document.write( "Question 181446: Joseph live 5 miles from the school and Maya lives 2 miles from the school. If Joseph walks from his house in the opposite direction from the school at a speed of 4 miles an hour and Maya walks from her house in the opposite direction from the school at a speed of 6 miles per hour, after how long will they be the same distance from the school? What is the distance? \n" ); document.write( "

Algebra.Com's Answer #136074 by ptaylor(2198) $\"\"$ $\"About$

You can put this solution on YOUR website!

\n" ); document.write( "Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
\n" ); document.write( "Let t=time that elapses 'till they are the same distance from school
\n" ); document.write( "Distance Joseph is from school=5+4t
\n" ); document.write( "Distance Maya is from school=2+6t
\n" ); document.write( "The above two distances are the same, so:\r
\n" ); document.write( "\n" ); document.write( "Our equation to solve is:
\n" ); document.write( "5+4t=2+6t subtract 2 and also 4t from each side
\n" ); document.write( "5+4t-4t-2=2-2+6t-4t collect like terms
\n" ); document.write( "3=2t divide each side by 2
\n" ); document.write( "t=1.5 hr------------time that elapses 'till they are the same distance from school\r
\n" ); document.write( "\n" ); document.write( "Distance Joseph is from school=5+4*1.5 =11 mi
\n" ); document.write( "or
\n" ); document.write( "Distance Maya is from school=2+6*1.5=11 mi\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Hope this helps----ptaylor
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