document.write( "Question 892762: Solve by the substitution method 3x+7y=40
\n" ); document.write( " -4x+y=19 \n" ); document.write( "
Algebra.Com's Answer #540764 by LinnW(1048)\"\" \"About 
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3x+7y=40
\n" ); document.write( "-4x+y=19
\n" ); document.write( "Solve for y in the second equation
\n" ); document.write( "-4x+y=19
\n" ); document.write( "add 4x to each side
\n" ); document.write( " y = 4x + 19
\n" ); document.write( "Now substitute (4x + 19) for y in 3x+7y=40
\n" ); document.write( "3x + 7(4x + 19) = 40
\n" ); document.write( "3x + 28x + 133 = 40
\n" ); document.write( "31x + 133 = 40
\n" ); document.write( " add -133 to each side
\n" ); document.write( " 31x = -93
\n" ); document.write( "divide by 31
\n" ); document.write( " x = -93/31
\n" ); document.write( " x = -3
\n" ); document.write( "Since y = 4x + 19 ,
\n" ); document.write( "y = 4(-3) + 19
\n" ); document.write( "y = -12 + 19
\n" ); document.write( "y = 7\r
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