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## Evaluating expressions with unknown variables

# Worked example: evaluating expressions using structure

CCSS.Math:

## Video transcript

I'm now going to do a
few classic problems that show up a lot at
math competitions and sometimes on
standardized tests, and they seem like very daunting
and intimidating problems. But hopefully over the
course of this video, you'll realize that if you kind
of see what they're asking for, it's really not that bad. So let's try this first
one right over here. They tell us a plus b
plus c is equal to 7. And then they ask us, what is
5a plus 5b plus 5c equal to? And the first
reaction's like, well, they just gave me one
equation right over here. One equation with
three unknowns, how do I solve for a, b, or c? Don't I have to be able
to solve for a, b, or c in order to figure out
what 5a plus 5b plus 5c is? So I'll let you think
about that for a second. So the big idea
here is to realize that 5a plus 5b
plus 5c, that's just the same thing as 5
times a plus b plus c. If you distribute the 5 here, 5
times a, 5 times b, 5 times c, you get 5a plus 5b plus 5c. Another way you could think of
it is we are factoring a 5 out. If you factor a 5
out, you're left with 5 times a plus b plus c. Now, how do we evaluate this? Well, that first equation
gave us all the information that we needed. They told us that a plus
b plus c is equal to 7. So 5 times a plus b plus
c is the same exact thing. I think you see where
this is going now. It's 5 times 7, and now this
becomes pretty straightforward. That's going to be equal to 35. Let's do one more of these. And this is going to be a
little bit more involved, but hopefully you
see the same idea. So we're told a plus b plus
c is equal to negative 1. And we have two
more variables here. We're told that x
plus y is equal to 7. Then they ask us this
big, hairy thing. What is this equal to? And I'll give you a
few seconds to give it a little bit of thought. Well, you could imagine. They've given three
variables with one equation up here, another two variables
with another equation. We have five unknowns
with two equations. There's no way we're going to be
able to individually figure out what a, or b, or
c, or x, or y is. But maybe we can use what
we saw in the last example to solve this. And so what we might want do
is rearrange it so that we have the x's and the y's kind
of grouped together, and the a's, b'c, and c's
kind of grouped together. So let's try that out. So let's focus first on
the a's, b's, and c's. So we have negative
9a, negative 9b-- I'll go in order, alphabetical
order-- negative 9b, and we have negative 9c, and I
think you see what's emerging. And then let's work on
the x's and the y's. And then we have
negative 7x, minus 7x, and then we have a minus 7y. So all I have done is
rearranged this expression here. But this makes it a
little bit clearer of what's going on here. These first three terms, I
can factor out a negative 9, and I get negative 9
times a plus b plus c. And these second two terms,
I can factor out a 7, minus 7 times x plus y. And just to verify, if you
wanted to go the other way, multiply this negative 7 times
x plus y, you'll get this. Multiply negative 9 times a
plus b plus c, distribute it, you'll get this right over here. And so this makes it
a little bit clearer. What is a plus b
plus c equal to? Well, they tell us
right over here. a plus b plus c is
equal to negative 1. This whole expression
is negative 1, at least in the parentheses. And what is x plus y equal to? Let me do this in a new color. What is x plus y equal to? Well, they tell us right over
here. x plus y is equal to 7. So this whole thing
simplifies to negative 9 times negative
1 minus 7 times 7. And so this is equal to--
we're in the home stretch-- negative 9 times negative 1. Well, that's positive 9. And then negative 7
times 7 is negative 49. So it's 9 minus 49, which
is equal to negative 40. And we are done.