document.write( "Question 141236: Well i have to solve the following system of equations by substitution\r
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document.write( "0.5x-2y=1
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document.write( "-3y+x=4\r
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document.write( "I understand that I must get of the the equations to equal to either x or y so i can substitute.\r
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document.write( "i'll try the second one\r
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document.write( "-3y+x=4\r
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document.write( "x=4-3y is that right? if so, what to do next? \n" );
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Algebra.Com's Answer #102892 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! 0.5x-2y=1 \n" ); document.write( "-3y+x=4 \n" ); document.write( "I understand that I must get of the the equations to equal to either x or y so i can substitute. \n" ); document.write( "i'll try the second one \n" ); document.write( "-3y+x=4 \n" ); document.write( "x=4-3y is that \n" ); document.write( "--------------- \n" ); document.write( "Substitute that into 5x-2y=1 to solve for \"y\": \n" ); document.write( "0.5(4-3y)- 2y = 1 \n" ); document.write( "2 - (3/2)y - 2y = 1 \n" ); document.write( "-7/2 y = -1 \n" ); document.write( "y = 2/7 \n" ); document.write( "-------------------- \n" ); document.write( "Substitute into x = 4-3y \n" ); document.write( "x = 4-3(2/7) \n" ); document.write( "x = 4-6/7 \n" ); document.write( "x = 22/7 \n" ); document.write( "----------------------------- \n" ); document.write( "Checking: \n" ); document.write( "0.5x-2y = 1 \n" ); document.write( "0.5(22/7) -2(2/7) = 1 \n" ); document.write( "11/7 - 4/7 = 1 \n" ); document.write( "7/7 = 1 \n" ); document.write( "1 = 1 \n" ); document.write( "=========== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |