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Current time:0:00Total duration:7:12

Equations with rational expressions (example 2)

CCSS.Math:

Video transcript

so we have a nice little equation here that has some rational expressions in it and and like always pause the video and see if you can figure out which X's satisfy this equation alright let's work through it together now when I see things in the denominator like this my instinct is to try to not have denominators like this and so what we could do is to get rid of this X minus 1 and the denominator on the left hand side we can multiply both sides of the equation times X minus 1 X minus 1 so we're gonna multiply both sides by X minus 1 and once again the whole point of doing that is so that we get rid of this X minus 1 in the denominator right over here and then to get rid of this X plus 1 in the denominator over here we can multiply both sides of the equation times X plus 1 so X plus 1 multiply both sides times X plus 1 and so what is that going to give us well on the left hand side that is going to X minus 1 divided by X minus 1 it's just going to be 1 for the X's where for the X's where that's defined for X not being equal to 1 and so we're going to have X plus 1 times negative 2x plus 4 so let me write that down so we have X plus I think I'm going to need some space so let me make sure I don't write too big X plus 1 times negative 2x negative 2x plus 4 is going to be equal to now if we multiply both of these times 3 over X plus 1 the X plus 1 is going to cancel with the X plus 1 and we're going to be left with 3 times X minus 1 so that is going to be 3 X minus 3 3 X minus 3 and then minus -1 times both of these so 1 times X minus 1 times X plus 1 so minus 1 times X minus 1 times X plus 1 all I did is I multiplied it took the X minus 1 times X plus 1 multiply times each of these terms when I multiply it times this first term the X plus 1 and the plus one cancelled so I just have to multiply 3 times X minus 1 and then for the second term I just multiplied it times both of these and now you might recognize this if you have something X plus 1 times X minus 1 that's going to be x squared minus 1 so I could rewrite all of this right over here as being equal to as being equal to x squared minus 1 and once again that's because this is the same thing as x squared minus 1 and since I'm subtracting an x squared minus 1 actually let me just I don't want to do too much on one step so let's go to the next step so I could multiply this out so I can multiply x times negative 2x which would give us negative 2x squared x times 4 which is going to give us plus 4x and I can multiply 1 times negative 2x so I'm going to subtract 2x and then 1 times 4 which is going to be plus 4 and then that is going to be equal to that is going to be equal to we have 3x minus 3 and then we can distribute this negative sign so we could say minus x squared plus 1 and over here we can simplify it a little bit this is going to be that is 4x minus 2x is going to be let me make sure I didn't write over yep 4x minus 2x so that would be 2x and so this is simplified to let's see well this is we have a negative 3 and a 1 so those two together are going to be equal to subtracting it to so we can rewrite everything as I'll do it in a neutral color now negative 2x squared plus 2x plus 4 is equal to negative x squared plus 3x plus 3x minus 2 now we can try to get all of this business on to the right hand side so let's subtract it from both sides so we're subtract or we will add x squared to both sides add x squared that gets rid of this white X this white negative x squared we subtract 3x from both sides subtract 3x from both sides add 2 to both sides add 2 and we will be left with we are going to end up with C negative 2x squared plus x squared is negative x squared 2x minus 3x is negative x and then 4 plus 2 is 6 is going to be equal to well that's going to cancel with that that that is equal to 0 I don't like having this negative on the x squared so let's multiply both sides times negative 1 and so if I do that so if I just take the negative of both sides so I just multiply that times negative 1 same thing is taking the negative both sides I'm going to get knit I'm going to get positive x squared plus X minus 6 is equal to 0 and we're making some good progress here so we can factor this and actually let me just do it right over here so that we can see the original problem so if I were to factor this what two numbers their product is negative 6 they're gonna have different signs since their product is negative and they add up to 1 the coefficient on the first degree term well positive 3 and negative 2 work so I can rewrite this as X plus 3 times X minus 2 is equal to 0 did I do that right yeah 3 times negative 2 is negative 6 3 X minus 2 X is positive x all right so I just factored I just wrote this in this quadratic in factored form and so the way that you get this equaling 0 is if either one of those equals 0 X plus 3 equals 0 or X minus 2 is equal to 0 well this is going to happen if you subtract 3 from both sides you get that's going to happen if X is equal to negative 3 or over here if you add 2 to both sides X is equal to 2 so either one of these will satisfy but we want to be careful we want to make sure that our original equation isn't going to be undefined for either one of these and negative 3 does not make either of the denominators equal to 0 so that's cool and positive 2 does not make either of the denominators equal to 0 so it looks like we're in good shape there's two solutions to to that equation if one of them made the knot either any of the denominator is equal to 0 then they wouldn't have then they would have been extraneous solutions it would've been solutions for some of our intermediate steps but not for the actual original equations with the expressions as they were written but this we can feel good about because neither of these make any of these denominators equal to 0