Alissa Fong

**MA, Stanford University**

Teaching in the San Francisco Bay Area

Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts

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This problem is more difficult from the ones we've already seen because it’s her original rate and then we need to take or her time is going to change by one and a half hours. Let's try to set us some equations.

Well we know usually distance is equal to rate times time, so Alison's distance usually is 12, her rate usually we are just going to call R and her time usually we'll call T. We don't know either of those. That's like what we know that Alison usually does, but then it tells us if she doubles her usual speed, okay doubles her speed. So I'm going to call that 2R now. If she doubles her usual speed she can complete the course in one and a half hours less than her usual speed.

Okay so her time instead of being just T, it’s going to be T take away 1½ because now she is going one and a half hours more quickly. She's still going on a 12 mile course, that's my system of equations that I can go ahead and solve.

I got these two equations this one represents her original situation, 12 miles some rate some time, we don’t know what they are and I’m going to be solving that with the same distance but twice her usual speed and then her time take away an hour and a half.

So from here you guys have a choice using substitution, elimination, maybe matrices if you learnt about that you can maybe graph these nasty guys if you want to, up to you. I'm going to go ahead and solve this using substitution that's my own personal favorite.

So if I was to for example solve this first equation for R, divide both sides by T I would get R is equal to 12/T. Then I’m going to substitute that equation right here so I only have one equation with one variable. I'm going to be solving 12 equals 2 times 12/T times T take away 1.5. Once I have this all set up, this is just your standard straight forward solving and simplifying that I know you guys know how to do.

12 is equal to 24/T when I distribute it there times T take away 1.5, draw a little arrow move over here, so if I go through and if I distribute this into both of these quantities I have 12 is equal to 24/T times T/1, those Ts are going to be eliminated, I’ll just have 24 there take away, 24 times -1.5 is -36 and I need to put that on top of T, -36/T. What I’m going to do is subtract 24 from both sides -12 equals -36/T. If you want to you could multiply both sides by T in this case to get T out of the denominator or you could turn this into a fraction and cross multiply it’s the same process written a little bit differently, I’m going to multiply both sides by T so that I have this. Oops! What happened to my T, there it is and when I divide both sides by -12, I’ll get that T equals 3 hours.

Now let me just think about what that means before I decide that I’m done, I want to decide, I want to figure out if I did this all correctly. This tells me that R represents her original speed I chose R to represent her original speed, so if I go through and I know that T is three hours that's how long it usually took her. It doesn't tell me how fast she goes we need to find her speed.

It's told me how long it took her so in order to find her speed I need to go back to this original first problem I know that 12 is equal rate which I’m trying to find times 3, because 3 is my T value, her original time, and it tells me she used to walk 4 miles per hour that's what her rate used to be, R.

So this one was kind of tricky because we had some Ts in the denominator we had some fractions to deal with but as long as you guys can set up this first step, if you can get this systems of equations problem ready to go I’m confident you guys can do these problems.

These rate problems are tricky, they are difficult but you guys can do them if you just put in a little bit extra time and make sure you are focusing when you’re doing these problems.