Question 1197435
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<table border = "1" cellpadding = "5"><tr><td></td><td>Distance (km)</td><td>Rate (kph)</td><td>Time (hr)</td></tr><tr><td>1st Part</td><td>120</td><td>x</td><td>120/x</td></tr><tr><td>2nd Part</td><td>100</td><td>x-10</td><td>100/(x-10)</td></tr><tr><td>Total</td><td>220</td><td></td><td>4</td></tr></table>
Notes:
distance = rate*time
time = distance/rate


The two time durations of 120/x and 100/(x-10) must add to the 4 hour total time. 
We'll use this fact to set up our equation and solve for x.


120/x + 100/(x-10) = 4
x(x-10)( 120/x + 100/(x-10) ) = x(x-10)*4
120(x-10) + 100x  = 4x^2-40x
120x-1200 + 100x  = 4x^2-40x
0 = 4x^2-40x+1200-220x 
4x^2-260x+1200 = 0
In the second step, I multiplied both sides by x(x-10) to clear out the fractions.


We'll turn to the quadratic formula to solve for x.
Use a = 4, b = -260, c = 1200
{{{x = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{x = (-(-260)+-sqrt((-260)^2-4(4)(1200)))/(2(4))}}}


{{{x = (260+-sqrt(67600-19200))/(8)}}}


{{{x = (260+-sqrt(48400))/(8)}}}


{{{x = (260+-  220)/(8)}}}


{{{x = (260+220)/(8)}}} or {{{x = (260-220)/(8)}}}


{{{x = (480)/(8)}}} or  {{{x = (40)/(8)}}}


{{{x = 60}}} or  {{{x = 5}}}


The solution x = 5 isn't possible since x-10 = 5-10 = -5. It doesn't make sense to have a negative speed.
We'll ignore it and go with x = 60 as the only possible value of x.


If x = 60, then x-10 = 60-10 = 50


Check:
Traveling 120 km at a speed of 60 kph means you take 120/60 = 2 hours.
Traveling 100 km at a speed of 50 kph means you take 100/50 = 2 hours.
That's a total of 2+2 = 4 hours to confirm our answers.




Answers:
Speed on the 1st part: <font color=red>60 kph</font> 
Speed on the 2nd part: <font color=red>50 kph</font> 
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