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

Question

Solve each rational inequality. Give the solution set in interval notation. See Examples 8 and 9. 4/(2-x)≥3/(1-x)

Accepted Answer

Everyone. Here we are asked to solve the following inequality and express the answer in interval notation. And here we are given the inequality where we have 10 divided by the quantity of three minus X is greater than or equal to eight divided by the quantity of one minus X. Now our first step in solving this in equality is to get the right side of the inequality to just be zero. So first, we just need to subtract eight divided by the quantity of one X from both sides of our inequality. And now rewriting our inequality, we now have 10 divided by the quantity of three minus X minus eight divided by the quantity of one minus X. And this expression is now greater than or equal to zero. So now our next step is to simplify the left side of our, our inequality by combining our fractions. Well, to combine the fractions, we need to first go get a common denominator. So our first fraction needs to be multiplied by the expression in the denominator of our second fraction. So our first fraction 10 divided by the quantity of three minus X gets its numerator. And denominator multiplied by the quantity of one minus X. And then looking at our second fraction, well, eight divided by the quantity of one minus X needs to be multiplied by the expression of the denominator of the first fraction. So eight divided by the quantity of one minus X gets the numerator and denominator multiplied by the quantity of three minus X. So now we just need to simplify this expression. And so our inequality becomes 10 times the quantity of one minus X divided by the quantity of three minus X times the quantity of one minus X minus eight times the quantity of three minus X all divided by the quantity of three minus X Times the quantity of 1 -1. And this expression is still greater than or equal to zero. So now we can see that both fractions have common denominators. So we can simplify the left side further by combining their new. So combining the new Raiders, we need to factor N D positive 10 and the negative eight. So we get 10 minus 10, X minus 24 plus eight X is all divided by the quantity of three minus X times the quantity of one minus X. And this expression is still greater than or equal to zero. Now, we need to finish simplifying our inequality by combining the like terms in the numerator. So combining the like terms, our final simplified inequality becomes the quantity of negative 14 minus two X all divided by the quantity of three minus X times the quantity of one minus X. And now this expression is still greater than or equal to zero. So now with this simplified inequality, we need to solve for values of X or solutions for X. So to do this, we need to set the numerator and the denominator of our inequality individually equal to zero. So we can start with the numerator where we have negative minus two X. And again, we set this equal to zero. So we can solve for X. So first, we can add 14 to both sides to get X alone. So we have negative two, X is equal to 14. And now we need to get rid of the coefficient for X by dividing both sides by negative two. And we see that one of our solutions for X is negative seven. Now we can repeat this with the denominator of our inequality. So we take the quantity of three minus X Times the quantity of one -X and we can set this expression equal to zero. And now we can take each, each part of the expression, each set of parentheses and individually set them equal to zero to find our values for. So we can start with our first set of parentheses which is 3 -1. And again, we set this expression equal to zero. Now solving for X, we can just add X to both sides and we get X is equal to positive three. And now we can repeat this with the other part of our expression one minus X. Again, we set this expression equal to zero as well. Now solving for X, we can just add X to both sides and we get X is equal to positive one. So now we have found three different values for X and we can use them to write intervals. And so we, we'll see that the X axis can be separated into intervals from negative infinity to first the smallest solution which is negative seven. And both of these terms get closed by parentheses for now. And then we go from the solution of negative seven to the next closest solution which is one and then we go from one to the next solution which is three. And lastly, we go from the highest solution which is three to positive infinity. And again, each of our terms for each interval are closed by parentheses. So now we need to assess each of these intervals that we found to see which of them serve as a to our inequality. So first, we need to let F of X be equal to the left side of our simplified inequality where we have the quantity of negative 14 minus two X all divided by the quantity of three minus X times the quantity of one minus X. So now with our function identified and our intervals identified as well, we can go ahead and assess each of our intervals. So we need to set up a table where we first take our interval and we then choose a test value that fits within the interval. And then we take this test value and we substitute it, we substitute it into our function F of X. And then from the value we get from F of X, we are able to form a conclusion based on the sign a conclusion based off of the sign of the value. So we can begin with our first interval where we go from negative infinity to negative seven. And now again, we choose a test value within this interval, we can choose negative eight. Now we find F of negative eight, F of negative eight gives us negative 14 minus two times negative eight all divided by the quantity of three minus negative eight times the quantity of one minus negative eight. And now simplifying, we see that F of negative eight gives us two divided by 99. And now, since we obtained a positive value from our function, we can conclude that F of X is going to be greater than zero for all values of X within this interval from negative infinity to negative seven. So now we finished assessing our first interval. So we can move on to the next interval where we go from negative 7 to 1. And now we need to choose a test value within this interval, we can choose zero. Now finding F of zero, we get negative 14 minus two times zero all divided by the quantity of three minus zero times the quantity of one minus zero. And now simplifying we get negative 14 3rd. And since we obtained a negative value, we can conclude that F of X is going to be less than zero for all X values within the interval from negative 7 to 1. And now we repeat this for our next interval where we go from 1-3 and we need to choose a test value. So here we can just choose to. And now we find f of two. Well, from F of two, we have negative 14 minus two times two, all divided by the quantity of three minus two times the quantity of one minus two. And now simplifying we get the value of positive 18. While since we obtained a positive value, we see that F of X is going to be greater than zero for all X values within the interval from 123. Now we can move on to our fourth and final interval where we go from three to positive infinity. Now a test value within this interval, we can choose is four. Now F of four gives us negative minus two times four all divided by the quantity of three minus four times the quantity of one minus four. Simplifying we get negative 22. Divided by three. And from this negative value, we can conclude that for this interval, we will have X be less than zero for all X values again, within this interval from three to infinity. So now taking a look at all of our conclusions that we found we can summarize and we see that F of X is going to be greater than zero for all X values within the intervals from negative infinity to negative seven and from 1 to 3. So originally, we were trying to find F of X being greater than, or equal to zero. And so now we just need to adjust the intervals that we found to be greater than zero. So that we also include when these values are equal to zero. So F of X is going to be greater than or equal to zero for all X within the intervals from negative infinity to negative seven. Well, negative infinity gets closed by a parenthesis. Since we know that infinity is never included and then negative seven does return us F of X being equal to zero. So we need to include this value by closing it by a parentheses. And next for our other interval, we go from 1 to 3. Well, one and three give us a negative denominator from F of X. And since we do not want to divide by zero, we will still keep both of these values closed by parentheses since we want them excluded. From our solution. So with this, we have found two intervals that satisfy or express our solution for our given inequality and interval notation. So with this, we are left here with answer choice. A 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. See Examples 8 and 9. 4/(2-x)≥3/(1-x)

Key Concepts

Rational Inequalities

Finding a Common Denominator and Combining Fractions

Sign Analysis and Interval Testing

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