As long as Loren drove, the law of motion was
[tex]200 = st_1 \implies t_1 = \dfrac{200}{s}[/tex]
As long as Loid drove, the law of motion was
[tex]100 = (s-10)t_2 \implies t_2 = \dfrac{100}{s-10}[/tex]
So, the total time they took is
[tex]t_1+t_2=\dfrac{200}{s}+\dfrac{100}{s-10}[/tex]
Had Loren driven the whole time, the law of motion would have been
[tex]300=st_3 \implies t_3 = \dfrac{300}{s}[/tex]
And we know that this time would have been 30 minutes (i.e. 0.5 hours) faster. So, we have
[tex]t_3 = t_1+t_2-0.5[/tex]
This translates into
[tex]\dfrac{300}{s}=\dfrac{200}{s}+\dfrac{100}{s-10}-\dfrac{1}{2}[/tex]
If we subtract 200/s from both sides, we have
[tex]\dfrac{100}{s}=\dfrac{100}{s-10}-\dfrac{1}{2}[/tex]
We can simplify the right hand side by summing the two fractions:
[tex]\dfrac{100}{s-10}-\dfrac{1}{2} = \dfrac{200-(s-10)}{2(s-10)}=\dfrac{210-s}{2(s-10)}[/tex]
So, we have to solve
[tex]\dfrac{100}{s}=\dfrac{210-s}{2(s-10)}[/tex]
If we cross multiply the denominators, we have
[tex]200(s-10)=s(210-s) \iff 200s-2000=210s-s^2 \iff s^2-10s-2000=0[/tex]
Which yields the solutions
[tex]s=-40,\quad s=50[/tex]
We accept the positive solution, because the negative would mean to travel backwards, so Loren's rate was 50mph
