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.3/(x+1) = 5/(x−1)

Accepted Answer

Welcome back. I am so glad you're here. We're asked to find the solution to the following rational equation. Otherwise choose no solution. Our rational equation is on the left side of the equal sign in the numerator, we have one and that's divided by, in the denominator, the sum of X plus six. And that's equal to, on the right hand side of the equation in the numerator, we have four and that's divided by, in the denominator, the difference of X minus six. And our answer choices are answer choice. A no solution, answer choice B X equals negative six, answer choice C X equals negative eight and answer choice D X equals negative 10. All right, we have two fractions equal to each other. And we're trying to solve for X. In order to get the XS out of the fractions, we want to get rid of these fractions. When two fractions are equal to each other, we can cross multiply. So we'll multiply the numerator from the left hand side of the equation by the denominator from the right hand side. And likewise, we'll multiply the numerator from the right hand side of the equation by the denominator from the left hand side of the equation. So now we can rewrite this as one multiplied by the difference of X minus six and that is equal to four, multiplied by the sum of X plus six. And now we can distribute in order to get rid of those parentheses on the left hand side, multi multiplying one by the two terms inside of the parentheses will just give us the two terms inside of the parentheses. So we'll just have X minus six on the left hand side. But on the right hand side, distributing the four to the X and to the six will give us four, multiplied by X is four, X and four multiplied by six is a positive 24. Now, we want to get all of the X S together on one side of the equation and all of the constant terms together on the other side, when possible, I like to make the X terms positive. And in this case, the larger X term is on the right hand side. So I'm going to subtract the X term from the left hand side, there's just one X so minus X and what we do to one side, we do to the other because on the left X minus X is zero, and then I want to bring the constant terms over to the left hand side. So on the right, we have a positive 24. So I'm going to subtract 24. And what we do from one side, we need to do to the other as well on the right, positive 24 minus 24 is zero. So now let's see what we've got left. On the left hand side, a negative six minus 24 is negative 30. That's equal to on the right four X minus X equals three X. Now X is being multiplied by three. In order to do undo that multiplication, we need to divide. So we will divide by three. And what we do to one side, we do to the other on the right three, divided by three is one. So we have one X which is just X, which is exactly what we want. Yay on the left negative 30 divided by three is negative 10. And we have solved for X X is equal to negative 10. We look at our answer choices and this matches with answer choice. D well done. We'll catch you on the next one.

In Exercises 1–34, solve each rational equation. If an equation has no solution, so state.3/(x+1) = 5/(x−1)

Key Concepts

Rational Equations

Cross Multiplication

Checking for Extraneous Solutions

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