Apply the seven-step problem-solving strategy for systems of linear equations to a uniform motion problem involving wind, using two variables for the still-air speed and the wind speed and solving the resulting system by elimination.

Problem: A private jet can fly $1{,}095$ miles in three hours with a tailwind but only $987$ miles in three hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

1. Read the problem and draw a diagram showing the jet traveling with the wind (tailwind, $1{,}095$ miles in $3$ hours) and against the wind (headwind, $987$ miles in $3$ hours).

2. Identify: The jet's speed in still air and the wind speed.

3. Name: Let $j$ = the speed of the jet in still air (mph) and $w$ = the speed of the wind (mph). With a tailwind the wind assists the jet, so the effective rate is $j + w$. With a headwind the wind opposes the jet, so the effective rate is $j - w$. Fill in the rate–time–distance table:

	Rate (mph)	Time (hrs)	Distance (miles)
Tailwind	$j + w$	$3$	$1{,}095$
Headwind	$j - w$	$3$	$987$

4. Translate into a system of equations. Since distance equals rate times time:

$\left\{\begin{array}{l} 3(j + w) = 1{,}095 \\ 3(j - w) = 987 \end{array}\right.$

5. Solve by elimination. Distribute to convert both equations to standard form:

$3j + 3w = 1{,}095$ $3j - 3w = 987$

Because both trips take the same amount of time ($3$ hours), the $w$-coefficients are already opposites ($+3w$ and $-3w$). Add the equations directly:

$6j = 2{,}082$

Divide both sides by $6$: $j = 347$.

Substitute $j = 347$ into the first equation to find $w$:

$3(347 + w) = 1{,}095$

$1{,}041 + 3w = 1{,}095$

$3w = 54$

$w = 18$

6. Check: Tailwind rate: $347 + 18 = 365$ mph, and $365 \times 3 = 1{,}095$ miles ✓. Headwind rate: $347 - 18 = 329$ mph, and $329 \times 3 = 987$ miles ✓.

7. Answer: The jet travels at $347$ mph in still air, and the wind speed is $18$ mph.

This example demonstrates how the tailwind/headwind framework — where the effective rates are $j + w$ and $j - w$ — produces a system of two linear equations in two unknowns when combined with the distance formula $d = rt$. Because both trips have equal travel times, distributing the same time factor across each equation yields $w$-coefficients that are already opposites, so the equations can be added immediately without any preliminary multiplication — making elimination especially efficient. Compare this with the downstream/upstream river problem where unequal travel times produce different coefficients that require a multiplication step before the equations can be added.