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.6/x − x/3 = 1

Accepted Answer

Hello, today we're going to be solving the following rational equation. So what we are given is 10 over X minus X over eight and that is going equal to two. Now, for simplicity, we're going to represent two as a rational value by putting it over one. Now, all of the terms in the current equation are in its simplest form. But the easiest way to work with this equation is to go ahead and eliminate the denominators. So what we can go ahead and do is find the lowest common multiple between the three denominators. Now, in the denominators, we have X eight and one. Now the lowest common multiple is going to be the product of the different terms of each of the denominators. So in this case here, we're going to have the lowest common multiple of eight X. Technically, this will be one times eight X but anything times one is going to be itself. So the lowest common to multiple will just be eight X. Now, what we're going to do is we're going to multiply each term in the equation by the lowest common multiple that is going to give us 10 over X multiplied by eight X over one minus X over eight, multiplied by eight X over one. And that is going to equal to 2/1, multiplied by eight X over one. Now, using the method of cross reduction, we can go ahead and eliminate the denominators. For example, in the first term, we have X in the denominator and we have X in the numerator. So we can just go ahead and cancel the X from both the numerator and denominator. In the second term, we have eight in both the numerator and denominator. So we can just go ahead and cancel those out. And in the last term, there's nothing to, to reduce. So we're just gonna multiply both the numerators and denominators together that is going to leave us with 10 times eight minus X times X and that is going to equal to two times eight X. So multiplying the terms together 10 times eight is going to give us 80 and X times X is going to give us X squared and two times eight X is going to give us 16 X. Now, what we're going to do is we want to set this equation equal to zero. In order to solve for X, we're gonna go ahead and move 16 to the other side by subtracting both sides by 16. And we're gonna reorder the terms from le highest power to least power. So what we are left with is negative X squared minus X plus 80 is equal to zero. And we do want this leading coefficient to be positive. So we can go ahead and divide everything by negative one that is going to give us X squared plus 16, X minus 80 is equal to zero. Now, what we are left with is a trinomial in a factor form. And using the trial and error method, this is going to factor to give us X plus 20 times X minus four. Now, what we can go ahead and do is use the zero property and set both factors of X equal to zero. So we're going to end up with two equations. We're going to have X plus 20 is equal to zero and we're going to have X minus four is equal to zero. On the left hand side, we can go ahead and subtract 20 to both sides of the equation. And that is going to give us X is equal to negative 20 which is the first solution of X. And on the right hand side, we can go ahead and add four to both sides of the equation. And that is going to give us X is equal to positive four, which is going to be the second solution. So that means that we end up with two solutions for X X is equal to negative 20 and X is equal to positive four. 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.6/x − x/3 = 1

Key Concepts

Rational Equations

Finding a Common Denominator

Checking for Extraneous Solutions

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