Uniform motion question

The equation of motion for the biker is $x_1(t)=(2v+15) t$. The equation of motion of the skater is $x_2(t)=vt.$ We know that $x_1(1.5)-x_2(1.5)=35.$ Thus $$(2v+15)cdot 1.5-1.5v=35$$ $$1.5(2v+15-v)=35$$ $$v+15=frac{70}{3}$$ $$v=frac{25}{3}~text{km}/text{hr}.$$

You are correct in the first part. If we let the biker's speed = $S_1$ and the skater's speed = $S_2$, then $S_1 = 2S_2 + 15$. Note the distance travelled in $1.5$ hours by the biker is:

$$1.5times{(2S_2 + 15)}$$

Similarly, the distance travelled in $1.5$ hours by the skater is:

$$1.5times{S_2}$$

The differnce after $1.5$ hours between the biker and skater is $35$km, i.e.:

$$1.5times{(2S_2 + 15)}-1.5times{S_2}=35$$ $$1.5times{(2S_2 + 15-S_2)}=35$$ $${S_2+15}=frac{35}{1.5}$$ $${S_2}=frac{35}{1.5}-15$$ $${S_2}=8frac{1}{3}$$

Your main error was adding the distances rather than subtracting them. I hope this helps!

The difference in the positions is 35 km.

Subtract $1.5S_2$ in your second equation instead of adding.