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

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. (x+3)/(x+4)<0

Accepted Answer

Welcome back. I am so glad you're here. We are given the rational inequality. X plus seven divided by X plus nine is less than zero. And we're asked to solve this and then graph this on the number line. Well the first step in solving a rational inequality is to set one side equal to zero and yay that's already been done for us. So the next step is to set both the numerator, X plus seven and the denominator, X plus nine equal to zero. Then we are going to solve for our boundaries. So solving for X plus seven equals zero. We're solving for X. So will subtract seven from both sides. That will give us X equals negative seven. And solving for X plus nine equals zero will subtract nine from both sides and that will give us another boundary at X equals negative nine. Now we can draw our number line and we'll have boundaries at negative seven. Negative nine comes first negative nine is closer to negative infinity and then negative seven. So we have three intervals now that we're looking at from negative infinity. All the way up to negative nine. From negative nine to negative seven. And from negative seven all the way up to positive infinity. And what we need to do is select a test point from each of those three intervals. Then we're going to plug the test point value in for X in our original inequality and we'll see if this is still true. If the left side is still less than zero. When we plug those test points in and if it's true then those values are included for that interval. So let's choose our test points for negative infinity to negative nine. I'm going to choose negative 10 as a test point for negative nine to negative seven. I'm going to choose negative eight and for negative seven to positive infinity I'm going to choose zero. Now we'll plug those test points in for X. So we have negative 10 plus seven divided by negative 10 plus nine is less than zero, negative 10 plus seven is negative three over negative 10 plus nine is one is less than zero, negative three divided by negative one is a positive three is positive three, less than zero, nope it is not. So for values from negative infinity to negative nine. Our inequality is not true. So those are not included. Now let's test negative eight, negative eight plus seven. In the numerator divided by negative eight plus nine is less than zero. Negative eight plus seven is going to be a negative one. Negative eight plus nine is a positive one. Negative one divided by positive one is negative, one is negative one less than zero. Yes. So from negative nine to negative seven. All of those values are true for our inequality. Alright let's test zero, we've got zero plus seven divided by zero plus nine is less than 00 plus seven is 70 plus nine is nine so is seven. Nineths less than zero, nope, that value is positive, it is not less than zero. So for values from negative seven all the way up to infinity, those are not true for our inequality. Now how about negative nine and negative seven themselves? Well we take a look at our original inequality and this is less than we found negative nine and negative seven. When we found for negative seven, that's when it was equal to zero, the numerator was equal to zero so it can't include negative seven. And for negative nine we found that when we set the denominator equal to zero and that's when the whole fraction would be undefined because the denominator of a fraction cannot be equal to zero. Otherwise it's undefined. So neither negative nine nor negative seven can be included. How do we denote that? We put parentheses at negative nine and negative seven to show those can't be included and then we draw a line in between them to show all the numbers in between can be included. How do we show that? An interval notation? It's very similar. We put parentheses negative nine, comma negative seven parentheses. Now we look at our answer choices and this matches with answer choice A while 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. (x+3)/(x+4)<0

Key Concepts

Rational Inequalities

Critical Points and Sign Analysis

Interval Notation and Graphing Solutions

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