document.write( "Question 572581: solve by the substitution method
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document.write( "4x-4y=0
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document.write( "7x+54=y \n" );
document.write( "
Algebra.Com's Answer #368589 by mathsmiles(68)![]() ![]() You can put this solution on YOUR website! When you solve by substitution, you're looking for one of the variables to be alone on one side of one of the equations. In this case, you have Y by itself in the 2nd equation. What we do then is \"substitute\" for Y in the first equation.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "So, \n" ); document.write( "4X - 4Y = 0 \n" ); document.write( "But we know from the 2nd equation that Y is 7X+54 \n" ); document.write( "So every time they use Y in the first equation, substitute this instead. \n" ); document.write( "4X - 4Y = 0 \n" ); document.write( "4X - 4(7X+54) = 0\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Multiplying to get rid of the parens: \n" ); document.write( "4X - 28X - 216 = 0 \n" ); document.write( "-24X - 216 = 0 \n" ); document.write( "Add 24X to both sides: \n" ); document.write( "-216 = 24X \n" ); document.write( "Divide both sides by 24: \n" ); document.write( "-9 = X\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "But we still need Y: \n" ); document.write( "7X+54 = Y \n" ); document.write( "7(-9)+54 = Y \n" ); document.write( "-63+54=Y \n" ); document.write( "-9=Y\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Checking: \n" ); document.write( "4(-9)-4(-9) = 0 \n" ); document.write( "-36 + 36 = 0 Correct \n" ); document.write( "Next equation: \n" ); document.write( "7(-9) + 54 = -9 \n" ); document.write( "-63 + 54=-9 \n" ); document.write( "-9 = -9 Correct! \n" ); document.write( " |