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.2/(x+3) − 5/(x+1) = (3x+5)/(x²+4x+3)

Accepted Answer

Hello, today we're gonna be finding the solution to the given rational equation. So what we are given is five over X plus six minus 15, over X plus two is equal to four, X plus four over X squared plus eight, X plus 12. Now, before we continue with solving the rational equation, we must always check to make sure that our denominators are in its simplest form, X plus six is in its simplest form and X plus two is in its simplest form, but X squared plus eight, X plus 12 is in the form of a factor trinomial. So we're going to need a factor X squared plus eight X plus 12. Now, through the trial and error method, this is going to factor to give us the quantities of X plus six times X plus two. And if we res substitute this into the equation, we get five over X plus six minus 15 over X plus two is equal to four, X plus four over X plus six times X plus two. Now, what we wanna do is we want to figure out a lowest common multiple to multiply each term by, in order to eliminate the denominators. While the lowest common multiple is going to be the product of the different combinations of terms from the denominators. In the denominator, we have X plus six, X plus two and X plus six times X plus two. So the lowest common multiple in this case is going to be X plus six times X plus two. And so what we're going to do is we're going to multiply each term in the equation by this multiple. So that is going to give us five over X plus six, multiplied by X plus six times X plus 2/1. Then we're gonna subtract 15 over X plus two times X plus six times X plus 2/1. And that is going to equal to four, X plus four over X plus six times X plus two, multiplied by X plus six times X plus 2/1. Now, using the method of cross elimination, we can go ahead and eliminate the denominators from each of the terms. In the first term, we have X plus six in the denominator and in the numerator. So we can go ahead and eliminate those quantities and repeat this process for the other two terms. In the second term, we have X plus two in both the numerator and denominator. So we can go ahead and eliminate those terms. And in the final term, we have the product of X plus six times X plus two in both the numerator and denominator. So we can just go ahead and eliminate these quantities that is going to leave us with five times the quantity of X plus two minus 15 times the quantity X plus six. And that is going to equal to four X plus four. Now, what we want to go ahead and do is eliminate the parentheses. And in order to do that, we're going to distribute five into the quantity X plus two. And we're gonna distribute negative 15 into X plus six. That is going to give us five X plus minus 15, X minus 90 and that is going to equal to four X plus four. Now, what we wanna do is we want to go ahead and combine the like terms on the left hand side, we can combine negative 15 X and five X to give us negative 10 X and we can combine the constants of negative 90 and positive 10 to give us negative 80. And that is going to equal to four X plus four. Now, in order to solve for X, we're going to move all the X terms to the left hand side of the equation and all the constants to the right hand side. In order to do that, we're going to subtract four X to both sides of the equation. And that is going to give us the left hand side of negative X. And if we add 80 to both sides of the equation that is going to give us 84 on the right hand side. Now, all we need to do is we need to divide X or both sides of the equation by negative 14. And that is going to give us X is equal to negative six. Now, yes, this is a solution to X. However, there is an issue with the solution. Remember that first term is five over X plus six. If we were to plug in the value of X, which is negative six. In this case, we get five over negative six plus six and negative six plus six is going to give us zero. So we end up with five divided by zero. And any value that is divided by zero is going to produce an undefined value. So that means that the whole equation is going to end up as an undefined value when X is equal to six. And that means that there is going to be no solution to this given 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.2/(x+3) − 5/(x+1) = (3x+5)/(x²+4x+3)

Key Concepts

Rational Equations

Finding Common Denominators

Factoring Polynomials

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