In Exercises 1–34, solve each rational equation. If an equation h...

Question

In Exercises 1–34, solve each rational equation. If an equation has no solution, so state.4/(x²+3x−10) + 1/(x²+9x+20) = 2/(x²+2x−8)

Accepted Answer

Hello, today we're gonna be finding the solution to the given rational equation. So what we are given is six over the quantity X squared plus two, X minus three plus two over the quantity X squared plus nine X plus 18, equal to five over the quantity X squared plus five, X minus six. Now, there's a bit of terms to keep track of in this given equation. However, whenever we are work working with rational equations, we want to go ahead and make sure that our denominators are in its simplest form. Currently, each of our denominators are in the forms of factor trinomial. So let's go ahead and factor each of the given denominators. In the first denominator, we have the quantity X squared plus two, X minus three. And through the trial and error method, this is going to factor to give us X plus three times the quantity X minus one. For the second denominator, we have the quantity X squared plus nine X min uh plus 18. And through the trial and error method, this is going to factor to give us X plus three times X plus six. And for the final denominator we have the quantity X squared plus five, X minus six and this is going to factor to X plus six times X minus one. So now that we have the factored forms of the given denominators, let's go ahead and res substitute these back into the original equation that is going to give us six over the product of X plus three times X minus one plus two over the product of X plus three times X plus six, equal to five over the product of X plus six times X minus one. Now, what we want to go ahead and do is we wanna try to simplify this by eliminating the denominators. Well, in this case here, we can go ahead and use a lowest common multiple from the denominators in order to eliminate them, the lowest common multiple is going to be a product of the different combinations of quantities in our denominators. So we have combinations of X plus three, X minus one and X plus six. And that means that the lowest common multiple is going to be X plus three times X minus one times X plus six. So now what we want to go ahead and do is we want to multiply each of the quantities in our equation by the lowest common multiple. That is going to give us six over X plus three times X minus one multiplied by X plus three times X minus one times X plus six. And we're gonna represent this as a rational quantity by putting it over one. Then we are adding two over X plus three times X plus six, multiplied by X plus three times X minus one times X plus six. And this is going to be equal to five over X plus six times X minus one multiplied by the levels common multiple, which is X plus three times X minus one times X plus six. Now, the reason why we're multiplying each of the terms by the lowest common multiple is because we can go ahead and eliminate the denominators by using cross multiplication. So in the first term, we have quantities of X plus three and X minus one in both the numerator and denominator. So we can just go ahead and divide those out and we can do the same thing in the second term and eliminate X plus three and X plus six and repeat this process for the third term which allows us to eliminate X plus six and X minus one. That is going to leave us with six times the quantity X plus six plus two times the quantity X minus one. And that is going to be equal to five times the quantity X plus three. Now, we wanna go ahead and solve for X, but in order to do that, we need to first eliminate the parentheses. In order to eliminate the parentheses, we're going to distribute six into X plus six. We're going to distribute two into X minus one and we're going to distribute five into X plus three. That is going to give us six X plus plus two, X minus two is equal to five X plus 15. On the left hand side, we're going to go ahead and combine like terms, we can combine six X and two X to give us eight X and we can combine 36 with negative two to give us positive 34. And that is going to be equal to five X plus 15. Now, what we wanna do is we want to move all the constants to one side of the equation and all the X variables to the other side. So what we can do is we can subtract five X to both sides of the equation. And that is going to give us the left hand side of three X. And on the, on the right hand side, we can subtract 34 to both sides of the equation. And that is going to give us the quantity of negative 19. Now, in order to solve for X, we're just gonna divide both sides by three and that is going to leave us with X is equal to negative 19/3, which is going to be the solution to the rational equation. And with that being said, the answer to this problem is going to be B so I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

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In Exercises 1–34, solve each rational equation. If an equation has no solution, so state.4/(x²+3x−10) + 1/(x²+9x+20) = 2/(x²+2x−8)

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