If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Video transcript

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.