document.write( "Question 324946: 2y + 3x = 12
\n" ); document.write( "-4y + 5x = -2\r
\n" ); document.write( "\n" ); document.write( "How do I solve the above linear equations using the elimination method? I don't understand
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Algebra.Com's Answer #232675 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
Solve by eliminiation:
\n" ); document.write( "\"2y%2B3x+=+12\"
\n" ); document.write( "\"-4y%2B5x+=+-2\"
\n" ); document.write( "Here you have two equations in two unknowns. The idea behind the \"elimination\" method is to eliminate one of the variables (either x or y) then solve for the remaining variable. Once you know the value of one of the variables, you can find the value of the other one by substitution.
\n" ); document.write( "Let's start:
\n" ); document.write( "1) \"2y%2B3x+=+12\" Multiply this equation by 2.
\n" ); document.write( "2) \"-4y%2B5x+=+-2\"
\n" ); document.write( "--------------------
\n" ); document.write( "1a) \"4y%2B6x+=+24\"
\n" ); document.write( "2a) \"-4y%2B5x+=+-2\" Add these two equations to eliminate the y-variable.
\n" ); document.write( "----------------------
\n" ); document.write( "3) \"11x+=+22\" Divide both sides by 11.
\n" ); document.write( "3a) \"highlight%28x+=+2%29\" Now substitute this value of x into either one of the two original equations to solve for y. Let's use equation 1).
\n" ); document.write( "1) \"2y%2B3x+=+12\" Substitute x = 2.
\n" ); document.write( "1a) \"2y%2B3%282%29+=+12\"
\n" ); document.write( "1b) \"2y%2B6+=+12\" Subtract 6 from both sides.
\n" ); document.write( "1c) \"2y+=+6\" Divide both sides by 2.
\n" ); document.write( "1d) \"highlight%28y+=+3%29\"
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