document.write( "Question 468973: solve the system of equation using substitution.
\n" ); document.write( "y= 2/3x+1
\n" ); document.write( "y= -3/2x+40 \n" ); document.write( "

Algebra.Com's Answer #321760 by stanbon(75887) $\"\"$ $\"About$

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solve the system of equation using substitution.
\n" ); document.write( "y= 2/3x+1
\n" ); document.write( "y= -3/2x+40
\n" ); document.write( "-----------
\n" ); document.write( "Substitute for \"y\" to get:
\n" ); document.write( "(2/3)x+1 = (-3/2)x+40
\n" ); document.write( "---
\n" ); document.write( "[(2/3)+(3/2)]x = 39
\n" ); document.write( "----
\n" ); document.write( "[(4+9)/6]x = 39
\n" ); document.write( "(13/6)x = 39
\n" ); document.write( "----
\n" ); document.write( "x = (6/13)39
\n" ); document.write( "---
\n" ); document.write( "x = 6*3
\n" ); document.write( "x = 18
\n" ); document.write( "---
\n" ); document.write( "Solve for \"y\":
\n" ); document.write( "y = (2/3)x+1
\n" ); document.write( "y = (2/3)18 + 1
\n" ); document.write( "y = 12+1
\n" ); document.write( "y = 13
\n" ); document.write( "==============
\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H. \n" ); document.write( "

\n" );