Solve each rational inequality. Give the solution set in inter...

Question

Solve each rational inequality. Give the solution set in interval notation. 3/(2x-1)>-4/x

Accepted Answer

Hey, everyone here we are asked to solve the following inequality and express the answer in interval notation. Now, here we are given the inequality nine divided by the quantity of three X minus one is greater than negative two divided by X. Now our first step in solving this inequality is to get the right side of the inequality to just be zero. So we just, we need to add two divided by X to both sides of our inequality. And now we can rewrite our inequality as nine divided by the quantity of three X minus one plus two divided by X is greater than zero. Now, we want to work to simplify the left side of our inequality further. So that means we need to combine both of our fractions. But first, before we can combine their numerator, we need to get common denominators. So we need to multiply our first fraction nine divided by the quantity of three X minus one by the expression in the denominator of our second fraction. So the numerator and the denominator of our first fraction gets multiplied by X. So our fraction it's multiplied by X divided by X and now, with our second fraction, we need to multiply the numerator and the denominator by the expression in the denominator of the first fraction. So two divided by X gets multiplied by the quantity of three X minus one divided by the quantity of three X minus one. And now from here, we just need to start simplifying our expression. So we have nine X all divided by X times the quantity of three X minus one plus two times the quantity of three X minus one, all divided by X times the quantity of three X minus one. And this is still greater than zero. Now, we see that we have common denominators before both of our fractions. So we can simplify further by combining the values in the numerator. So combining the terms we have nine X and now distributing the positive two, we have plus six X minus two all divided by X times the quantity of three X minus one. And this expression is still just greater than zero. And now our next step in simplifying our last step here is to just combine the like terms in the numerator. So our final simplified inequality becomes X minus two all divided by X times equality of three X minus one and this is greater than zero. So now we have found our simplified inequality. So next, we need to find solutions for X by setting the numerator and the denominator individually equal to zero. So starting with our numerator, which we have 15 X minus two. Again, we set this expression equal to zero. So now we can solve for X. So first we need to add two to both sides to get our X term alone. So we have 15 X is equal to two. And now we need to get rid of our coefficient for X. So we need to divide both sides by 15. And we see that one of our solutions for X is two divided by 15. And now we can repeat this process with our denominator where we have X times the quantity of three, three X minus one. And again, we set this expression equal to zero. And now to find values for X, we can set both parts of our expression individually equal to zero. So we can start with our first part of the expression where we just have X. Well, X is just going to be equal to zero. So we see that zero is another solution. And next we take three X minus one and set it equal to zero. Solving for X, we first add one to both sides and we get three, X is equal to one. Now we divide both sides by three to get X completely alone. And we see a third solution for X is one third. So now with these three solutions, we found for X, we know that the X axis can be separated into the following intervals where we first go from negative infinity to the smallest solution that we found, which is zero. And then we go from zero to the next solution which is to 15th. And then we go from, to fit 15th to our next solution, which is one third. And lastly, we go from our largest solution of one third all the way to positive infinity. And now with these intervals identified, we can, we need to assess each of these intervals to see which of them serve as a solution to our given inequality. So first, we need to let the function F of X be equal to the left side of our simplified inequality where we have the quantity of 15 X divided by two or minus two, we have the quantity of 15 X minus two divided by X times the quantity of three X minus one. So now with our function clearly identified, we can move on to testing hour intervals to again see which are actual solutions. So to test our intervals, we need to set up a table where we first take an interval, we take an interval and then we choose a test value that is within the bounds of that interval. And then with that test value, we substitute the value into our function F of X. And lastly from the value we get from F of X, we are able to form a conclusion. So first we can choose our first identified interval to test where we went from negative infinity to zero. And then we need to choose a test value within this interval. And here we can just choose negative one. And now we need to find em of negative one while F of negative one, it gives us 15 times negative one minus two, all divided by negative one times the quantity of three times negative one minus one. And now simplifying, we see that F of negative one gives us negative 17 4th. And now, since we obtained a negative value from our function, we can conclude that F of X is going to be negative or less than zero for all values of X within the interval where again, we go from negative infinity to zero. So now we can move on to testing our next interval. So our next interval, we have 0 to 2/15 and choosing a test value here, we can just choose 1/10. And now with our positive 1/10 we need to find F of 1/10 and substituting in this value, we have 15 times 1/10 minus two all divided by 1/10 times the quantity of three times 1/10 minus one. And now it's simplifying from this function, we get positive 50 divided by seven. And since we obtained a positive value, we can conclude that our function F of X is it's going to be greater than zero or positive for all X values within our interval that we're retesting, which is from 0 to 2/15. And now we can repeat this with our next interval where we go from 2/15 to 1 third. And now choosing a test value here, we can choose 1/4. And now we need to find F of 1/4 F of 1/4 gives us 15 times 1/4 minus two, all divided by 1/4 times the quantity of three times 1/4 minus one. And now simplifying this expression here, we get negative 28. And with this negative value, we will, we can conclude that F of X is going to be less than zero for all X values within the interval from 2 to 1 third. And now we repeat this process for our fourth and final interval where we go from one third to positive infinity. Now a test value within this interval, we can choose, we can just chew use one. So now we find F of one, well, F of one gives us 15 times one minus two, all divided by one times the quantity of three times one minus one. And now simplifying we get positive 13 halves. Well, from this positive value, we can conclude that F of X is going to be greater than zero for all X values for all X values within the interval that we were testing, which is from one third to positive infinity. Now, if we recall from our simplified inequality. From earlier, we were looking for where F of X, where our function F of X is going to be greater than zero. And now looking at our conclusions that we found we c we obtain a positive F of X value from our second interval and from our fourth tested interval. So now we can conclude that overall F of X is going to be greater than zero for all X values for all X values within the interval from 0 to 2/15 and from the interval of one third to positive infinity. So with this, we have found the solution set in interval notation for our given inequality. So here we are left with answer choice B thanks for watching. I hope you found this video helpful and I'll see you again next time.

Solve each rational inequality. Give the solution set in interval notation. 3/(2x-1)>-4/x

Key Concepts

Rational Inequalities

Finding a Common Denominator and Combining Terms

Sign Analysis and Interval Testing

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