# A jet plane, flying 165 mph faster than a propeller plane, travels 2970 miles in 3 hours less. How fast is each plane?

A jet plane, flying 165 mph faster than a propeller plane, travels 2970 miles in 3 hours less time than the propeller plane takes to fly the same distance. How fast does each plane fly?

Is there a formula I can use to start this problem? Specifically, what can I do to begin finding the MPH of one of the planes?

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For this type of problem, you are creating the formula. In this case, you have a couple unknowns: the total amount of time travelled and the speed. If $x$ is the speed of the faster plane and $y$ is the speed of the slower plane, then $xt_x - yt_y=0$. Can you come up with two more equations relating $x$ and $y$, and $t_x$ and $t_y$? – abiessu Sep 11 '13 at 23:32
Well, maybe. But before I try, can I ask what the smaller x to the right of the "t" indicates? (Sorry, I am so slow when it comes to this stuff). – Robin Sep 11 '13 at 23:34
Or, what it means in the context of the formula? – Robin Sep 11 '13 at 23:34
$t_x$ indicates the time it takes the plane associated with speed $x$ to fly the given distance. – abiessu Sep 11 '13 at 23:35
So what can I do to begin finding the MPH of one of the planes? – Robin Sep 11 '13 at 23:40

Distance = Rate $\times$ Time.

So, for the propeller plane, $2970 = x * y$, where $x$ is how fast the propeller plane flies, and $y$ is how long it flies for.

Then, for the jet plane, we know $2970 = (x + 165) * (y - 3)$.

If we use the first equation, we could represent $x$ in terms of $y$, or vice versa. Let's represent $x$ in terms of $y$ as $x = \frac{2970}{y}$. Then, for the jet plane, we know $2970 = (\frac{2970}{y} + 165)*(y-3)$.

Hence, $\frac{2970}{y-3} = \frac{2970}{y} + 165$. We could make this easier by saying, $\frac{2970}{y} + 165 = \frac{2970}{y} + \frac{165*y}{y} = \frac{2970 + 165*y}{y}$

This leads us to cross-multiplication of $\frac{2970}{y-3} = \frac{2970 + 165*y}{y}$. Hence, $2970*y = 2970 + 165*y*(y-3) \implies 2970*y = 2970*y - 2970*3+165*y^2 - 165*3*y$. Hence, 0 = -2970*3 + 165*y^2 - 165*3*y. From this point, you could use the quadratic equation to figure out y, and then use 2970 = x*y to solve for x.

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How would I set up the quadratic equation if for instance what would normally just be "165y" is now "165y^2"? – Robin Sep 11 '13 at 23:59
Well, the quadratic equation would be $165*y^2 - 165*3*y - 2970*3$, hence you just plug in the respective coefficients into $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. – David Sep 12 '13 at 0:07
Argh, I'm sorry. I don't get it. I'll keep working with it. – Robin Sep 12 '13 at 0:15
Do you understand that the coefficients are 165, -165*3, -2970*3 in order? – David Sep 12 '13 at 0:16
Yes, and I got the solutions 9 and -6 but I'm not sure where those would be relevant in continuing to find the solution... – Robin Sep 12 '13 at 0:17