document.write( "Question 82571: Solve the system by substitution:\r
\n" ); document.write( "\n" ); document.write( "2x+2y=0
\n" ); document.write( "y=4x+0 \n" ); document.write( "
Algebra.Com's Answer #59165 by tutorcecilia(2152)\"\" \"About 
You can put this solution on YOUR website!
2x+2y=0 [substitute (y=4x+0) for the y-term and solve for the x-term]
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\n" ); document.write( "2x+2(4x+0)=0
\n" ); document.write( "2x+2(4x)=0
\n" ); document.write( "2x+8x=0
\n" ); document.write( "10x=0
\n" ); document.write( "10x/10=0/10
\n" ); document.write( "x=0
\n" ); document.write( ".
\n" ); document.write( "solve for the y-term by plugging x=0 back into one of the original equations:
\n" ); document.write( "2x+2y=0
\n" ); document.write( "2(0)+2y=0
\n" ); document.write( "2y=0
\n" ); document.write( "2y/2=0/2
\n" ); document.write( "y=0
\n" ); document.write( ".
\n" ); document.write( "check by plugging the values for x and y back into either eiquation and solve:
\n" ); document.write( "2x+2y=0
\n" ); document.write( "2(0)+2(0)=0
\n" ); document.write( "0=0 [checks out].
\n" ); document.write( "and
\n" ); document.write( "y=4x+0
\n" ); document.write( "0=4(0)+0
\n" ); document.write( "0=0 [also checks out]\r
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