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## College Algebra

### Unit 8: Lesson 2

Modeling with rational expressions# Rational equation word problem

CCSS.Math:

Sal models a context that concerns the number and price of pizza slices. The model turns out to be a rational equation. Created by Sal Khan.

## Want to join the conversation?

- at4:51how do I know which set of numbers to add/subtract from each side? Real confused about it :/ and how do you know weather to add/subtract?(7 votes)
- It doesn't really matter. Just get everything on one side however you want to do it. Go back and view the "Quadratic equations" playlist before these, if you need review.(7 votes)

- In the upcoming practice section, Modeling with one-variable equations and inequalities, I stumbled across a problem which, given the following equation, required us to solve for t:

1024*(1/2)^(t/29)=32

solve for t.

The solution given in the hints went:

(1/2)^(t/29)=32/1024

(1/2)^(t/29)=1/32

(1/2)^(t/29)=(1/2)^5

therefore

t/29=5

t=29*5=145

Now, although I fully understand the logic when I see it, the passage that went from (1/2)^(t/29)=1/32 to (1/2)^(t/29)=(1/2)^5 would never have occurred to me on my own, not in a million years. Is there any previous section I should study harder, any method of handling bases and exponents I should assimilate perhaps, or am I just not smart enough to see what should be obvious at this stage? Thanks in advance for any answer.(5 votes)- If you have a variable or constant in an exponent that you need to solve for, you only have a few options available to you at this level of study (there are some much more advanced techniques, but those a few years ahead of Algebra 2). And, actually, the two methods are really just different applications of the same method.

Method 1: use logarithms to bring the constant or variable out of the exponent, so you can solve for it.

Method 2: get both sides of the equation to be the same base raised to some exponent. If you can do that, then you know that the two exponents must be equal to each other.

You've seen method 2, but it is not always particularly easy to use (and is just a different way of doing method 1 anyway). So, let us look at how to do this same problem with method 1:

1024(½)^(t/29)=32

(½)^(t/29)=32/1024

(½)^(t/29)= 1 /32

log [(½)^(t/29)] = − log (32) ←**This uses the property log (1/a) = − log (a**)

(t/29) log (½) = − log (32) ←**This uses the property log (a^b) = b log (a**)

(t/29)[− log 2] = − log (32)

(t/29) = − log (32)] / (− log 2)

t/29 = 5

t = 29*5

t = 145(6 votes)

- Why is it that making more pizzas per day would result in her expenses not changing when the ingredients started costing more? Maybe I'm just overthinking it but I don't really get it.(4 votes)
- Her total expenses will go up, since the ingredients cost more and she is using more of them. But the expense
*per pizza*is the same. Expense per pizza is (total expenses)/(number of pizzas). Total expenses have increased the numerator, so we must also increase the denominator to keep expense per pizza the same.(4 votes)

- A slow train traveling from Tashkent to Samarkand arrives 9 minutes late when traveling at 36km/h. If it travels at 27km/h it arrives 39 minutes late. What is the distance between Tashkent and Samarkand?

Help me, please, simulate this problem, if anyone can!(3 votes)- This is a simultaneous equation problem.

Call the*scheduled*journey time, in hours, t, and the distance d (km)

The first statement tells us that the journey takes t + 9/60 hours at 36 km/h

So d = (t + 9/60)*36

Convert the second statement into a similar equation. You now have two equations in two variables, which you should be able to solve.

There are many videos on simultaneous equations, here's one:

https://www.khanacademy.org/math/algebra-home/alg-system-of-equations/alg-equivalent-systems-of-equations/v/solving-systems-of-equations-by-elimination(3 votes)

- At3:00, for the equation on the right (purple), can't it be simplified to 8+2?

My reasoning: 8+2(p+8)/p+8, the p+8 cancel out and left with 8+2.

Also at6:30, does the coefficient must be 1?(2 votes)- No becasue it is (8+2(p+8)/(p+8), so you are not canceling equivalents. If you broke this fraction down into 2 parts, it would be 8/p+8 + 2. This makes it worst than where we started.

To your second question, a coefficient of 1 can become invisible, so p^2 and 1 p^2 are actually the same thing. If you see a variable without a coefficient, get used to the idea that there is an invisible 1 in front that will almost never show up in final answers.(2 votes)

- At1:36, why does Sal divide the whole "8 + 1.5p"

expression by**p**and why did we not divide just the "1.5p" by**p**to get the cost for per pizza?(2 votes)- 1.5p/p only tells us the price of toppings per pizza, which we already know. We want the total price per pizza in one day. So for 1 pizza running the stove is 8 dollars and the price of toppings is 1.5, this means we can estimate that one pizza cost 9.50 dollars. similarly, 2 pizzas would be 8 + 1.5*2 = 11 dollars. this means 11/2 = 5.50 dollars per pizza. It not TECHNICALLY the actual price per pizza, but it is a helpful way to express it. Does this help?(2 votes)

- p=-16 is called extraneous, isn't it ? Also shouldn't we assume P doesn't equal 0 or -8 ? because if it's we would've divided by 0 also would've multiplied by zero which loose information(2 votes)
- why do we need to find per pizza?

Especially in this equation

Dominic from "Dominic's Pizza" always bakes p pizzas each day. Currently, it costs him $10 per day to use a brick oven and $2 per pizza for the ingredients.

One day Dominic realized that if he switched to an electric oven, the use of the oven would cost him $20 per day, but the ingredients would be cheaper, only $0.80 per pizza. This way, his total expenses for each pizza (including shared oven costs and ingredient costs) would be reduced by $1(2 votes) - Intuitively I feel like there should be a variable for # of days in this model. I know it would cancel out somehow, but I can't figure out where.(1 vote)
- Since the calculations are for only one day, the number of days does not need to be included as part of the equation.(3 votes)

- Why did he find the price per pizza? What would prompt one to find the price per item?(2 votes)

## Video transcript

Dominique from Dominique's
Pizza bakes the same amount of pizza every day. She used to spend $8 each
day on using the oven, and $1.50 on ingredients
for each pizza. So $8 each day on
using the oven, and $1.50 on ingredients
for each pizza. One day the price for the
ingredients increased from $1.50 to $2 per pizza. Dominique made some
calculations and found that she should bake
8 pizzas more each day so the expenses for a single
pizza would remain the same. And assume they're
saying the total expenses for a single pizza, because
clearly the ingredients cost is not the same. We're talking about
the total ingredients. So if we were to spread
the cost of the oven across all of the pizzas. Write an equation to
find-- or the total cost for the oven per day, to spread
that across all the pizzas. Write an equation to find out
how many pizzas Dominique baked each day before the
change in price. Use p to represent
the number of pizzas. So let's just think about her
total cost per pizza before and then her total
cost per pizza after if she bakes
8 more pizzas. So before, we're
going to use p to say that's the number
of pizzas she baked per day before the
change in price. So before the change in
price, on a given day, she would spend $8 on the oven
and then $1.50 on ingredients for each pizza. So 1.5, or $1.50, times
the number of pizzas. This would be her total cost
on all the pizzas in that day. It's the oven cost plus
it's the ingredients cost. So if you wanted this
on a per pizza basis, you would just divide
by the number of pizzas. Now let's think
about what happens after the change in price. After the change in price,
her cost per day for the oven is still $8. But now she has to spend $2
per pizza on ingredients. So $2 per pizza. And instead of saying that
she's baking p pizzas, let's say that she's now
baking 8 more pizzas each day. So it's going to be p plus 8. And so this is going to
be her total cost for all of the pizzas she's now baking. And so if you want it on
a per pizza basis, well, she's now making
p plus 8 pizzas, you would divide by p plus 8. And the problem tells
us that these two things are equivalent. Here you had a higher cost
in ingredients per pizzas, but since you are now
baking more pizzas, you're spreading the oven cost
amongst more and more pizzas. So let's think about
what p has to be. p has to be some number,
some number of pizzas, so that these two
expressions are equal. Her total cost per pizza
before, when she only made p, is going to be the same as
her total cost per pizza when she's making
p plus 8 pizzas. So these two things
need to be equal. So we did that first part,
or we did what they asked us. We wrote an equation to find out
how many pizzas Dominique baked each day before the
change in price. And we used p to represent
the number of pizzas. But now for fun, let's
actually just solve for p. So let's just simplify
things a little bit. So this part right over here. Actually, let's just cross
multiply this on both sides. Or another way of thinking
is multiply both sides times p plus 8 and multiply
both sides times p. So if we multiply by p plus
8, and we multiply by p, we multiply by p plus
8, and we multiply by p, that cancels with that. That cancels with that. On the left-hand
side-- so let's see. We have to just do the
distributive property twice right over here. What is p times 8 plus 1.5p? Well, that's going to be 8p. I'm just multiplying the
p times this stuff first. Plus 1.5p squared. And now let's multiply the
8 times both of these terms. So plus 64 plus-- 8 times
1.5, that is 12-- plus 12p. And that's going to be equal
to-- let's see, let's multiply p times all of this business. So that's going to be equal
to 8p-- 8 times p is 8p-- and let's see, I could
distribute these terms and then multiply by p. So 2 times p is 2p, times p is
2p squared, plus 2p squared. And then 2 times 8
is 16 times p is 16p. So now we have--
well, we essentially end up with a
quadratic equation, but let's simplify
it a little bit so that we can either factor it
or apply the quadratic formula. So let's see, let's subtract
1.5p squared from both sides. So subtract 1.5p squared. Actually, let me
just put everything on the left-hand side
just because that might be a little
bit more intuitive. So let's subtract 2p
squared from both sides. Let's subtract 16p
from both sides. We have an 8p and a
12p, and then we're going to subtract a
16p from both sides. And then, actually, let's
subtract an 8p as well from both sides. We have a 16p and an 8p, so that
actually works out quite well. So now we've subtracted
8p from both sides, 16p from both sides. So we've essentially
subtracted all of this stuff from both sides. And we are left with--
let's see, I'll do it in degree order. 1.5p squared minus 2p squared
is negative 0.5p squared. Now let's see, these cancel out. 12p minus 16p is minus 4p. And then we have plus 64. And then that is going
to be equal to 0. And just to simplify
this a little bit, or just to make this
a little bit cleaner, let's multiply both sides of
this equation by negative 2. I want the coefficient
over here to be 1. So then we get p
squared plus 8p. p squared plus 8p is going
to be equal to-- let's see, negative times negative 2. So minus 128-- is
going to be equal to 0. So let's see if we
can factor this. Can we think of two numbers
where if we take their product, we get negative 128? And if we were to add them
together, we get positive 8. So they're going to
have different signs right over here. So let's see, if we say 12
times-- well, let's see. What numbers could this be? So if we were to think about
128 is the same thing as-- 16, let's see, 16 goes into 128. Let me work through this. 16 goes into 128,
does it go 8 times? 8 times 6 is 48. 8 times 10 is 80,
plus 40 is 128. Yep, it goes 8 times. So 16 and 8 seem to work. So if you have positive
16 and negative 8, their product would
be negative 128. So we can factor this out
as p plus 16 times p minus 8 is equal to 0. Now, this is going to be equal
to 0 if at least one of these is going to be equal to 0. So we have two solutions. Either p plus 16 is going to
be equal to 0, or p minus 8 is equal to 0. This one right over here,
subtract 16 from both sides, you get p is equal
to negative 16. Here, you get p is equal to
8, if you add 8 to both sides. Now, we're talking about
a number of pizzas made. So this one doesn't apply. This would be like
Dominique eating 16 pizzas, or somehow destroying
16 pizzas a day. We're not interested
in that solution. So if we want the solution
to the original question, the number of pizzas she made
before the increase in price, she made 8 pizzas per day. So p right over here
needs to be equal to 8. So before the change in price,
she made 8 pizzas a day. After the change in price,
she made 8 more pizzas a day, or 16 pizzas per day.