Solve each equation. See Example 1. 3/(x^2+x-2) - 1/(x^2-1) = 7/(...

Question

Solve each equation. See Example 1. 3/(x^2+x-2) - 1/(x^2-1) = 7/(2x^2+6x+4)

Accepted Answer

Hey everyone in this problem for the following equation were asked to solve for x. And the equation we're given is two over X squared plus x minus six minus one over x squared minus four is equal to 3/2 X squared plus 10 X plus 12. So let's just rewrite what we're given. A two over X squared plus x minus six minus one over X squared minus four is equal to three over two, X squared plus 10 X plus 12. Now when we have an equation like this and we're asked to solve for x, what we wanna do is we want to try to combine all three of these terms were given into one and we want to have one rational equation and that's going to make it easier to solve rags. So in order to do that we need to find a common denominator. So let's just factor these denominators we were given and see what factors we have. Okay, and then we can see what's missing in each of these terms. So in the first one we have two over. Okay, X plus three times x minus two K. You can factor whichever way you like to factor the best, we're gonna have minus one over. We have x squared minus four. Okay, this is a difference of squares. So this is going to give us x plus two times x minus two and then we have 3/2 X squared plus 10 X plus 12. And you'll notice that there's a common factor of two in this denominator. Okay, so we can factor out the two. So we're gonna have 3/2 and then we have X squared plus five. X plus six. And if we factor that we get X plus three Times X-plus two. Yeah. Alright. So what we can do instead of writing 3/2 like this in the denominator I'm going to move this to and just rewrite it in the numerator. So we have three have divided by these two factors. Okay then we don't have to worry about that too. In our common denominator it's only part of this numerator. Alright, so we see we have factors of X plus three X minus two and X plus two. Okay, Amongst these three terms. So that's that's going to be our common denominator. X plus three X minus two. X plus two. So in the first term we have to, the denominator is X plus three over x minus two right now. So the factor it's missing is X plus two. So we want to go ahead and multiply the denominator By X-us two. If we multiply the denominator by something, we have to do the same to the numerator. Okay, so that we're multiplying by X plus two over X plus two which is equal to one. Multiplying by one means our equation stays the same for the next term. We're going to do the same. Our denominator is X plus two times x minus two right now. So the factor we're missing is X plus three. So we're going to multiply numerator and denominator by X plus three and we get X plus three divided by X plus two times X minus two times X plus three. Okay? And then on the right hand side we have three halves. Our denominator is X plus three times X plus two. So what we're missing is the x minus two term. So we're going to multiply the numerator and denominator by X minus two. And we get three halves times X minus two divided by X plus three times X plus two times x minus two. So now we have this common denominator. All three terms have the same denominator which means that we can combine them. Okay? It's just like working with common denominators when you're dealing with fractions that just have numbers. Okay? So let's move everything over to the left hand side. Okay and work with this common denominator, so we have two X plus four. Okay? We have two times X plus two that gives us two X plus four. Then we have minus X plus three. So that's gonna be minus x minus three. And then we're gonna subtract this term from the right hand side so we're gonna get -3/2. Okay and then we're gonna have minus negative three halfs times two. Okay so that's gonna be plus three. And all of this is divided by that common denominator X plus three X minus two X plus two and it's all equal to zero because we've moved everything from the right to the left. Alright so let's simplify this numerator and see what that gets us. So for our X terms we have two X minus X which gives us one X. And then we have minus three half X. So we're going to have negative a half X. And then in terms of the constants we have four minus three plus three. So this is going to be plus four. So minus a half X plus four. In our denominator stays the same X plus three X minus two X plus two equal to zero. Okay now the only way we can have this equal to zero is that the numerator is equal to zero? If the numerator is equal to zero we have zero divided by something that's gonna give us zero. So we want to find the X values when minus one half X plus four is equal to zero. Okay but we have to be careful when we do this. Okay we have some restrictions on this denominator, we know we can't divide by zero. So values of X. Where this denominator is equal to zero cannot be solutions. So working from the left we have X plus three. If this is equal to zero we have X is equal to negative three. So if X is negative three we have no solution. If X is equal to two, same thing or if X is equal to negative two. Okay, we cannot have these X values. These are restrictions that come from our denominator. Alright, so now solving negative one half X plus four is equal to zero. We get that one half X is equal to four, which tells us that X is equal to eight is a solution. And if we look X equals eight is not one of our restrictions, that means that it is a valid solution. So we have X equals eight. And if we go back to our answer choices, we see that that is going to correspond with answer choice. B. Thanks everyone for watching. I hope this video helped see you in the next one.

Solve each equation. See Example 1. 3/(x^2+x-2) - 1/(x^2-1) = 7/(2x^2+6x+4)

Key Concepts

Factoring Polynomials

Common Denominator

Rational Equations

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