document.write( "Question 144579: Steve traveled 200 miles at a certain speed. Had he gone 10mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle. \n" ); document.write( "
Algebra.Com's Answer #105328 by shahid(44)\"\" \"About 
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let x be his speed to cover given distance d=200miles in time t hours then
\n" ); document.write( "d=xt
\n" ); document.write( "200=xt .........(1)
\n" ); document.write( "if he runs 10mph faster so his speed becomes (x+10)mph and this takes 1 hour less so now time to complete this distance becomes (t-1)hour.again
\n" ); document.write( "d=(x+10)(t-1) by formula distance= speed*time
\n" ); document.write( "200=xt-x+10t-10
\n" ); document.write( "200=200-x+10t-10 from equ:1 xt=200
\n" ); document.write( "0=-x+10t-10 200 be canceled
\n" ); document.write( "x=10t-10 putting this in equ:1
\n" ); document.write( "200=(10t-10)t
\n" ); document.write( "200=10t^2-10t
\n" ); document.write( "10t^2-10t-200=0
\n" ); document.write( "t^2-t-20=0 taking 10 common
\n" ); document.write( "t^2-5t+4t-20=0
\n" ); document.write( "t(t-5)+4(t-5)=0
\n" ); document.write( "(t-5)(t+4)=0
\n" ); document.write( "t-5=0 i-e t=5 we neglect t+4=0 that gives negative time
\n" ); document.write( "putting t=5hours in equ:1
\n" ); document.write( "200=x(5)
\n" ); document.write( "x=200/5
\n" ); document.write( "x=40mph \n" ); document.write( "
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