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. (x+2)/(2x+3)≤5

Accepted Answer

Hey, 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 the quantity of X plus eight divided by the quantity of five X plus one is less than or equal to two. Now our first step here in solving this inequality is to make the right side of our inequality zero. So we just need to subtract two from both sides. In now rewriting our inequality, we have the quantity of X plus eight divided by the quantity of five X plus one minus two is less than or equal to zero. Now, we want to simplify our expression on the left side by combining the whole number with our fraction. So we first need to get a common denominator which means we just need to multiply our whole number here by the expression in our denominator of the fraction. So we multiply two by the quantity of five X plus one divided by the quantity of five X plus one. Now we just need to see simplify. So our expression becomes the quantity of X plus eight divided by the quantity of five X plus one minus two times the quantity of five X plus one all divided by the quantity of five X plus one. And this expression is still less than or equal to zero. Now that we have found common denominators between both of our terms, we can simplify our expression further by combining our numerators. So combining the numerators, we get X plus eight and now we need to distribute negative two into the second set, the second fraction that we have. So we have minus 10 X minus two and this is all divided by the quantity of five X plus one. And this expression is still less than or equal to zero. Now, we can simplify the left side further by combining our like terms in the numerator. So combining like terms, our simplified inequality becomes the quantity of negative nine X plus six divided by the quantity of five X plus one. And this expression is less than or equal to zero. Now, we want to find values for X by setting the numerator and the denominator individually equal to zero. So we can start with the expression in our numerator where we have negative nine X plus six. And again, we want to set this equal to zero. And now we can just solve for X. So first, we need to just subtract six from both sides to get X alone. And then we have negative nine X is equal to negative six. And now we need to get rid of the coefficient of X by dividing both sides by negative nine. And now simplifying, we see that one solution for X is positive two thirds. And now we can repeat this process with the denominator from our or a equality. So we have five X plus one. And again, we set this equal to zero. Now solving for X first, we just need to subtract one from both sides which gives us five X is equal to negative one. Now to get X completely alone, we need to get rid of the coefficient by dividing both sides by five. And here we see the other solution for X is negative 1/5. And now with these two values that we found for X, we can use them to write our solution as intervals. So we see that the X axis can be separated into the following intervals where we first go from negative infinity to the smallest solution of X that we found which is negative 1/5. And both of these values for now are going to be closed by enthesis. And then we go from negative 1/5 to our next solution which is two thirds. And again, both of these terms are going to be closed by parentheses. And lastly, we go from our largest solution of two thirds to positive infinity. And again, both terms are closed by parentheses. So now we just need to test each of these intervals to see which of them serve as a solution to our original inequality. So we need to first let the function F of X be equal to the left side of our inequality where we had the quantity of negative nine X plus six divided by the quantity of five X plus one. So now that we have identified our intervals and our F of X function, we can now assess our intervals by making a table. So first, we will select 11 of our inter intervals and then we will choose a test value that is within the interval, a value within the interval. And then we take this te test value and substitute it into our F of X function. And from the value we get from F of X, we are able to form a conclusion based off of the sign of the value. So we can start with our first identified interval where we go from negative infinity to negative 1/5. And now we just want to choose a test interval that is within or a test value. Sorry, that's within this interval. And so here we can choose negative one. So now we need to find F of negative one. So we will have negative nine times negative one plus six, all divided by five times negative one plus one. And now simplifying we get negative 15 4th. And so since we obtained a negative value, we can conclude that F of X will be less than zero for all X values within the interval from negative infinity to negative 1/5. So now we can move on to testing our next interval where we go from negative 1/5 to positive two thirds. And now we need to choose a test value that is in between this interval. So we can choose, let's say zero. So now we need to find F of zero. Well, substituting in zero, we get negative nine times zero plus six all divided by five times zero plus one. And now simplifying we just get positive six. And since we obtained a positive value, we know that at of X will be greater than zero for all X values within the interval from negative 1/5 to positive two thirds. So now we can test our last interval where we go from two thirds to, to positive infinity. And a test value within this interval, we can choose as one. So now we find F of one, F of one gives us negative nine times one plus six all divided by five times one plus one. And now simplifying we get negative one half. And since our value is negative, we can conclude that F of X will be less than zero for all X values for all X values within this interval from two thirds to infinity. So now, from our conclusions, we can see that F of X is going to be less than zero for all X values from negative infinity to negative 1/5 and from negative or sorry, from positive to thirds to positive infinity. And so if we recall our original inequality, we were looking for when F of X is less than or equal to zero. So now we just need to adjust our intervals here so that they also include zero. So we see that F of X is less than or equal to zero for all X values from first, from negative infinity to negative 1/5. And now we know infinity is always excluded. So it will still get closed by a parentheses, but negative 1/5 is also going to be excluded because it causes the denominator to be zero. And we don't want to divide by zero. So this also needs to be excluded and closed by a parentheses. But for our second interval where we go from two thirds to positive infinity, well, again, infinity is excluded. So it will keep the parentheses but two thirds does cause our numerator to be zero. Thus causing F of X to be zero. So we need to include two thirds by putting a bracket before the value. So with this, we have found our solution to our given inequality and express the solution in interval notation. So with this, 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. See Examples 8 and 9. (x+2)/(2x+3)≤5

Key Concepts

Rational Inequalities

Critical Points and Sign Analysis

Interval Notation

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