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.1/x−1 + 1/x+1 = 2/x²−1

Accepted Answer

Hello, today we're going to be finding a solution to the following rational equation. So what we are given is four over X minus four plus four, over X plus four is equal to over X squared minus 16. Now, in order to approach this problem, we need to first check and make sure that all of our denominators are in its simplest form. In the first term, we have X minus four. And in the second term, we have X plus four, which are both in its simplest forms. However, we can further simplify X squared minus 16. The term X squared minus 16 can be simplified using the difference of squares. And so if we use the difference of squares equation on X squared minus 16, this is going to factor to give us X plus four times X minus four. And so if we res substitute this value into the equation, we get four over X minus four plus four, over X plus four is equal to over X plus four times X minus four. Now, what we want to go ahead and do is we want to try to eliminate the denominators in order to do that, we can find a lowest common multiple between the three denominators. Now, the lowest common multiple is going to be a combination of the products of all the different terms and denominators. So for example, we have X minus four and X plus four and we have a repeat of those two values in the third denominator. So that means that the lowest common multiple in this case is going to be X plus four times X minus four or in other words, this is the product of all the different terms in all of the denominators. So, what we're going to do now is we're going to multiply each term by the lowest common multiple. That is going to give us four over X minus four times, the lowest common multiple, which is X plus four times X minus 4/ plus four, over X plus four times X plus four times X minus four. And that is going to equal to 32 over X plus four times X minus four, multiplied by X plus four times X minus four over one. Now, if we use the method of cross reduction, we can go ahead and eliminate the denominators from the three terms. For example, in the first term, we have the quantity X minus four in the denominator. And we also have it in the numerator. Therefore, we can go ahead and eliminate these quantities from both the numerator and denominator then we're gonna repeat this process for the other two terms. In the second term, we have X plus four in the denominator and in the numerator. So we can just go ahead and cross eliminate those. And in the last term, the product of X plus four times X minus four are in both the numerator and denominator. So we can just go ahead and eliminate those. And that is going to leave us with four times X plus four plus four times X minus four is equal to 32. Now, one thing to note is that we have multiples of four for each of the coefficients and constants in the equation. So we can further simplify this by dividing everything by four. And if we divide everything by four, that is going to leave us with X plus four plus X minus four and that is going to equal to eight. Now, what we're going to do is we're going to combine like terms on the left hand side of the equation, we can combine both of the X to give us two X and four minus four is just going to cancel each other out. So what we are left with is two, X is equal to eight. And finally, in order to solve for X, we're going to divide both sides of the equation by two. And that is going to give us X is equal to four. And now we need to check to make sure that this value actually works. And there is going to be an issue see in the first term of the equation recall that the first term is four over X minus four. If we plug in the value of X, which in this case is four into the first term, what we are left with is 4/4 minus four and four minus four is going to give us zero. So what we end up getting is 4/0, which is going to produce an undefined value. So since we got an undefined value from the only solution of X, that means that there is going to be no solution to this rational equation. And with that being said, the answer to this problem is going to be D. 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.

In Exercises 1–34, solve each rational equation. If an equation has no solution, so state.1/x−1 + 1/x+1 = 2/x²−1

Key Concepts

Rational Equations

Finding a Common Denominator

Identifying Extraneous Solutions

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