document.write( "Question 335790: driving from town A to town D, you pass thru town B then town C. town A to town B is 10 miles farther than from town B to town C. town B to town C is 10 miles farther than from town C to town D. if town A is 390 miles from town D, how many mile is town A to town B??\r
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\n" ); document.write( "\n" ); document.write( "A-------B-------C-------D
\n" ); document.write( ".-------------------------.=390
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\n" ); document.write( "390/3=130
\n" ); document.write( "answer is 140 mi; just can't explain or show method; could you help?
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Algebra.Com's Answer #240805 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
driving from town A to town D, you pass thru town B then town C.
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\n" ); document.write( "town A to town B is 10 miles farther than from town B to town C.
\n" ); document.write( "town B to town C is 10 miles farther than from town C to town D.
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\n" ); document.write( "if town A is 390 miles from town D, how many mile is town A to town B?? \r
\n" ); document.write( "\n" ); document.write( "A-------B-------C-------D
\n" ); document.write( ".-------------------------.=390
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\n" ); document.write( "Let dist. from B to C be \"x\" miles.
\n" ); document.write( "Then dist. from A to B is \"x+10\" miles.
\n" ); document.write( "And dist. from C to D is \"x -10\" miles
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\n" ); document.write( "Equation:
\n" ); document.write( "x+10 + x + x-10 = 390
\n" ); document.write( "3x = 390
\n" ); document.write( "x = 130 miles
\n" ); document.write( "----
\n" ); document.write( "A to B = x+10 = 140
\n" ); document.write( "B to C = x = 130
\n" ); document.write( "C to D = x-10 = 120
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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