document.write( "Question 1026157: Jaime and Alex leave the same location and travel in opposite directions. Traffic conditions enabled Alex to average 14 miles per hour faster than Jaime. After 1½ hours they are 159 miles apart. Find the speed at which each was able to travel. \n" ); document.write( "
Algebra.Com's Answer #641415 by josmiceli(19441)\"\" \"About 
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Let \"+d+\" = distance in miles Jaime travels in 1.5 hrs
\n" ); document.write( "\"+159+-+d+\" = distance Alex travels in 1.5 hrs
\n" ); document.write( "Let \"+s+\" = Jaime's speed in mi/hr
\n" ); document.write( "\"+s+%2B+14+\" = Alex's speed in mi/hr
\n" ); document.write( "--------------------------------
\n" ); document.write( "Jaime's equation:
\n" ); document.write( "(1) \"+d+=+s%2A1.5+\"
\n" ); document.write( "Alex's equation:
\n" ); document.write( "(2) \"+159+-+d+=+%28+s+%2B+14+%29%2A1.5+\"
\n" ); document.write( "----------------------------
\n" ); document.write( "Substitute (1) into (2)
\n" ); document.write( "(2) \"+159+-+1.5s+=+%28+s+%2B+14+%29%2A1.5+\"
\n" ); document.write( "(2) \"+159+-+1.5s+=+1.5s+%2B+21+\"
\n" ); document.write( "(2) \"+3s+=+159+-+21+\"
\n" ); document.write( "(2) \"+3s+=+138+\"
\n" ); document.write( "(2) \"+s+=+46+\"
\n" ); document.write( "and
\n" ); document.write( "\"+s+%2B+14+=+46+%2B+14+\"
\n" ); document.write( "\"+s+%2B+14+=+60+\"
\n" ); document.write( "Jaime's speed was 46 mi/hr
\n" ); document.write( "Alex's speed was 60 mi/hr
\n" ); document.write( "------------------------
\n" ); document.write( "check:
\n" ); document.write( "(1) \"+d+=+s%2A1.5+\"
\n" ); document.write( "(1) \"+d+=+46%2A1.5+\"
\n" ); document.write( "(1) \"+d+=+69+\" mi
\n" ); document.write( "and
\n" ); document.write( "(2) \"+159+-+d+=+%28+s+%2B+14+%29%2A1.5+\"
\n" ); document.write( "(2) \"+159+-+d+=+%28+46+%2B+14+%29%2A1.5+\"
\n" ); document.write( "(2) \"+159+-+d+=+60%2A1.5+\"
\n" ); document.write( "(2) \"+d+=+159+-+90+\"
\n" ); document.write( "(2) \"+d+=+69+\" mi
\n" ); document.write( "
\n" );