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

Question

Solve each inequality. Give the solution set in interval notation. (x+7)/(2x+1)<0

Accepted Answer

Welcome back. I am so glad you're here. We are asked to provide the solution set for the given rational inequality. In interval notation. Our rational inequality is in the numerator X plus 19 divided by the denominator seven X plus one, which is all less than zero. Our answer choices are answer choice. A the solution set open parentheses 19 to negative 1/7 closed parentheses. Answer choice B the solution set open parentheses negative to 1/7 closed parentheses. Answer choice C the solution set open parentheses negative to negative 1/7 closed parentheses. In answer choice D the solution set 19 to 1/ and with open and closed sets of parentheses. All right. So how do we solve this? Well, we want to make sure when we have a rational inequality that one side is equal to zero. Yep, we've got that and that the other side is a single fraction. Yes, we have that. Now, what we do is we set both our numerator and our denominator equals to zero and solve. So for the numerator X plus 19 setting it temporarily equal to zero. Solving for X will subtract 19 from both sides and we get a value of X equals negative 19. We'll hold on to that. And next we set the denominator equal to +07 X plus one equals zero. Solving for X, we subtract one from both sides. We have seven, X equals negative one, divide both sides by seven. And we have another X value of X equals negative 1/7. So what does that tell us? Well, we have just found the end points for our intervals for our inequality. So we can draw a number line from negative infinity to positive infinity. And we can label our end points that we have found the first endpoint, the more negative one was negative 19. And the next end point is negative 1/7 and we have three intervals, one from negative infinity to negative 19. The next interval from negative 19 to negative 1/7 and the final interval from negative 17 to positive infinity. And now we need to find out for which intervals this given inequality is true. So we choose test points from each of the three intervals from negative infinity to negative 19. I'm choosing negative 20 from negative 19 to negative 1/7. I'm choosing negative one and from negative 1/7 to positive infinity. I'm choosing zero. Now we plug each of those test points in for X in our original inequality and see if it's true. So we can start off instead of X for X plus 19. We have negative 20 plus 19 in the numerator all divided by seven multiplied not by X but by negative 20. And then that'll be plus one is all supposedly less than zero. Simplifying negative 20 plus 19 is negative one. All divided by seven multiplied by negative 20 is negative 140 plus one. Supposedly less than zero in the numerator. We still have negative one in the denominator. We have negative 140 plus one which is negative 139 but negative one, negative 139th would be a positive one, 139th, which is not less than zero. So the interval is not true for the inequality is not true from the interval from negative infinity to negative 19. Now let's test negative one. So instead of X, we have negative one plus 19 all divided by seven multiplied not by X but by negative one and then plus one supposedly less than zero. Simplifying negative one plus 19 is equal to a positive 18 all divided by seven multiplied by negative one is negative seven plus one. Simplifying negative seven plus one is negative six. So we have 18 divided by negative six which is equal to a negative three is negative three, less than zero. Yes, it is. So this is true for that interval. The inequality is true for that interval. Now we can test for zero for the last interval. So we'll have zero plus 19, all divided by seven multiplied by zero plus one, all supposedly less than zero in the numerator zero plus 19 is 19 7 multiplied by zero. And the denominator is zero. So we have zero plus one which is one while 19 divided by one is is 19, less than zero. Nope, it is not. So the interval is not true. The inequality is not true there. So we're talking about the interval from negative 19 to negative 1/7. Is that including negative 19 and negative 1/7? No, it's not because this is just less than not less than or equal to. So nothing could be included for our end points. So how do we write this in interval notation? Well, since negative 19 is not included, it's open with an open parentheses, then we have negative 19 comma negative 1/ close parentheses because negative 1/7 is not included. We look at our answer choices and this matches with answer choice. C well done. We'll catch you on the next one.

Solve each inequality. Give the solution set in interval notation. (x+7)/(2x+1)<0

Key Concepts

Solving Rational Inequalities

Critical Points and Sign Analysis

Interval Notation

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