Question 1197435
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please help me in my homework thank you so much!!
Car at a certain speed travels 120 km during the first part. 
On the second part, it reduces its speed by 10 kph when it travels 100 km.
Find the speed on each part of the trip in the trip lasted for 4 hours
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<pre>
Let r be the faster speed, (r-10) be the slower speed.


Time moving with the speed "r" is  {{{120/r}}}  hours.

Time moving with the speed (r-10) is  {{{100/(r-10)}}}  hours.


Total time equation is

    {{{120/r}}} + {{{100/(r-10)}}} = 4  hours.


To solve, multiply both sides by r*(r-10).  You will get

    120*(r-10) + 100r = 4r*(r-10).


Simplify and find r

    120r - 1200 + 100r = 4r^2 - 40r

    220r - 1200        = 4r^2 - 40r

     55r -  300        =  r^2 - 10r

     r^2 - 65r + 300  = 0


It is factorable

    (r-5)*(r-60) = 0.


There are two roots:  r = 5  and  r = 60.

We reject the small positive root (since then (r-10) is negative speed) and accept the greater root r = 60.


<U>ANSWER</U>.  The faster rate is 60 km/h.  The slower rate is 60-10 = 50 km/h.


<U>CHECK</U>.  Total time is {{{120/60}}} + {{{100/50}}} = 2 + 2 = 4 hours.   ! Correct !
</pre>

Solved.


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The method to solve this and many other similar problems is to write "time" equation: it is your setup.


Then reduce this time equation to the standard form quadratic equation.


It can be solved by using the quadratic formula, or by factoring, if it is a lucky case.


At the end, you should be accurate, when you accept or decline one of the two possible roots:

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;if you decline, you should explain WHY.