Answer:
Initial speed:
Step-by-step explanation:
Both equations are concerned about the time differences between A and B. To avoid unknowns in the denominators,
First equation:
"A takes 3 hours more than B to walk 30 km."
[tex]x = y + 3[/tex].
[tex]x - y = 3[/tex].
When A doubles his pace, he takes only 1/2 the initial time to cover the same distance. In other words, now it takes only [tex]x/2[/tex] hours for A to walk 30 km.
Second equation:
"[A] is ahead of B by 3/2 hours [on their 30-km walk.]"
[tex]\displaystyle \frac{x}{2} + \frac{3}{2} = y[/tex].
[tex]\displaystyle \frac{1}{2}x - y = -\frac{3}{2}[/tex].
Hence the two-by-two linear system:
[tex]\left\{\begin{aligned}&x - y = 3\\&\frac{1}{2}x - y = -\frac{3}{2}\end{aligned}\right.[/tex].
Solve this system for [tex]x[/tex] and [tex]y[/tex]:
Subtract the second equation from the first:
[tex]\displaystyle \frac{1}{2}x = \frac{9}{2}[/tex].
[tex]x = 9[/tex].
[tex]y = 6[/tex].
It initially takes 9 hours for A to walk 30 kilometers. The initial speed of A will thus be:
[tex]\displaystyle v = \frac{s}{t} = \rm \frac{30\; km}{9\; h} = \frac{10}{3}\; km/h[/tex].
It takes 6 hours for B to walk 30 kilometers. The speed of B will thus be:
[tex]\displaystyle v = \frac{s}{t} = \rm \frac{30\; km}{6\; h} = 5\; km/h[/tex].
