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# Write a system of two equations in two variables to solve the problem?

How do I solve this world problem?

Write a system of two equations in two variables to solve the problem.

Flying with a tailwind, a pilot flew an airplane 560 miles in 3.5 hours. Flying into a headwind, the return trip took 4 hours. Find the speed of the plane in calm air and the speed of the wind.

speed in calm air mph:

speed of wind mph:

### 4 Answers

- la consoleLv 712 months agoFavorite Answer
Recall: s = d/t → where s is the speed, d is the distance, t is the time

s₁: speed of the plane in calm air

s₂: speed of wind

Flying with a tailwind, a pilot flew an airplane 560 miles in 3.5 hours. You can add the 2 speeds together.

s = s₁ + s₂ = d/t

s₁ + s₂ = d/t → where: d = 560 miles

s₁ + s₂ = 560/t → where: t = 3.5 hours

s₁ + s₂ = 560/3.5

s₁ + s₂ = 160 ← equation (1)

Flying into a headwind, the return trip took 4 hours. You must subtract the 2 speeds.

s = s₁ - s₂ = d/t

s₁ - s₂ = d/t → where: d = 560 miles

s₁ - s₂ = 560/t → where: t = 4 hours

s₁ - s₂ = 560/4

s₁ - s₂ = 140 ← equation (2)

You calculate (1) + (2)

(s₁ + s₂) + (s₁ - s₂) = 160 + 140

s₁ + s₂ + s₁ - s₂ = 300

2.s₁ = 300

s₁ = 150 mph ← speed of the plane

Recall (2)

s₁ - s₂ = 140

s₂ = s₁ - 140 → we've just seen that: s₁ = 150

s₂ = 10 mph ← speed of the wind

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- KrishnamurthyLv 712 months ago
Flying with a tailwind, a pilot flew an airplane 560 miles in 3.5 hours.

Flying into a headwind, the return trip took 4 hours.

Find the speed of the plane in calm air and the speed of the wind.

With Wind:

distance = 560 miles ; time = 3.5 hrs ; rate = 560/3.5 = 160 mph

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Against Wind:

distance = 560 miles ; time = 4 hrs ; rate = 140 mph

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p + w =160

p - w = 140

-----------------

2p = 300

p = 150 mph (speed of the plane in still air)

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Since p + w = 160, w = 10 mph (speed of the wind)

speed in calm air = 150 mph:

speed of wind = 10 mph:

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- alexLv 712 months ago
speed in calm air mph=a

speed of wind mph= w

3.5(a+w)=560

4(a-w)=560

solve for a , w

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