Question 370575
Let r = original speed.


and let t = time to get to work driving that original speed.



So if "Liz commutes 30 mi. to her job each day", this means that {{{30=rt}}} because we're using the formula {{{d=rt}}} where the distance is {{{d=30}}}



So the first equation is {{{30=rt}}}



In addition, because "she drives 10 mi/h faster it takes her 6 minutes less to get to work", we can say that {{{30=(r+10)(t-1/10)}}}


Notes: since she drives 10 mph faster, her new speed is r+10. Also, because 60 min = 1 hour, this means that 6 min = {{{6/60=1/10}}} hours. So if it takes 6 mins or {{{1/10}}} or an hour less, then the new time is {{{t-1/10}}}



So the second equation is {{{30=(r+10)(t-1/10)}}}



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Now let's use both equations to solve for t and r



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



{{{30/t=r}}} Divide both sides by t to isolate r.



{{{r=30/t}}} Flip the equation.



{{{30=(r+10)(t-1/10)}}} Move onto the second equation.



{{{30=(30/t+10)(t-1/10)}}} Plug in {{{r=30/t}}}



{{{30=(30/t+10t/t)(t-1/10)}}} Multiply 10 by {{{t/t}}}



{{{30=((30+10t)/t)(t-1/10)}}} Combine the fractions.



{{{30=((30+10t)/t)(t)-((30+10t)/t)(1/10)}}} Distribute



{{{30=30+10t-(3+t)/t}}} Multiply.



Note: the 't' terms cancel in the first fraction while in the second, we're dividing each term by 10.



{{{30t=30t+10t^2-(3+t)}}} Multiply EVERY term by the LCD 't' to clear out the fractions.



{{{30t=30t+10t^2-3-t}}} Distribute.



{{{0=30t+10t^2-3-t-30t}}} Subtract 30t from both sides.



{{{0=10t^2-t-3}}} Combine like terms.




Notice that the quadratic {{{10t^2-t-3}}} is in the form of {{{At^2+Bt+C}}} where {{{A=10}}}, {{{B=-1}}}, and {{{C=-3}}}



Let's use the quadratic formula to solve for "t":



{{{t = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{t = (-(-1) +- sqrt( (-1)^2-4(10)(-3) ))/(2(10))}}} Plug in  {{{A=10}}}, {{{B=-1}}}, and {{{C=-3}}}



{{{t = (1 +- sqrt( (-1)^2-4(10)(-3) ))/(2(10))}}} Negate {{{-1}}} to get {{{1}}}. 



{{{t = (1 +- sqrt( 1-4(10)(-3) ))/(2(10))}}} Square {{{-1}}} to get {{{1}}}. 



{{{t = (1 +- sqrt( 1--120 ))/(2(10))}}} Multiply {{{4(10)(-3)}}} to get {{{-120}}}



{{{t = (1 +- sqrt( 1+120 ))/(2(10))}}} Rewrite {{{sqrt(1--120)}}} as {{{sqrt(1+120)}}}



{{{t = (1 +- sqrt( 121 ))/(2(10))}}} Add {{{1}}} to {{{120}}} to get {{{121}}}



{{{t = (1 +- sqrt( 121 ))/(20)}}} Multiply {{{2}}} and {{{10}}} to get {{{20}}}. 



{{{t = (1 +- 11)/(20)}}} Take the square root of {{{121}}} to get {{{11}}}. 



{{{t = (1 + 11)/(20)}}} or {{{t = (1 - 11)/(20)}}} Break up the expression. 



{{{t = (12)/(20)}}} or {{{t =  (-10)/(20)}}} Combine like terms. 



{{{t = 3/5}}} or {{{t = -1/2}}} Simplify. 



So the possible solutions are {{{t = 3/5}}} or {{{t = -1/2}}} 

  
  
However, since a negative time isn't possible, this means that the only solution for 't' is {{{t=3/5}}}



Remember that 't' is the time in hours. So convert to minutes to get {{{(3/5)*(60) = 180/5 = 36}}} minutes



So the time it takes to travel 30 miles at the original speed is 36 minutes.



Recall that we made {{{r=30/t}}}. So plug {{{t=3/5}}} into the equation to get {{{r=30/(3/5)=(30/1)(5/3)=150/3=50}}}



So her original speed is 50 mph. Add 10 mph to this speed to get 50+10 = 60 mph



So her new speed is 60 mph.



I'll leave the check to you. Remember to use the formula {{{d=rt}}}



If you need more help, email me at <a href="mailto:jim_thompson5910@hotmail.com?Subject=Algebra%20Help">jim_thompson5910@hotmail.com</a>


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Jim