Let the distance between points a and b = d
For the women starting from a her velocity be [tex]V_1[/tex] and for the women starting from b her velocity be [tex]V_2[/tex]
Let the rising time of sun = x AM
We have [tex]V_1[/tex] = [tex]\frac{d}{12+4-x}[/tex]
and we also have [tex]V_2[/tex] = [tex]\frac{d}{12+9-x}[/tex]
At they meet means, they both travel a combined distance of d
So we have [tex]V_1*(12-x)+V_2*(12-x)=d[/tex]
Substituting velocity values we will get
[tex]\frac{d}{12+4-x}*(12-x)+\frac{d}{12+9-x}*(12-x)=d\\ \\ \frac{12-x}{12+4-x}+\frac{12-x}{12+9-x}=1\\ \\ \frac{12-x}{16-x}+\frac{12-x}{21-x}=1\\ \\ 252-21x-12x+x^2+192-12x-16x+x^2=336-16x-21x+x^2\\ \\ x^2-24x+108=0\\ \\ (x-6)(x-18)=0\\ \\[/tex]
x= 6 AM or 18 AM, but 18 AM is not possible,
Sun rise time is 6 AM