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

Video transcript

we're now ready to solve for 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 two so that I can cancel out the C terms so it's negative 2a minus 2b minus 2c is equal to negative two and then if I if I perform 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 it's equal to three and on the right hand or 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 27 a plus 9b plus 3c is equal to is equal to 14 and here let's see I can multiply by negative three if I want to get the C's to cancel out again and my whole goal is to have two equations in two unknowns that have leveraged that have leveraged these two constraints in tandem and these two constraints in tandem so now I'm going to do these two so I multiply this equation times negative three I get negative 3a minus 3b minus 3 C is equal to negative three and now I can perform the subtraction and I get 24 a 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 three I should get the B's to cancel out so let's do that so if I multiply this times negative three this constraint or this term 6a times negative three is negative 18 a 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 on the left-hand side we have let's see 24 a minus 18 a is 6a these cancel is equal to 11 minus 9 is to divide both sides by 6 we get a is equal to 2 over 6 which is the same thing as 1 over 3 and so now we can substitute back to solve for B so let's see we have 6 6 times 1/3 RA is 1/3 plus 2 B is equal to 3 6 times 1/3 is 2 2 plus 2 B is equal to 3 subtract 2 from both sides to be 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 there's some space let me do it right over here so I have 1/3 plus 1/2 plus C is equal to 1 now let me find ourselves a common denominator so C 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 over 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 figured out a formula we have figured out a formula for the sum of the first n squares so we can rewrite this this this formula is now going to be a is 1/3 so it's one third and to the third power plus one half plus one half N squared let me make sure you don't look like this look like that n is it one half n squared plus one over six n so so pretty pretty handy this is a pretty this right over here is a pretty handy formula if now you wanted to find zero squared plus 1 squared plus 2 squared plus three squared all the way to 100 squared instead of having to add up a squaring a hundred numbers and then adding together you can figure out one third of a hundred to the third power plus one half times 100 squared plus one sixth times 100 or you could do that for any number so this is kind of a this right over here is a pretty neat little formula