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Megan traveled 165 miles to visit friends. On the return trip she was delayed by construction and had to reduce her average speed by 22 miles per hour. The return trip took 2 hours longer. What was the time and average speed for each part of the trip?

Initial trip: 55 $\mathrm{mph}$ and 3 hoursReturn trip: 33 $\mathrm{mph}$ and 5 hours

Algebra

Chapter 4

RELATIONS AND FUNCTIONS

Section 10

Inverse Variation

An Introduction to Geometry

Functions

Linear Functions

Polynomials

Oregon State University

Harvey Mudd College

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for this problem. There's a pursued use traveling 165 miles total and were given that on the way back there speed decreases by 22 their time increases by two hours. So we can set up a equation to equations actually, better going to be ours equal to ours. And we'll get that by multiplying miles per hour, times, hours, and then just setting. Um then oh, we're setting actually miles, two miles. So it would just be like distance is equal to distance. So we multiply in miles per hour by ours, we get miles, and then we're giving a miles. So let me start with 1 65 which is are known, and we know that that's just equal to the initial speed. I'll call initial Speed s Times the initial time that it took. So that's t. And on the other equation that we use is 1 65 is equal to like the way back. So this is the way back equation. So we do s minus 22 times T plus two. So if we were to substitute the first equation into the second equation, we that probably the best way to do is with us. So those two s is equal to 1 65 over T. And so if I was to put that in, I get 1 65 is equal to 1 65 T minus 22 times T plus two. It's probably easiest to multiply both sides, but a t So if I multiply them by T, I'll get 1 65 t is equal to I'll put the tea into this first term here, so I'll get 1 65 Why is 22 t And since I distributed the tea in the first term, I don't have to distribute it in the second term cause of on the community property, so it just stays the same. So if I was to distribute this hour, I get 1 65 t sequel to 1 65 T minus 22 t squared. Um, plus, I'm actually minus 20 Li aa plus. So I did first Terrorism, tens. First term. I did second term trans first term. I mean, he first turned time second terms. That's two times 1 65 and then second term physic in term is minus 40 40. So these two terms. Cancel out. And if I'm a bit to the right side, I get 22 t squared plus 44 t minus, um, 3 30 which is two times 1 65 is equal to zero. And so there's a common devices here that is 22. So if I pull that, I get t squared plus to t minus 15 this equal to zero. And so if I use, um, factoring, I'll get three and five as the numbers of monies. And in order to fix the signs, I'll use this one and this one it is equal to zero. So you get that The initial time taking was three hours. We can't use the other one because it will be negative. So this is tthe e real answer for initial time. So it's a t T. Initial three hours s initial is 1 65 divided by three, which comes out to 55 uh, Mph. And then for a T final, it's ah, two more. So be five hours and because it's in the same proportion s as final will be 33 and ph

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