Solve each inequality. Give the solution set using interval no...

Question

Solve each inequality. Give the solution set using interval notation. x+7 / 2x+1 ≤1

Accepted Answer

Hey, everyone here, we are asked to find the solution set of the following inequality and express it in interval notation. And we're given the inequality where we have the quantity of X plus 15 divided by the quantity of four X plus nine. And this is less than or equal to two. So now our first step in solving this problem is to solve our inequality. So we first want to get the right side of the inequality to be zero. So we just need to subtract two from both sides. And now rewriting our inequality, we have the quantity of X plus 15 divided by the quantity of four X plus nine. And now we have minus two. And so this expression is less than or equal to zero. So now we want to combine our two terms here on the left side of our inequality, which means we need to find a common denominator. So all we have to do is multiply our whole number by the expression in our denominator. So we just multiply two by the quantity of four X plus nine divided by the quantity of four X plus nine. And so now we just need to simplify our inequality. So we have the quantity of X plus 15 divided by the quantity of four X plus nine minus two times the quantity of four X plus nine all divi guided by the quantity of four X plus nine. And this expression is still less than or equal to zero. So now that we have found a common denominator for both of our terms, we can simplify this expression by combining the terms in the numerator. So doing this, our expression becomes X plus 15. And now factoring in the negative two, we get minus eight, X minus 18 is all divided by the quantity of four X plus nine. And now this rewritten expression is still less than or equal to zero. So now we need to continue to simplify our inequality. And now we can do this by combining the like terms in our numerator. So combining the like terms, we get negative seven X minus three all divided by the quantity of four X plus nine and this is less than or equal to zero. So now we want to find our X solutions to this inequality by setting the numerator and the denominator individually equal to zero. So we can start with the expression in the numerator where we have negative seven X negative seven X minus three. And again, we set this equal to zero. And now we just want to solve for X. So first we need to add three to both sides and we get negative seven, X is equal to positive three. And now to get rid of the coefficient for X, we divide both sides by negative seven. And we see that one of our solutions for X is negative 37. So now we just need to repeat this process with our expression in the denominator. So we take four X plus nine and again, set this expression equal to zero. Now we need to solve for X. So first, we just subtract nine from both sides and we get four X is equal to negative nine. Now we need to get rid of the coefficient of X. So we divide both side by four. And we see that the other solution for X that we find is negative 9/4. So now with these two values of X that we found we can write a set of intervals to express how the X axis can be separated. So we know the X axis can be separated into the following intervals where we first go from negative infinity to the smallest solution which is negative 9/4. And for now, we just close both of these terms by parentheses and then we go from the solution negative 9/4 to the next solution which is negative three sevens and close both of these terms by parentheses. And lastly, we go From the largest solution -37s all the way to positive infinity. And once again, both of these terms are closed by parentheses. So now with these identified intervals, we want to test each of these intervals to see which of them form the correct solution set for our given inequality. So first, we need to let the function F of X be equal to the expression on the left side of our inequality, which is the quantity of negative seven X minus three divided by the quantity of four X plus nine. So now we can begin testing our intervals. So first, we need to take one of our identified intervals and then we need to choose a test value that is within the interval. And next, we take the test value and we substitute it into the function F of X. And from the value that we get from F of X, we form some sort of conclusion based on the sign of the value. So we can begin with the first identified interval where we go from negative infinity to negative 94. And now we need to choose a test interval within this interval or sorry, a test value within this interval. And so here we can just choose negative three. And now we need to find F of negative three and now F of negative three gives us negative seven times negative three minus three, all divided by four times negative three plus nine. And now simplify, we see that F of negative three gives us negative six. So now, since we have obtained a negative value from our function, we see that F of X is going to be less than zero for all X values within the interval that we were testing, which is from negative infinity to negative 9/4. So now we can move on to testing our next interval where we go from negative 9/4 to negative three sevens. So again, we need to choose a test value within this interval. We can go ahead and choose negative one. Now we need to find F of negative one. So we have negative seven times negative one minus three all divided by four times negative one plus nine. And now simplifying, we see that F of negative one gives us positive forfeits. So 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 X values in the interval that we tested, which is from negative 9/4 to negative three sevens. So now we have tested our first two intervals. We can move on to our final interval where we go from negative three sevens to positive infinity. So now again, we choose a test interval within the interval or a test value within the interval that we are testing. So we can choose zero. And now we need to find f of zero F of zero gives us negative seven times zero minus three, all divided by four times zero plus nine. And now simplifying F of zero gives us negative one third. So now from this solution, we see that F of X is going to be less than zero for all of the X values within the interval where we go from negative three sevens to positive infinity. So now summarizing our conclusions, we see that we have two of the same conclusions for our first interval and our third interval. So we see that F of X is going to be less than zero for all X values within the intervals where we go from negative infinity to negative 9/4 and then from negative three sevens to positive infinity. So if we recall our original inequality, we were testing if F of X was less than or equal to zero. So we need to just adjust our inequalities that are less than zero so that they also can be equal to zero. So we can look at the first interval where we go from negative infinity to negative 9/4 when, well, we know that infinity is always excluded. So it gets closed by a parenthesis and now negative 9/ gives us a zero in the denominator. So it also needs to be excluded and therefore is closed by a parenthesis. But now for our next interval where we go from negative three sevens to positive infinity, well, once again infinity is excluded. So it gets closed by a parenthesis. And then negative three sevens does in fact give us a zero in our numerator and therefore gives us our function to be equal to zero. So this value negative three sevens is included and gets closed by a bracket. So we have found the solution set for when our inequality is less than or equal to zero. So we have found the solution set for our original inequality. And with this, we are left with answer choice. C Thanks for watching. I hope you found this video helpful and I'll see you again next time.

Solve each inequality. Give the solution set using interval notation. x+7 / 2x+1 ≤1

Key Concepts

Solving Rational Inequalities

Interval Notation

Sign Analysis and Testing Intervals

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