Apply the distance, rate, and time problem-solving strategy together with the effect of wind on speed to find an unknown vehicle speed when different distances are covered in equal time — one segment with a headwind and one with a tailwind — producing a rational equation that is solved by multiplying both sides by the LCD.

Problem: An airplane can fly $200$ miles into a $30$ mph headwind in the same amount of time it takes to fly $300$ miles with a $30$ mph tailwind. What is the speed of the airplane?

1. Read and draw: Sketch two parallel paths — one showing $300$ miles with the wind (rate $r + 30$) and one showing $200$ miles against the wind (rate $r - 30$). Create a rate–time–distance table.

2. Identify: The speed of the airplane in still air.

3. Name: Let $r$ = the speed of the airplane. With a $30$ mph tailwind the effective rate is $r + 30$; with a $30$ mph headwind the effective rate is $r - 30$. Because $D = r \cdot t$, solving for time gives $t = \frac{D}{r}$. Divide each distance by its rate to fill in the time column:

	Rate (mph)	Time (hrs)	Distance (miles)
Headwind	$r - 30$	$\frac{200}{r - 30}$	$200$
Tailwind	$r + 30$	$\frac{300}{r + 30}$	$300$

4. Translate: The travel times are equal, so set the two time expressions equal:

$\frac{200}{r - 30} = \frac{300}{r + 30}$

5. Solve: Multiply both sides by the LCD, $(r + 30)(r - 30)$:

$(r + 30)(r - 30) \cdot \frac{200}{r - 30} = (r + 30)(r - 30) \cdot \frac{300}{r + 30}$

Cancel the matching factors:

$200(r + 30) = 300(r - 30)$

Distribute: $200r + 6{,}000 = 300r - 9{,}000$.

Subtract $200r$ from both sides: $6{,}000 = 100r - 9{,}000$.

Add $9{,}000$: $15{,}000 = 100r$.

Divide by $100$: $r = 150$.

6. Check: Is $150$ mph reasonable for an airplane? Yes. Tailwind: $150 + 30 = 180$ mph, and $\frac{300}{180} = \frac{5}{3}$ hours. Headwind: $150 - 30 = 120$ mph, and $\frac{200}{120} = \frac{5}{3}$ hours. The times are equal. $\checkmark$

7. Answer: The airplane was traveling at $150$ mph.

This example demonstrates the equal-time with wind scenario. Unlike earlier uniform motion problems where the distances or total times were known (producing linear equations), here the distances differ and the rates contain the variable in the denominator. Using $t = \frac{D}{r}$ to express each time as a rational expression and then setting those expressions equal produces a rational equation. Multiplying both sides by the LCD — the product $(r + 30)(r - 30)$ — clears the denominators and reduces the equation to a linear one that is solved with basic algebra.