document.write( "Question 1015007: Thanks!\r
\n" ); document.write( "\n" ); document.write( "In which line is a mistake first made when solving the following system of equations by substitution?
\n" ); document.write( "3x + y = 9
\n" ); document.write( "-2x + y = 4\r
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\n" ); document.write( "\n" ); document.write( "Line 1
\n" ); document.write( "y = -3x + 9
\n" ); document.write( "-2x + y = 4
\n" ); document.write( "Line 2
\n" ); document.write( "3x + (-3x + 9) = 9
\n" ); document.write( "Line 3
\n" ); document.write( "3x - 3x + 9 = 9
\n" ); document.write( "Line 4
\n" ); document.write( "9 = 9
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Algebra.Com's Answer #631327 by ValorousDawn(53)\"\" \"About 
You can put this solution on YOUR website!
Line 2. You don't plug in a variable's identity back into the original equation. \r
\n" ); document.write( "\n" ); document.write( "Otherwise, you get a useless statement, as above with 9=9 Given a linear equation \"ax%2Bby=c\", and one isolates a term, for example, the y as such \"by=-ax%2Bc\" then \"y=-ax%2Fb%2Bc%2Fb\". If you were to plug this back into the equation, you'd get \"ax%2Bb%28-ax%2Fb%2Bc%2Fb%29=c\", then \"ax-ax%2Bc=c\" which is then \"c=c\". \r
\n" ); document.write( "\n" ); document.write( "In this case, you'd plug in the equation for y \"y=-3x%2B9\" into the OTHER equation. This gives you \"-2x%2B%28-3x%2B9%29=4\" which gives you \"-5x=-5\" or \"x=1\". You then plug this into your equation for y, and solve. \"y=-3%281%29%2B9\" \"y=6\"
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