document.write( "Question 159536: if x+2y+3z=53 and 3x+2y+z=52 what is the value of x+y+z? \n" ); document.write( "
Algebra.Com's Answer #117656 by ankor@dixie-net.com(22740)\"\" \"About 
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if x+2y+3z=53 and 3x+2y+z=52 what is the value of x+y+z?
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\n" ); document.write( "Add 'em up:
\n" ); document.write( " x+2y+3z=53
\n" ); document.write( "3x+2y+ z=52
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\n" ); document.write( "4x + 4y + 4z = 105
\n" ); document.write( "Divide equation 4
\n" ); document.write( "x + y + z = \"105%2F4\"
\n" ); document.write( "x + y + z = 26\"1%2F4\"
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