# A tourist travels miles using two planes. the second plane averages miles per hour faster than the first plane. the tourist uses the slower plane for the first miles and the faster plane for the next miles. the total flying time is hours. what is the speed of the first plane?

Question: A tourist travels miles using two planes. the second plane averages miles per hour faster than the first plane. the tourist uses the slower plane for the first miles and the faster plane for the next miles. the total flying time is hours. what is the speed of the first plane?

If you enjoy solving word problems, you might be interested in this one: a tourist travels miles using two planes. The second plane averages miles per hour faster than the first plane. The tourist uses the slower plane for the first miles and the faster plane for the next miles. The total flying time is hours. What is the speed of the first plane?

This problem can be solved by setting up a system of equations and using algebra to find the unknown speed. Let x be the speed of the first plane and y be the speed of the second plane. Then we have:

y = x + m (the second plane is m miles per hour faster than the first plane)

t = d/x + d/y (the total flying time is t hours, where d is the distance traveled by each plane)

Substituting the values given in the problem, we get:

y = x + m

h = m/x + m/y

We can eliminate y by plugging in y = x + m into the second equation and simplifying:

h = m/x + m/(x + m)

h(x + m) = m(x + x + m)

hx + hm = mx + mx + mm

hx - mx - mx = mm - hm

x(h - m - m) = mm - hm

x = (mm - hm)/(h - m - m)

This is the speed of the first plane in terms of m, which is unknown. To find m, we can use the fact that the distance traveled by each plane is equal to its speed times its time. Since the slower plane travels for d/x hours and covers d miles, we have:

d = dx/x

d = x

Similarly, since the faster plane travels for d/y hours and covers d miles, we have:

d = dy/y

d = y

Substituting y = x + m, we get:

d = (x + m)d/y

d = xd/y + md/y

d - xd/y = md/y

y(d - x) = md

y = md/(d - x)

Plugging in y into the first equation, we get:

md/(d - x) = x + m

md = x(d - x) + m(d - x)

md = xd - xx + md - mx

xx - mx - md + mx = 0

xx - md = 0

x(m - x) = 0

This equation has two solutions: x = 0 or x = m. However, x cannot be zero because that would mean the first plane has no speed, which is impossible. Therefore, x must be equal to m. This means that the second plane is twice as fast as the first plane.

Now that we know that m = x, we can plug it into the formula for x that we derived earlier:

x = (mm - hm)/(h - m - m)

x = (xx - hx)/(h - x - x)

x = (x^2 - hx)/(h - 2x)

This is a quadratic equation that can be solved by using the quadratic formula or by factoring. We will use factoring here. First, we multiply both sides by (h - 2x):

x(h - 2x) = x^2 - hx

hx - 2xx = xx - hx

-3xx + 2hx = 0

x(-3x + 2h) = 0

Again, we have two solutions: x = 0 or x = 2h/3. As before, x cannot be zero, so we choose x = 2h/3. This is the speed of the first plane in terms of h, which is given in the problem as hours. Therefore, the speed of the first plane is:

x = 2h/3

x = 2( hours)/3

x =

This is our final answer. The speed of the first plane is miles per hour.