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.8/x²−9 + 4/x+3 = 2/x−3

Accepted Answer

Hello, today we're gonna be solving the following rational equation. So what we are given is nine over X squared minus 25 plus six over X plus five is equal to three over X minus five. Now, before we start solving this rational equation, we always want to check and make sure that all of our denominators are in its simplest form. See X plus five and X minus five are both in its simplest form. However, we can further simplify X squared minus 25. The quantity X squared minus 25 can be simplified by factoring using the, using the difference of squares. So if we factor X squared minus 25 using the difference of squares, this is going to factor to give us X plus five times X minus five. So if we go ahead and resue this value into the given equation, that is going to give us nine over X plus five times X minus five plus six over X plus five is equal to three over X minus five. Now, in order to solve this equation, we want to work with it in an easier form. So what we can go ahead and do is figure out a lowest common multiple between the three denominators. Now, in this case here, the lowest common multiple is going to be the product of the different groups in the denominators. So for example, we have X plus five, we have X minus five and we have both X plus five times X minus five in the denominators. The different values that we see in the denominators are going to be X plus five and X minus five, which means that the lowest common multiple is going to be X plus five times X minus five. So now what we're going to do is we're going to multiply each term by the lowest common multiple in order to eliminate the denominate needs. So what we're going to get is nine over X plus five times X minus five multiplied by the lowest common multiple, which is going to be X plus five times X minus 5/ plus six over X plus five, multiplied by X plus five times X minus 5/1. And that is going to equal to three over X minus five, multiplied by X plus five times X minus 5/1. Now, we can go ahead and cross reduce each of the following denominators. For example, in the first term, we have the product of X plus five and X minus five in both the numerator and denominator. And using cross reduction, we can just go ahead and eliminate those from both the numerator and denominator, then we're going to repeat this process for the other two terms. So in the second term, we have X plus five in the denominator and we have X plus five in the numerator. Therefore, we can eliminate these from both the numerator and denominator. And the same thing applies with the last term we have X minus five in both the numerator and denominator. So we can just go ahead and eliminate those. That is going to leave us with the equation nine plus six times the quantity X minus five is equal to three times the quantity X plus five. Now, what we wanna go ahead and do is simplify the equation by getting rid of the parentheses. In order to do that, we're going to distribute six into X minus five and we're going to distribute three into X plus five that is going to leave us with nine plus six, X minus 30 is equal to three X plus 15. On the left hand side, we're going to go ahead and combine like terms, we can go ahead and combine nine and negative 30 and that is going to give us negative 21. So we have six X minus 21 is equal to three X plus 15. Now, what we're going to do is we're gonna go ahead and solve for X. So what we can do is we can go ahead and subtract three X to both sides of the equation. And we can add 21 to both sides of the equation that is going to leave us with three X equal to positive 36. And in order to solve for X, if we divide both sides by three, that is going to give us X is equal to 12, which is going to be the solution to this given equation. And with that being said, the answer to this problem is going to be a. 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.8/x²−9 + 4/x+3 = 2/x−3

Key Concepts

Rational Equations

Finding Common Denominators

Identifying Extraneous Solutions

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