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Current time:0:00Total duration:5:12

We're now ready to
solve for A, B, and C, given these three equations
with three unknowns. So let's first start with
these top two constraints. So I have this one
right over here in blue. Let me just copy and
paste it right over here. And then let me rewrite
this constraint, but let me multiply
it by negative 2 so that I can cancel
out the C terms. So it's negative 2A
minus 2B minus 2C is equal to negative 2. And then, if I add the left-hand
sides and the right-hand sides, on the left-hand sides, I get
6A plus 2B-- these cancel out-- is equal to 3. And then the other
constraint I could use is this orange one
and the green to get the C's to cancel out here. So I have 27A plus 9B
plus 3C is equal to 14. And here, let's see. I can multiply it
by negative 3 if I want to get the C's
to cancel out again. And my whole goal is to have two
equations and two unknowns that have leveraged these two
constraints in tandem and these two
constraints in tandem. So now I'm going do these two. So I've multiplied this
equation times negative 3. I get negative 3A minus 3B
minus 3C is equal to negative 3. And now I can perform
the subtraction. And I get 24A plus 6B-- these
cancel-- is equal to 11. Now I have two equations
in two unknowns. And let's see. If I multiply this
equation right over here times
negative 3, I should get the B's to cancel out. So let's do that. So if I multiply this
times negative 3, this constraint, or this
term, 6A times negative 3, is negative 18A. 2B times negative 3, minus 6B,
is equal to 3 times negative 3 is equal to negative 9. And now if we add both
sides, on the left-hand side, we have-- let's see. 24A minus 18A is 6A-- these
cancel-- is equal to 11 minus 9 is 2. Divide both sides by 6. We get A is equal to 2/6,
which is the same thing as 1/3. And so now we can substitute
back to solve for B. So let's see. We have 6 times 1/3. Our A is 1/3. Plus 2B is equal to 3. 6 times 1/3 is 2. 2 plus 2B is equal to 3. Subtract 2 from both sides. 2B is equal to 1. Divide both sides by 2. B is equal to 1/2. So A is 1/3. B is 1/2. Now we just have to solve for C. So we can go back to
this original equation right over here. So we have 1/3 plus 1/2
plus C is equal to 1. Actually, let me do that over
here so I have some space. So I have some space, let
me do it right over here. So I have 1/3 plus 1/2
plus C is equal to 1. Now we find ourselves
a common denominator. So let's see. A common multiple
of 3, 2, and-- I guess you could say 1-- this
is 1 over 1-- is going to be 6. So I can rewrite
this as 2/6 plus 3/6 plus C is equal to 6/6. 1 is the same thing as 6/6. So this is 5/6 plus
C is equal to 6/6. Subtract 5/6 from both sides. We get C is equal to 1/6. So C is equal to 1/6. And there we are. We deserve a drum roll now. We've figured out a formula for
the sum of the first n squares. So we can rewrite this. This formula is now
going to be-- A is 1/3. So it's 1/3 n to the third
power plus 1/2 n squared-- let me make sure it doesn't
look like that n is a-- 1/2 n squared plus 1/6 n. So this is a pretty
handy formula. If now you wanted to find
0 squared plus 1 squared plus 2 squared plus 3 squared,
all the way to 100 squared, instead of squaring 100 numbers
and then adding them together, you could figure out 1/3
of 100 to the third power plus 1/2 times 100 squared
plus 1/6 times 100. Or you could do
that for any number. So this right over here is a
pretty neat little formula.