Question

two cyclists leave towns 120 miles apart at the same time and travel toward each other. one cyclist travels 6 \\(\frac{\text{mi}}{\text{h}}\\) faster than the other. if they meet in 3 hours, what is the rate of each cyclist? rate of the faster cyclist: \\(\square \frac{\text{mi}}{\text{h}}\\) rate of the slower cyclist: \\(\square \frac{\text{mi}}{\text{h}}\\)

Explanation:

Step1: Define variables

Let the rate of the slower cyclist be \( x \) miles per hour. Then the rate of the faster cyclist is \( x + 6 \) miles per hour.

Step2: Use distance formula

The distance traveled by the slower cyclist in 3 hours is \( 3x \) miles, and the distance traveled by the faster cyclist in 3 hours is \( 3(x + 6) \) miles. Since they start 120 miles apart and meet, the sum of their distances is 120 miles. So we have the equation:
\( 3x + 3(x + 6) = 120 \)

Step3: Simplify and solve the equation

First, expand the left side: \( 3x + 3x + 18 = 120 \)
Combine like terms: \( 6x + 18 = 120 \)
Subtract 18 from both sides: \( 6x = 120 - 18 = 102 \)
Divide both sides by 6: \( x = \frac{102}{6} = 17 \)

Step4: Find the rate of the faster cyclist

The rate of the faster cyclist is \( x + 6 = 17 + 6 = 23 \)

Answer:

Rate of the faster cyclist: \( \boldsymbol{23} \) \( \frac{\text{mi}}{\text{h}} \)
Rate of the slower cyclist: \( \boldsymbol{17} \) \( \frac{\text{mi}}{\text{h}} \)