Question 1200264
<pre>
Kiran drove from Tortula to Cactus, a distance of 204 mi. She increased her speed by 13 mi/h for the 308 mi trip from Cactus to Dry Junction. If the total trip took 6 h, what was her speed from Tortula to Cactus?

Let speed from Tortula to Cactus, be S
Then speed from Cactus to Dry Junction is S + 13
Since it took 6 hours to complete the entire journey, we get TIME equation: {{{matrix(1,3, 6, "=", 204/S + 308/(S + 13))}}}
                     {{{matrix(1,3, 3, "=", 102/S + 154/(S + 13))}}} ---- Factoring out GCF, 2, in numerators
            3S(S + 13) = 102(S + 13) + 154S ----- Multiplying by LCD, S(S + 13) 
{{{matrix(4,3, 3S^2 + 39S, "=", 102S + 13(102) + 154S, 3S^2 + 39S - 256S - "1,326","=", 0, 3S^2 - 217S - "1,326", "=", 0, 3S^2 - 234S + 17S - "1,326", "=", 0)}}}
3S(S - 78) + 17(S - 78) = 0
S - 78 = 0             or             3S + 17 = 0____3S = - 17 (ignore)

<font size = 4><font color = blue><b>Speed from Tortula to Cactus</font></font></b>, or S = <font size = 4><font color = blue><b>78 mph</font></font></b></pre>