document.write( "Question 427061: Solve the following system of equations using substitution.\r
\n" ); document.write( "\n" ); document.write( "2x+4y=12
\n" ); document.write( "4x+16y=8\r
\n" ); document.write( "\n" ); document.write( "I substituted 2x+4y for x in the second equation and I ended up with 8x+32y=8. I've tried so many of these problems and I just can't understand them.
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Algebra.Com's Answer #297009 by Gogonati(855)\"\" \"About 
You can put this solution on YOUR website!
Solution:First we simplify the system dividing the first equation by 2 and the second by 4,and the system can be written:\r
\n" ); document.write( "\n" ); document.write( "\"system%28x%2B2y=6%2C+x%2B4y=2%29\", Express x in the first equation with respect to y and substitute in the second equation:\r
\n" ); document.write( "\n" ); document.write( "\"system%28x=6-2y%2C+6-2y%2B4y=2%29\"\r
\n" ); document.write( "\n" ); document.write( "The second equation is a linear equation of one variable. We solve this equation with respect to y:\r
\n" ); document.write( "\n" ); document.write( "6+2y=2 => 2y=-4 => y=-2. Substitute this value of y in the first equation and find x: x=6-2(-2) => x=10. The ordered pair (10, -2) is the solution of our system.\r
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\n" ); document.write( "\n" ); document.write( "Check: \"+system%282%2A10%2B4%2A%28-2%29=12%2C+4%2A10%2B16%2A%28-2%29=8%29\"\r
\n" ); document.write( "\n" ); document.write( " \"+system%2820-8=12%2C+40-32=8%29\"\r
\n" ); document.write( "\n" ); document.write( " \"system%2812=12%2C+8=8%29\"
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