Let the cyclist's normal speed be $x$ km/h.

Given:

1. Distance to the destination is 35 km.
2. If the speed increases by 2 km/h, it takes 2 hours less to reach the destination.

Let's set up equations to solve this problem.

Step 1: Formulate the Equations

The time taken to reach the destination at the normal speed is: $\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{35}{x}$

If the speed is increased by 2 km/h, the speed becomes $x + 2$ km/h, and the time taken is: $\text{New time} = \frac{35}{x + 2}$

According to the problem, this new time is 2 hours less than the original time: $\frac{35}{x} - \frac{35}{x + 2} = 2$

Step 2: Solve the Equation

Multiply both sides by $x(x + 2)$ to eliminate the denominators: $35(x + 2) - 35x = 2x(x + 2)$

Expanding both sides: $35x + 70 - 35x = 2x^2 + 4x$

Simplify the equation: $70 = 2x^2 + 4x$

Divide by 2 to simplify further: $x^2 + 2x - 35 = 0$

Step 3: Solve the Quadratic Equation

Now we solve $x^2 + 2x - 35 = 0$ using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = 2$, and $c = -35$.

Calculating the discriminant: $x = \frac{-2 \pm \sqrt{2^2 + 4 \cdot 1 \cdot 35}}{2 \cdot 1} = \frac{-2 \pm \sqrt{144}}{2}$ $x = \frac{-2 \pm 12}{2}$

This gives two solutions: $x = \frac{-2 + 12}{2} = 5$ or $x = \frac{-2 - 12}{2} = -7$

Since speed cannot be negative, we have: $x = 5$

Answer

The cyclist's normal speed is 5 km/h.

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Would you like a breakdown of each step, or do you have any questions?

Here are some related questions to deepen understanding:

1. How would the equation change if the distance were different, say 50 km?
2. What if the time saved was 3 hours instead of 2 hours? How would we solve it?
3. How would you interpret the solution if one of the roots had been negative in a different scenario?
4. Can we solve this problem by another method, such as graphing?
5. How would the formula change if we knew only the time and wanted to find the distance?

Tip: Always check the units of speed and time to make sure they align properly in word problems.