/23\r \n" ); document.write( "\n" ); document.write( "Then add and keep the first one for next steps \n" ); document.write( "3x+2y=40 \n" ); document.write( "13y=80-150/23 \n" ); document.write( "so \n" ); document.write( "3x+2y=40 \n" ); document.write( "13y=(1840-150)/23 \n" ); document.write( "so \n" ); document.write( "3x+2y=40 \n" ); document.write( "13y=1690/23 \n" ); document.write( "We divide the second equation by 13 (because 169=13*13) \n" ); document.write( "3x+2y=40 \n" ); document.write( "13y/13=1690/13*23 \n" ); document.write( "so \n" ); document.write( "3x+2y=40 \n" ); document.write( "y=130/23 \n" ); document.write( "then we replace the y value in the first equation. \n" ); document.write( "3x+2*130/23=40 \n" ); document.write( "y=130/23 \n" ); document.write( "so \n" ); document.write( "3x=40-260/23 \n" ); document.write( "y=130/23 \n" ); document.write( "so \n" ); document.write( "3x=(920-260)/23 \n" ); document.write( "y=130/23 \n" ); document.write( "so \n" ); document.write( "3x=660/23 \n" ); document.write( "y=130/23 \n" ); document.write( "We divide the first equation by 3 \n" ); document.write( "x=220/23 \n" ); document.write( "y=130/23\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |