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Current time:0:00Total duration:6:59

CCSS.Math:

so I have a rational expression here and what my goal is is to simplify it but while I simplify it I want to make the simplified expression be algebraically equivalent so if there are certain x-values that would make this thing undefined that I have to restrict my simplified expression by those X values so you could pause this video and take a go at it and then we'll do it together all right so let's just think real quick what would what X values would make this expression undefined well it's not defined if we try to divide by zero so if x is 0 14 times 0 is 0 it's going to be undefined and so we could say we could say this is we could say X does not cannot be equal to 0 for any other X we can evaluate this expression now let's actually try to simplify it so when you look at the numerator in the denominator every term we see is divisible by X and every term is divisible by 7 so it looks like we can factor 7 X out of the numerator and a 7 X out of the denominator so the numerator we can rewrite as 7 x times if you factor 7 x out of 14 x squared you're going to be left with the 2x and then you factor a 7 X out of 7 X you're going to be left with a 1 and one way to think about it is we did the distribute distributive property in Reverse if you do it again 7x times 2x is 14x squared 7x times 1 is 7x all right and now let's factor 7x out of the denominator so 14x could be re-written as 7 x times 2 times 2 and remember I want to keep this algebraically equivalent so I want to keep the constraint that X cannot be equal to 0 and so we divide the numerator and the denominator by 7x or you one way to think well you could've I'd 7x by 7x and just get 1 and we are left with 2x plus 1 over 2 now this was the original expression X could take on any value but what if we want it to be algebraically equivalent to our original expression it has to have the same constraints on it so we're going to have to say X X does not equal 0 and this is a real it's a very subtle but really important thing for example if you find a function by this right over here the domain of the function could not include zero and so if you simplified how you how you define that function to this if you want that function to be the same it needs to have the same domain has to be defined for the same inputs and so that's why we're putting the exact same constraints for them to be equivalent if you got rid of this constraint these these two would be equivalent everywhere except for x equals zero this one would have been defined for x equals zero this one would not have and so they wouldn't been algebraically equivalent this makes them algebraically equivalent and of course you could write this in different ways you could divide each of these terms by two if you like so you could divide 2x by two and get X and then divide one by 2 and get 1/2 but once again we would want to keep the X cannot be equal to zero let's do another one of these so it's a slightly hairier expression but let's do the same drill and see if we could simplify it but as we simplify we really want to be conscientious of of restricting the Z's here so that we get an algebraically equivalent expression so let's think about where this is undefined we could think about where it's undefined by factoring the denominator here and actually let me just so this is going to be equal to actually let me just let me just do the first step where I could see well what's a common factor in the numerator in the denominator every term here is divided by Z squared and every term is also divided by 17 so it looks like 17 Z squared can be factored out so 17 Z squared can be factored out of the numerator and then we would be left with you factor out a 17 Z squared out of 17 Z to the third and you're going to be left with just a Z 17 Z squared you factor 17 Z squared out of that you're going to be left with just a 1 and once again you can you can distribute the 17 Z squared multiply it times Z you get 17 C to the third 17 Z squared times 1 is 17 Z squared alright all of that is going to be over we want to factor out a 17 Z squared out of the denominator 17 Z squared times and so 34 is Z to the third divided by 17 Z squared 34 divided by 17 is 2 and as Z to the 3rd divided by Z squared is Z and then we have minus 51 divided by 17 is 3 and Z squared divided by Z squared is just 1 so we'll just leave it like that and so over here it becomes a little bit clearer of well what how we're going to simplify it we're just going to divide 17 Z squared by 17 Z squared but let's be careful to restrict the domain here so we can tell if Z is equal to 0 then this then 17 Z squared is going to be equal to 0 to make the denominator equal to 0 and we could see that even looking here so we could say Z cannot be equal to 0 and what else Z cannot be equal to whatever makes this 2 Z minus 3 equals 0 so let's think about what makes 2 Z minus 3 equal to 0 2 Z minus 3 is equal to 0 you can add 3 to both sides and you would get 2 Z is equal to 3 divide both sides by 2 and you would get Z is equal to 3 halves so Z cannot be equal to 0 and Z cannot be equal to 3 halves so that's how we're going to restrict our domain but now let's simplify so if we simplify it these two cancel out and we are going to be left with we're going to be left with Z plus 1 over 2 Z minus 3 and we want to keep that constraint Z Z cannot be equal to 0 and actually this second constraint is redundant because we still have the 2 Z minus 3 here someone were to just look at this expression like well the denominator can't be equal to 0 and so Z cannot be equal to three-halves so this is still if we just left it the way it is if we don't have to rewrite this that would be redundant since looking at this expression it's clearly not defined for Z equals 3 halves so there you go if someone asks you for what values is this expression not defined well then you could you would also include that actually let me just write it doesn't hurt to be redundant Z does not equal three halves but this this constraint right here is really important because it's not obvious by looking at this expression this expression by itself would be defined for Z equals zero but if we wanted to be algebraically equivalent to this one in that one it has to be constrained in the same in the same way