document.write( "Question 412009: i have to write my own linear system problem for school and solve it by using elimination and substitution but when i tried solving it i got two different solutions
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document.write( "the equations are y=1500+20x and y=3000+10x
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document.write( "when i used substitution i got (150,4500)and with elimination i got (-300,-4500)
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document.write( "is that correct or wrong? \n" );
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Algebra.Com's Answer #289458 by nerdybill(7384)![]() ![]() You can put this solution on YOUR website! We'll do it both ways here... \n" ); document.write( "Substitution method: \n" ); document.write( "y=1500+20x \n" ); document.write( "y=3000+10x \n" ); document.write( ". \n" ); document.write( "1500+20x=3000+10x \n" ); document.write( "1500+10x=3000 \n" ); document.write( "10x=1500 \n" ); document.write( "x = 150 \n" ); document.write( ". \n" ); document.write( "y=1500+20x \n" ); document.write( "y=1500+20(150) \n" ); document.write( "y=1500+3000 \n" ); document.write( "y=4500 \n" ); document.write( ". \n" ); document.write( ".\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Applying the elimination method to: \n" ); document.write( "y=1500+20x \n" ); document.write( "y=3000+10x \n" ); document.write( ". \n" ); document.write( "moving the x's to the left: \n" ); document.write( "y-20x=1500 \n" ); document.write( "y-10x=3000 \n" ); document.write( ". \n" ); document.write( "multiply the bottom equation by -1 to get add both together: \n" ); document.write( " y-20x=1500 \n" ); document.write( "-y+10x=-3000 \n" ); document.write( "--------------- \n" ); document.write( "-10x = -1500 \n" ); document.write( "x = 150 \n" ); document.write( ". \n" ); document.write( "substitute into top equation to find y: \n" ); document.write( "y=1500+20x \n" ); document.write( "y=1500+20(150) \n" ); document.write( "y=1500+3000 \n" ); document.write( "y=4500\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |