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## Elimination method for systems of equations

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# Systems of equations with elimination: potato chips

CCSS.Math:

## Video transcript

Everyone in the kingdom is very
impressed with your ability to help with the party
planning, everyone except for this gentlemen
right over here. This is Arbegla. And he is the
king's top adviser, and also chief party planner. And he seems somewhat
threatened by your ability to solve these otherwise
unsolvable problems, or at least from
his point of view, because he keeps over-ordering
or under-ordering things like cupcakes. And he says, king, that
cupcake problem was easy. Ask them about the
potato chip issue, because we could never get
the potato chips right. And so the king says,
Arbegla, that's a good idea. We need to get the
potato chips right. So he comes to you
and says, how do we figure out, on average, how many
potato chips we need to order? And to do that, we
have to figure out how much, on average,
does each man eat and how much each woman eats. You say, well, what
about the children? The king says, in
our kingdom, we forbid potato
chips for children. You say, oh, well,
that's all and good. Tell me what happened
at the previous parties. And so the king says, you might
remember, at the last party, in fact, the last two
parties, we had 500 adults. At the last party, 200 of
them were men, and 300 of them were women. And in total, they ate
1,200 bags of potato chips. And you say, what about
the party before that? He says, that one, we had a
bigger skew towards women. We only had 100 men,
and we have 400 women. And that time, we
actually had fewer bags consumed-- 1,100
bags of potato chips. So you say, OK, king
and Arbegla, this seems like a fairly
straightforward thing. Let me define some variables
to represent our unknowns. So you go ahead
and you say, well, let's let m equal the number
of bags eaten by each man. And you could think
of it on average, or maybe all the
men in that kingdom are completely identical. Or maybe it's the average number
of bags eaten by each man. And let's let w equal the number
of bags eaten by each woman. And so with these
definitions of our variables, let's think about
how we can represent this first piece of information,
this piece of information in green. Well, let's think about
the total number of bags that the men ate. You had 200 men. Let me scroll over a little bit. You had 200 men, and they each
ate m bags, m bags per man. So the men at this
first party collectively ate 200 times m bags. If m is 10 bags per man,
then this would be 2,000. If m was 5 bags per man,
then this would be 5,000. We don't know what m
is, but 200 times m is the total eaten by the men. Same logic-- total
eaten by the women is 300 women times the number
of bags eaten by each woman. And so if you add the total to
eaten by the men and the women, you get the 1,200 bags. So this is information,
written algebraically, given these variable
definitions. Now, let's do the same
thing with the second part of the information that they
gave us right over here. Let's think about how we can
represent this algebraically. Well, similar logic--
what was a total that the men ate at that party? It was 100 men times
m bags per man. And we're assuming that m
is the same across parties, that men, on average, always
eat the same number of bags. And how many did the women
eat at that second party? Well, you had 400 women. And on average, they
ate w bags per woman. So this is 400 times
w is the total number that the women ate. You add those two together,
you get the total number that all the adults ate. So this is going
to be 1,100 bags. So it looks pretty similar now. You have a system of two
equations with two unknowns. And so you try your
best to solve it. But when you solve it, you
see something interesting. Last time, it was
very convenient. You had a, I think it was
a 500 here, for 500 adults, and you had another 500. And so it seemed
like it was pretty easy to cancel out
one of the variables. Here it seems a little
bit more difficult. What's multiplying by the
m's, it's different here. The coefficient on the w
is different over here. You say, well, maybe I can
change one of these equations so it makes it a little
bit easier to cancel out with the other equation. So what if, for example, I
were to take this blue equation right over here and
multiply it by negative 2? And you might say,
well, Sal, why are we multiplying it by negative 2? Well, if were to multiply
it by negative 2, this 100m would become
a negative 200m. And if it was a
negative 200m, then that would cancel out
with a positive 200m when we add the two. So let's see what happens. So let's multiply this blue
equation by negative 2. We're going to
multiply by negative 2. Let me scroll over to
the left a little bit. So what happens? Remember, when we
multiply an equation, we can't just do one
side of the equation. We have to do the
entire equation in order for the equality to hold true. So negative 2 times
100m is negative 200m. Negative 2 times 400w-- there's
a positive right over there. So it becomes negative 800w. And then negative 2-- now,
we did the left hand side, but we also have to do
the right hand side. Negative 2 times 1,100
is negative 2,200. So just to be
clear, this equation that I just wrote
here essentially has the same information
we just manipulated. We just changed this equation,
multiplied both sides by negative 2. But it's kind of
the same constraint. But what makes this
interesting is, now, we can rewrite
this green equation. Let me do it over
here, this first one. 200m plus 300w is
equal to 1,200. And the whole reason why
I multiplied by negative 2 is, so that if I were
to add these two things, I might be able to get rid
of that variable over there. And so let's do that. Let's add the left hand sides,
and let's add the right hand sides. And you could literally
view it as, we're starting with this
blue equation. We're adding this
quantity, the left hand side of the yellow equation to
the left hand side of the blue. And then 1,200 is
the exact same thing that we're adding to
the right hand side. We know that this
is equal to this. So we can add this to the
left hand side and this to the right hand side. So let's see what happens. So the good thing is,
the whole reason we multiplied it by negative
2, so that these two characters cancel out. You add those two together. You just get 0m or just 0. You have negative
800w plus 300w. Well, that's negative 500w. And then on the
right hand side, you have negative 2,200 plus 1,200. So that's negative 1,000. And now this is pretty
straightforward-- one equation, one unknown,
a fairly straightforward equation. We divide both sides by
the coefficient of w, multiplying w. So divide by negative 500 on
the left, divide by negative 500 on the right. And we are left with
w is equal to 2. On average, women ate
two bags of potato chips at these parties. We're assuming that's
constant across the parties. So let's think about how
you would then figure out how many bags, on
average, each man ate. Well, to do that,
we just go back to either one of
these equations. In the last set of videos, I
went to the first equation. I'll show that the second
equation should also work. Either one should work. So let's substitute back
into the second equation. And you could either pick this
version of it or this one. But I'll pick the original one. So you have 100
times m, which we're trying to figure out,
plus 400 times-- well, we now know that w is
equal to 2-- 400 times 2 is equal to 1,100. So you have 100m plus
800 is equal to 1,100. And now, to solve for
m, we could subtract 800 from both sides. And we are left with
100m is equal to 300. And now, divide
both sides by 100. And we are left with m, which
is, on average, the number of bags of chips each
man eats is equal to 3. So you have solved
Arbegla's problem, what he thought was a
difficult problem, using the magical, mystical
powers of algebra. You were able to tell the king
in his party planning process that, on average,
the men will eat three bags of potato chips each. And on average,
the women will eat two bags of potato chips each.