document.write( "Question 82890: solve the system of equations by sustitution\r
\n" ); document.write( "\n" ); document.write( "3x+4y=-20
\n" ); document.write( "x+y=-6
\n" ); document.write( "

Algebra.Com's Answer #59468 by afphrodietic(3)\"\" \"About 
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let 3x+4y=-20 be the first equation
\n" ); document.write( "and x+y=-6 the second equation.\r
\n" ); document.write( "\n" ); document.write( "then, the second equation becomes x=-6-y.
\n" ); document.write( "plug it in the first equation.\r
\n" ); document.write( "\n" ); document.write( "3(-6-y)+4y=-20
\n" ); document.write( "-18-3y+4y=-20
\n" ); document.write( "-18+y=-20
\n" ); document.write( "y=-20+18
\n" ); document.write( "y=-2\r
\n" ); document.write( "\n" ); document.write( "and\r
\n" ); document.write( "\n" ); document.write( "x+(-2)=-6
\n" ); document.write( "x-2=-6
\n" ); document.write( "x=-6+2
\n" ); document.write( "x=-4\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Then, check:\r
\n" ); document.write( "\n" ); document.write( "3(-4)+4(-2)=-20
\n" ); document.write( "-12+(-8)=-20
\n" ); document.write( "-12-8=-20
\n" ); document.write( "-20=-20
\n" ); document.write( "
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