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. (3x+5)/(6−2x)≥0

Accepted Answer

Welcome back. I am so glad you're here. We are given the rational inequality two X plus seven divided by eight minus four X is greater than or equal to zero. And we are asked to solve this rational inequality and graph it 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 been done for us. So now the next step is that we're going to find our boundaries. We do that by setting the numerator two, X plus seven and the denominator eight minus four X both equal to zero and then solve for X. So for the first one we're going to subtract seven from both sides and we'll have two X equals negative seven and we'll divide both sides by two and we have one boundary at X equals negative seven halves. Now let's find our other boundary. We will subtract eight from both sides and we have negative four X equals negative eight, Dividing both sides by - and we have X equals negative eight divided by negative four is a positive two and there are two boundaries. So now we can draw a number line And it might be a little bit less curved than that. And we draw our two boundaries at -7/2 and a positive two. And now we have our intervals the first one is from negative infinity. All the way up to negative seven halves. The next is from negative seven halves to two and the third intervals from two. All the way up to positive infinity. And what we do is we choose a test point from each of those three intervals, plug it back into our original inequality and see where it's true that two X plus seven divided by eight minus four X. Is greater than or equal to zero. So for the test point that's less than negative seven has negative seven halves is negative 3.5. For the test point that's less than that I'm going to choose negative five. For the test point between negative seven halves and two I'm going to choose zero and for the test point that is greater than two. I'm going to choose positive three and now we'll plug those test points into the original inequality. So instead of two times X it'll be two times negative five plus seven in the numerator over eight minus four times X. But times negative five in the denominator is greater than or equal to 02 times negative five is negative 10 plus 7/8, negative four times negative five is plus 20 and give myself a little bit room and negative 10 plus seven is negative 3/8 plus 20 is 28. So we've got negative three 28th is greater than or equal to zero. Is that true? No negative 3 20 eight's is not greater than zero or equal to it. So we know that for numbers between negative infinity and negative seven halves. The inequality is not true. Now let's test out zero. We have two times zero plus 7/ minus four times zero are greater than or equal to 02 times 00 plus seven is seven in the numerator over eight minus four times 00. So that's just eight in the denominator. So is 7/8 greater than or equal to zero. Yes it is. That's a positive number. So it is true for numbers between negative seven halves and two that those values make our inequality true. Now let's test numbers greater than two. We'll test it with three. So we've got two times three plus seven in the numerator over eight minus four times three. In the denominator is greater than or equal to 02 times three is six plus seven over eight, negative four times three is negative 12. And so we've got six plus seven is 13 in the numerator over eight minus 12 is negative four. And the denominator. Is it true that 13 over negative four is greater than or equal to zero, nope. That is a negative number again. So for numbers greater than two. This inequality is not true. So what about at negative seven halves. And to themselves? Well, we found negative seven halves when we set the numerator equal to zero. And let's take a look at our original inequality. This is not strict. It includes numbers that are equal to zero. And when we plug in negative seven halves it will be equal to zero. We found that from doing the work in the numerator. So that means we can include negative seven halves. We show that we can include it by drawing a bracket at negative seven halves. How about two? We found two? When we found the zero for the denominator and when the denominator is equal to zero then the fraction is undefined. So it cannot be too. We show that it cannot be too by drawing a parentheses at two and then we can draw a line connecting our bracket are parentheses to indicate that it does include the numbers between negative seven halves and two. How do we write this? An interval notation? We write a bracket and then negative seven halves comma two parentheses. 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 rational inequality in Exercises 43–60 and graph the solution set on a real number line. Express each solution set in interval notation. (3x+5)/(6−2x)≥0

Key Concepts

Rational Inequalities

Critical Points and Sign Analysis

Interval Notation and Graphing Solutions

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