Two cyclists each cover 120 km. The first rides 3 km/h faster than the second and finishes 2 hours sooner. Find both speeds.

Step-by-step solution

1. 1

   Set up a table with one unknown. Let $x$ be the second cyclist's speed in km/h; then the first rides at $x+3$. Time is distance over speed:

   cyclist	speed (km/h)	distance (km)	time (h)
   second	$x$	$120$	$\dfrac{120}{x}$
   first	$x+3$	$120$	$\dfrac{120}{x+3}$

   Physically $x>0$, and the slower rider's time is the larger one.

2. 2

   Turn the time gap into an equation. The first cyclist takes $2$ hours less, so subtract the smaller time from the larger:

   $\frac{120}{x}-\frac{120}{x+3}=2$

3. 3

   Clear the denominators. Multiply both sides by $x(x+3)$, which is nonzero for $x>0$:

   $120(x+3)-120x=2x(x+3)$

   The left side collapses because the $120x$ terms cancel:

   $360=2x^{2}+6x$

4. 4

   Reduce to standard quadratic form. Divide by $2$ and move everything to one side:

   $x^{2}+3x-180=0$

5. 5

   Solve and discard the extraneous root. The discriminant is $D=3^{2}+4(180)=729$ with $\sqrt{D}=27$, so

   $x=\frac{-3\pm 27}{2}\ \Longrightarrow\ x=12\ \text{or}\ x=-15$

   A negative speed is meaningless here, so $x=12$ and the first cyclist rides at $x+3=15$ km/h.

6. 6

   Check against the original wording. Times are $\dfrac{120}{12}=10$ h and $\dfrac{120}{15}=8$ h. The speed gap is $15-12=3$ km/h ✓ and the time gap is $10-8=2$ h ✓.