Solve each rational inequality in Exercises 43–60 and graph the s...

Question

Solve each rational inequality in Exercises 43–60 and graph the solution set on a real number line. Express each solution set in interval notation. (4−2x)/(3x+4)≤0

Accepted Answer

Welcome back. I am so glad you're here. We are given the rational inequality eight minus four X divided by five x plus nine is less than or equal to zero. And we asked to solve this rational inequality and graph the solution on a number line. Well the first step to solving a rational inequality is to set one side equal to zero and yay that's already done for us. So we don't have to do that. The next step is to set both our numerator eight minus four X and our denominator five X plus nine equal to zero to find out where our boundaries are. So we can solve both of these. We'll start off with eight minus four. X will subtract eight from both sides and then we'll have a negative four X equals negative eight will divide both sides by negative four and we'll have x on the left side equals negative eight divided by negative four is a positive two. So one of our boundaries is at positive two. Now let's solve the other one will subtract nine from both sides. On the left will have five X equal to on the right, negative nine, dividing both sides by five. We will have X equals negative 9/5 and we have found our two boundaries. So now we'll have three intervals let's plot those two boundaries on the number line. The first one will be at negative 9/ and the second one will be at positive two. So that means they're intervals will be from negative infinity up to negative 9/5 from negative 9/5 to 2 And from 2 to positive infinity those almost look like infinity symbols of butterflies or something there. Okay now we are going to choose test points in each of those three intervals and then we're going to plug those test points in for X. In our original inequality and see where it is. True that eight minus four X divided by five X plus nine is less than or equal to zero. So for the test point that is between negative infinity and negative 9/ negative 9/5 is the same as negative one and 4/5. So I'm going to choose negative three. For the test point between negative 9/5 and two I'm going to choose zero. And for the test point greater than two, I'm going to choose three. Now plugging those in to our inequality we've got eight minus four. Not times X. But now times negative 3/5. Not times X. But times negative three plus nine is all less than or equal to zero. And in the numerator, eight minus four Times negative three. Well that's eight plus 12 over five. Times negative three will give us negative 15 plus nine. Saw less than or equal to +08 plus 12. In the numerator is 20 over negative 15 plus nine is negative three. Excuse me It's negative six. So we've got 20 over negative six. Is that less than or equal to zero. Yes, that's a negative number. So the test point negative three works. And therefore numbers between negative infinity and negative 9/ do make the inequality true. Now let's check zero. So we will take eight minus four times 0/5 times zero plus nine is less than or equal to 08 minus four times zero is eight minus zero which is 8/5 times 00 plus nine gives us nine In the denominator. Is it true that 8/9 is less than or equal to zero, nope. That positive number is not less than or equal to zero. So for numbers between negative 9/5 and to that inequality is not true. Alright, let's check three. So we've got eight minus four times three instead of x over five times three plus nine are less than or equal to 08 minus four times three will be eight minus 12/5 times three is 15 plus nine. So eight minus 12 will give us a negative four in the numerator and 15 plus nine will give us a 24 is less than or equal to zero. Is that true? That negative four, 24th is less than or equal to zero. Yes, that is also true. So from numbers from two all the way up to positive infinity. This inequality is true. So what about -9/5 and to themselves? Well we found negative 9/5. When we solved for setting the denominator equal to zero and when the denominator is equal to zero then the whole fraction is undefined. So it cannot be negative 9/5. We say that it cannot be negative 9/5 by putting a parentheses negative 9/5 and then we can draw a line in an arrow heading toward negative infinity saying that it does include the numbers below negative 9/5. Now, how about two? We found two By setting the numerator equal to zero. So too will get the whole fraction equal to zero. And let's look at our original inequality, the original inequality says less than or equal to. So it does include two because that will make it equal to zero. How do we indicate that? It includes two on the number line. We put a bracket at two and then we will draw our arrow heading up toward positive infinity to show that it does also include all the numbers greater than two. How do we indicate this? In interval notation. While we'll put a parentheses negative infinity, comma negative 9/5 and then we have a parentheses for negative 9/5 to say it can't be negative 9/5 and that's in union with bracket two all the way up to positive infinity and then a parentheses to end at positive infinity. And we look at our answer choices and that matches with answer choice. A Well done. We'll catch you on the next one

Solve each rational inequality in Exercises 43–60 and graph the solution set on a real number line. Express each solution set in interval notation. (4−2x)/(3x+4)≤0

Key Concepts

Rational Inequalities

Critical Points

Interval Notation

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