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 + 1)/(x + 3) < 2

Accepted Answer

Solve the following rational inequality. And graph the solution set on a number line X plus five divided by X plus one is less than three. So we can re write this as X plus five divided by X plus one. And since we have three on the right side, we can move it over to the left side by sub crafting, so we get minus three that should be less than zero. Now, to solve this, we need to combine our fraction and our number to do that, we can turn our negative three into a fraction to combine fractions. We need the same denominator. So we will multiply our negative three by our denominator. X plus one divided by X plus one. We will then get X plus five divided by X plus one minus three, multiplied by X plus one, divided by X plus one. We can now simplify X plus five divided by X plus one minus three, X plus three divided by X plus one, which we can now directly combine X plus five, distributing our negative gives us negative three, X minus three, followed by X first one. We can combine our X and our constant to get negative two X plus two divided by X plus one. And we know this is less than zero. So what we can do now is set our numerator equals to zero and our denominator equals to zero to find our solutions to negative two, X plus two equals zero and X plus one equals zero. Fact two get negative two, X equals negative two and divide both sides by negative two to get X equals one. For our other number subtract one to get X equals negative one which you now see we have three domains. Our first domain will be our negative affinity to our smaller number, negative one. Then we have native one to our next number one. And finally, our last number one to infinity. Now to stop this, we need to find some test values on each of these intervals can see what the value of the function is. If it is negative, then that is our answer. It's positive. We discard that interval. Let's say in our first interval, we use X is equals to negative two, which is in that domain, X equals zero and X equals two will work for the other ones. We then plug in our numbers. We plug in X equals negative two of the two equals negative six, F of zero equals two and F of two equals negative two thirds. So we have two intervals which are less than zero. Our first interval. And our final interval, meaning the answer to our problem will be negative infinity to negative one and one to infinity. The answer to our problem then turns out to be answer a OK. I hope to help you solve the problem. Thank you for watching. Goodbye.

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 + 1)/(x + 3) < 2

Key Concepts

Rational Inequalities

Critical Points and Sign Analysis

Interval Notation and Graphing Solutions

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