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.(x−2)/2x + 1 = (x+1)/x

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 in the numerator. On the left hand side, we have the difference of X minus 10 that's divided by three X and then we are adding two and then that's equal to, on the right hand side in the numerator, we have the sum of X plus 18. And that's divided by X. Our answer choices are answer choice A X equals two, answer choice B X equals four, answer choice C X equals 16 and answer choice D no solution. I'll write when we are adding fractions and we have this equal to fractions. And we're solving for X. What we want to do first is get rid of the fractions. And in order to do that, we need to multiply all of our fractions by the lowest common multiple. We look at all of our denominators to figure out what the lowest common multiple of those denominators is. In this second term here, this two that's being added, that's a fraction as well. Its denominator is just one. So our denominators are three X one and X, the lowest common multiple that they all multiply into, well, for the three and the one that's going to be a three. And for the X and the X, they're both with a degree of one. So that will just be X, we're going to multiply all three of these fractions by three X. And next, we will take a look at how that three X being multiplied is reduced by the denominator for each fraction. So for the first fraction, the three X in the numerator is divided by a three X in the denominator. And that completely cancels out. So that first fraction, all that's left is X minus 10. The numerator for the next one, we have a three X being multiplied in the numerator. And that's only being divided by one itself. So nothing cancels out there. We can bring down this plus and then it's a three X being multiplied by two. Then we can bring down the equal sign. Now taking a look at the fraction on the right hand side, we have the three X being divided by X and those XS cancel out. And all that's left is that three being multiplied by the sum of X plus 18 from the numerator. Now, we can multiply where applicable. So we can bring down this X minus 10 and then we have a positive three X being multiplied by two. That would be a positive six X and that's equal to, we can distribute this three on the right hand side to the X and the 18 that are inside of our parentheses and three multiplied by X is a positive three X and three multiplied by 18 is a positive 54. Now, we just want to get all of the XS together somewhere and all of the regular numbers together on the other side. So let's see how we can bring the XS together. On the left. We have an X plus A six X. So we can combine those. And then on the right, we have a three X, we can subs subtract three X from both sides of the equation so that we'll have all of the axes on the left hand side of the equation. Now we want to bring all of the other numbers to the opposite side. We have that negative 10 on the left and we can add 10 to both sides of the equation. And now we can simplify it a bit on the left, we have X plus six X which is seven X then minus three X which will bring us back down to a positive four X, the negative 10 plus 10, that's zero. Now, on the right hand side of the equal sign that three X minus three X, that's zero and 54 plus 10, that's a positive 64. Now we just have four being multiplied by X X is almost by itself. In order to undo that multiplication, we need to divide, we will divide both sides by four because on the left four, divided by four is one which leaves us with just a positive X which is what we wanted. X is by itself ya. And then on the right hand side, 64 divided by four is a positive 16. And we have solved for X X is equal to 16. We look at our answer choices and that matches with answer choice. See, 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.(x−2)/2x + 1 = (x+1)/x

Key Concepts

Rational Equations

Finding Common Denominators

Extraneous Solutions

Watch next