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# Equation with two rational expressions (old example)

## Video transcript

solve the equation and find excluded values and what they're talking about what about finding excluded values is we need to think about what values would make these either side of this equation undefined and the reason why we want to do that is because as we manipulate this we might lose things in the denominator and then we might get some answer but if it's one of those things that made the original the original expressions or either side of the original equation undefined then that wouldn't be a legitimate solution so that's what they're talking about the excluded values so what values do we have to exclude here right from the get-go well for my over P minus 1 won't be defined if P is 1 because if P is 1 then this then you're going to be dividing by 0 and that's undefined so we know that P cannot be equal to 1 and over here if p u is at negative 3 then this denominator would be 0 and it would be undefined and so P cannot be equal to 1 or negative 3 so these right here are our excluded values so now let's try to solve let's try to solve this equation and I'm going to rewrite it over here so if 4 over P minus 1 is equal to 5 over P plus 3 so the first thing we can do especially because we can assume now that neither of these expressions are 0 and this is going to be defined since we've excluded these values of P is to get the P minus 1 out of the denominator we can multiply the left-hand side by P minus 1 but remember this is an equation if you want them to continue to be equal anything you do to the left hand side you have to do to the right hand side so I'm multiplying by P minus 1 now I also want to get this P plus 3 out of the denominator here on the right hand side so the best way to do that is multiply the right-hand side by P plus 3 but if I do that to the right hand side I also have to do that to the left hand side P P plus 3 and so what happens we have a P minus 1 in the numerator P minus 1 in the denominator these cancel out so you have just a 1 in the denominator or you have no denominator anymore and the left hand side simplifies to 4 times P plus 3 or if you were to distribute the 4 4 times P plus 3 so that is 4 P plus 12 and then the right hand side you have AP plus three cancelling with the P plus three this is people's three divided by P plus three and all you're left with is five times P minus one if you distribute the five you get five P 5 P minus five and now this is a pretty straightforward linear equation to solve we just want to isolate the P's on one side and the constants on the other so let's subtract 5p from both sides I'll switch colors so let's subtract 5p from both sides and we get on the left-hand side 4p minus 5p is negative P plus 12 is equal to these cancel out is equal to negative 5 and then we can subtract 12 from both sides subtract 12 from both sides and we get these cancel out we get negative P is equal to negative 5 minus 12 is negative 17 and we're almost done we can multiply both sides by negative 1 or divide both sides by negative 1 depending on how you want to view it and we get negative 1 times negative P is so let me just scroll down a little bit so I have a little bit of real estate that's positive P is equal to 17 P is equal to 17 and let's verify that this really works well it wasn't one of our excluded values but just just in case let's verify that it really works if we go if we have P is 17 we get 4 over 17 minus 1 needs to be equal to 5 over 17 plus 3 I'm just putting 17 in for P because that's our solution so this is the same thing as 4 over 16 needs to be the same thing as 5 over 20 or 4/16 is the same thing as 1/4 and that needs to be the same thing as 5 twentieths which is the same thing as 1/4 so it all checks out so these are excluded values and lucky for us this wasn't one of them