Question 200501

Remember, the distance-rate-time formula is 


{{{d=rt}}}


Since he "traveled 600 miles at a certain speed", this means that the first equation is {{{600=rt}}}


The statemtent "Had gone 20mph faster, the trip would have taken 1 hour less", tells us that the next equation is {{{600=(r+20)(t-1)}}}



{{{600=rt}}} Start with the first equation.



{{{600/r=t}}} Divide both sides by r.



{{{t=600/r}}} Rearrange the equation



{{{600=(r+20)(t-1)}}} Move onto the second equation.



{{{600=(r+20)(600/r-1)}}} Plug in {{{t=600/r}}} 



{{{600=600-r+12000/r-20}}} FOIL



{{{600r=600r-r^2+12000-20r}}} Multiply every term by the LCD {{{r}}} to clear the fractions.



{{{0=600r-r^2+12000-20r-600r}}} Subtract {{{600r}}} from both sides.



{{{0=-r^2-20r+12000}}} Combine and rearrange the terms.



Let's use the quadratic formula to solve for r



{{{r = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{r = (-(-20) +- sqrt( (-20)^2-4(-1)(12000) ))/(2(-1))}}} Plug in  {{{a=-1}}}, {{{b=-20}}}, and {{{c=12000}}}



{{{r = (20 +- sqrt( (-20)^2-4(-1)(12000) ))/(2(-1))}}} Negate {{{-20}}} to get {{{20}}}. 



{{{r = (20 +- sqrt( 400-4(-1)(12000) ))/(2(-1))}}} Square {{{-20}}} to get {{{400}}}. 



{{{r = (20 +- sqrt( 400--48000 ))/(2(-1))}}} Multiply {{{4(-1)(12000)}}} to get {{{-48000}}}



{{{r = (20 +- sqrt( 400+48000 ))/(2(-1))}}} Rewrite {{{sqrt(400--48000)}}} as {{{sqrt(400+48000)}}}



{{{r = (20 +- sqrt( 48400 ))/(2(-1))}}} Add {{{400}}} to {{{48000}}} to get {{{48400}}}



{{{r = (20 +- sqrt( 48400 ))/(-2)}}} Multiply {{{2}}} and {{{-1}}} to get {{{-2}}}. 



{{{r = (20 +- 220)/(-2)}}} Take the square root of {{{48400}}} to get {{{220}}}. 



{{{r = (20 + 220)/(-2)}}} or {{{r = (20 - 220)/(-2)}}} Break up the expression. 



{{{r = (240)/(-2)}}} or {{{r =  (-200)/(-2)}}} Combine like terms. 



{{{r = -120}}} or {{{r = 100}}} Simplify.



Since a negative speed doesn't make any sense, this means that the only solution is {{{r = 100}}}



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Answer:



So the speed of his vehicle is 100 mph




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