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# Equations with rational expressions

CCSS.Math:

## Video transcript

so we have a nice little equation here you're dealing with rational expressions I encourage you to pause the video and see if you can figure out what values of X satisfy this equation all right let's work through this together so the first thing I'd like to do is just see if I can simplify this at all and maybe by finding some common factors between numerators and denominators or common factors on either side of the equal sign so let's factor all of these all of the numerators in the denominators all the ones on the right hand side are already done so this thing up here I could rewrite this as let's see what what's product is 21 what two numbers when I take their product is 21 positive 21 so they're going to have the same sign and when I add them I get negative 10 well negative 7 and negative 3 so this could be rewritten is X minus 7 times X minus 3 this over here both are divisible by 3 I could rewrite this as 3 times X minus 4 and these are already factored so the one thing that jumps out at me is I have X minus 4 in the denominator on the left hand side and on the right hand side and so if I were to multiply both sides by X minus 4 so actually let me just let me formally replace this with that and up here it's not so obvious that's going to be valuable for me to keep this factored form so I'm just going to keep it in this yellow form and the in the expanded out form so let me just scratch that out for now because once I well let me multiply by X minus 4 so if we multiply both sides by X minus 4 and once again why am i doing this so I get rid of the X minus 4 is in the denominator X minus 4 and then X minus 4 that and that cancels that and that cancels and then we're left with in the numerator we're left with our x squared minus 10x plus 21 and let's see divided by 3 divided by 3 is equal to X minus 5 let's see now what we could do an extra I could have done it in the last step is I can multiply both sides by 3 multiply both sides do that another color just sticks out a little bit more so I can multiply both sides by three so multiply both sides by three on the left hand side that and that cancels and I'll just be left with x squared minus 10x plus 21 and on and I don't have a denominator anymore my denominator is one so I don't need to write it it's going to be equal to three times let's just distribute the three 3 times X is 3x three times negative five is negative 15 and now I can get this into standard quadratic form by getting all of these terms onto the left hand side best way to do that let's subtract 3x from the right but I can't just do it from the right otherwise equality won't hold I have to do it from both sides if I want the Equality to hold and I want to get rid of this negative 15 so I can add 15 to both sides so let's do that and what we are left with scroll down a little bit so a little more space what we are going to be left with is x squared minus 13x and then plus what is this plus 36 plus 36 to do that right yet plus 36 is equal to is equal to 0 all right now let's see we have this a quadratic in a standard form how can we solve this so first thing can we factor this product of two numbers 36 if I add them I get negative 13 they're both going to be negative since they have to have the same sign to get their product to be positive and let's see 9 and 4 seem to do the trick so our negative 9 and negative 4 so X minus 4 times X minus 9 is equal to 0 well that's going to happen if either if either X minus 4 is equal to 0 or X minus 9 is equal to 0 well add 4 to both sides of this this happens when X is equal to 4 add 9 to both sides of this this happens when X is equal to 9 so we could say that this will Lucian's are x equals four or x equals nine so X is equal to four or x equals nine but we need to be careful because we have to remember in our original expression X minus four was a factor of both denominators and so if we actually try to test X minus four in the original equation not one of these intermediary steps in the original equation I would end up dividing by zero right over here and actually I did to end up dividing by zero right over there as well so the original equations if I try to substitute for they they don't make sense so this is actually an extraneous solution it's not going to be a solution to the original equation the only solution is X is equal to nine