document.write( "Question 717557: a to b is 5 km more than d to e, c to d is twice the distance from a to b, c is midway between b and d. if the total from a to e is 91 km find the distance from d to e \n" ); document.write( "
Algebra.Com's Answer #440560 by ankor@dixie-net.com(22740)\"\" \"About 
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a to b is 5 km more than d to e, c to d is twice the distance from a to b, c is midway between b and d. if the total from a to e is 91 km find the distance from d to e
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\n" ); document.write( "\n" ); document.write( "let w = ab distance
\n" ); document.write( "let x = bc
\n" ); document.write( "let y = cd
\n" ); document.write( "let z = de
\n" ); document.write( ":
\n" ); document.write( "a-----w------b------x-------c-------y------d-------z-------e
\n" ); document.write( ":
\n" ); document.write( "\"a to b is 5 km more than d to e,\"
\n" ); document.write( "w = z+5
\n" ); document.write( "z = w-5
\n" ); document.write( ":
\n" ); document.write( "\" c to d is twice the distance from a to b,\"
\n" ); document.write( "y = 2w
\n" ); document.write( ":
\n" ); document.write( "\"c is midway between b and d.\" (bc = cd)
\n" ); document.write( "x = y
\n" ); document.write( ":
\n" ); document.write( "\"the total from a to e is 91 km\"
\n" ); document.write( "w + x + y + z = 91
\n" ); document.write( "If y = 2w, then x = 2w, also we know z = w-5
\n" ); document.write( "w + 2w + 2w + (w-5) = 91
\n" ); document.write( "combine like terms
\n" ); document.write( "6w = 91 + 5
\n" ); document.write( "6w = 96
\n" ); document.write( "w = 96/6
\n" ); document.write( "w = 16
\n" ); document.write( ":
\n" ); document.write( "\"find the distance from d to e, (that's z)
\n" ); document.write( "z = 16 - 5
\n" ); document.write( "z = 11 is the distance from d to e
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\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "See if that checks out
\n" ); document.write( "16 + 2(16) + 2(16) + 11 = 91 \n" ); document.write( "
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