Solve each rational inequality. Give the solution set in interval...

Question

Solve each rational inequality. Give the solution set in interval notation. See Examples 4 and 5. (x - 1)/(x - 4) > 0

Accepted Answer

Hey, everyone in this problem for the given inequality we're asked to solve and to write the solution set in interval notation. Now, the expression we're given is X minus two divided by X minus 13 is greater than zero. And we have four answer choices. Options A through D option A is the interval from negative infinity to infinity. Option B is the union of two different intervals. Option C is the interval from 2 to 13. And option D is that there is no solution. We're gonna start by writing out what we were given. We have X minus two divided by X minus is greater than zero. Mm Now this is already one rational function that's been factored. OK? There's no simplify we can do on the left hand side, we have zero on the right hand side. So this is great. A lot of the work has already been done for us. What we want to do at this point is to identify any points of interest. OK? And the points of interest for an inequality like this are going to be any points where the numerator or the denominator is equal to zero. Now, the numerator is equal to zero. We have X minus two is equal to zero, which tells us that X is equal to two. And if the denominator is equal to zero, we get that X minus 13 is equal to zero, which tells us that X is equal to 13. So we have our two points of interest. X is equal to two and X is equal to 13. OK? We're gonna go ahead and draw a number line. We're gonna add those points of interest to our number line. So we have X is equal to two, we have X is equal to 13 and we're gonna evaluate the intervals on either side of these points. Now, I wanna make one comment before we go forward, you may have been inclined to multiply both sides of this equation or this inequality by X minus 13. You have to be really careful when you do that. The first thing you have to be careful about is the restriction. OK? We have a restriction that X cannot be equal to 30. If it were the denominator would be zero, dividing by zero is gonna give us an undefined value. So that is not going to be a solution is you have to consider the restriction. And you also have to consider the direction of our inequality. And remember that if you multiply an inequality by a negative value, you have to flip the inequality. And so you'd have to consider two cases when X minus 13 is positive and when X minus 13 is negative. OK. All right. So there's a lot of things to consider. If you want to multiply it by X minus 13, we're gonna stick with this method where we just consider these points of interest in the intervals all around them. And then that's gonna take into consideration those restrictions. Now, starting on the far left and on the left side of two, on our number line means that the X values are less than two. If X is less than two, you can imagine using a trial point of say X is equal to zero and X minus two is gonna be negative. It's gonna be less than zero. That means that X minus 13 is also gonna be less than zero. OK. In the left hand side of our inequality, we're gonna call L H S which is X minus two divided by X minus 13, we have a negative value divided by a negative value. So the left hand side is going to be positive. It's gonna be greater than zero. Now, that's exactly what we want to satisfy our inequality. We want that the left hand side is greater than zero. And so this interval satisfies our inequality. It is gonna be part of our solution set. So we're gonna give a green check mark to this interval. Now moving to the right, we hit this point at exactly X equals two. Now X equals two, we get that the numerator is equal to zero. That means the entire left hand side is equal to zero, but zero is not greater than zero. OK? Because we have a strict inequality here, we have greater than zero. We do not have greater than, or equal to 00 is not a solution here. OK? And so the point X equals two is going to be excluded from our interval. So we're gonna draw a, an open circle on our number line to indicate that we do not include the point exactly at X equals two. Now we're gonna keep going, we move to our next interval. That's between X equals two and X equals 30. You can pick any test point in there, say X is equal to five. If you substitute that in X minus two, it's gonna be positive, it's gonna be greater than zero, but X minus 13 is going to be less than zero. That means that the left hand side will be a positive value divided by a negative value which we know is gonna give us a negative. So the left hand side is going to be less than zero. Now, this does not satisfy your inequality. We want the left hand side bigger than zero. And so this interval is going to be excluded from our solution set. We're gonna draw a red X on that interval. We're gonna move to the right. And now we're at the point X equals 13. And we already mentioned this as being a restriction. If X is equal to 13, the denominator is equal to zero, we get that the left hand side is undefined. And so this is not a solution. So just like we did for X equals two, we're gonna draw an open circle to indicate that we do not include that point. And we move to our final interval where X is greater than 13. Again, you can choose a test point. Say X is equal to 15, X minus two is gonna be greater than zero, X minus 13 is now also greater than zero. And so the left-hand side is a positive value divided by a positive value which gives us a positive value. The left-hand side is greater than zero, which satisfies our inequality. So this interval is included in our solution set. We give that a green check. So we can see from our number line that we have these two intervals that satisfy the inequality and that are part of our solutions. So we can write these out. We have everything to the left of negative or to the left of positive too sorry. So that's gonna be from negative infinity up to two. We're using round brackets to indicate that we do not include the end points. We're gonna take the union of that interval with the second interval we found which is the interval from 13 all the way up to positive infinity. Again, using round brackets to indicate that we are not including end points. And that is the complete solution to this inequality. If we compare this to the answer choices, we can see that this corresponds with answer choice. And B thanks everyone for watching. I hope this video helped see you in the next one.

Solve each rational inequality. Give the solution set in interval notation. See Examples 4 and 5. (x - 1)/(x - 4) > 0

Key Concepts

Rational Inequalities

Critical Points

Interval Notation

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