Video: Finding the Solution Set of Linear Inequalities with Real Numbers

Find the solution set of the inequality (5π‘₯ βˆ’ 1)/10 < βˆ’2π‘₯ + 5 < (π‘₯ + 3)/2 in ℝ. Give your answer in interval notation.

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Video Transcript

Find the solution set of the inequality five π‘₯ minus one over 10 is less than negative two π‘₯ plus five which is less than π‘₯ plus three over two in the real numbers. Give your answer in interval notation.

Now, this is a relatively complex question. We have a double-sided inequality, we have some fractions, and we also have our variable π‘₯ appearing in each part of this inequality. Let’s first consider how we could solve this problem by treating the two parts of the inequality separately. So we have the single-sided inequality five π‘₯ minus one over 10 is less than negative two π‘₯ plus five and another single-sided inequality negative two π‘₯ plus five is less than π‘₯ plus three over two.

Through our first inequality, we’d begin by multiplying each side by 10 to eliminate the fractions. And we now have five π‘₯ minus one is less than negative 20π‘₯ plus 50. We can then add one to each side of this inequality. Now, we still have π‘₯ terms appearing on each side. So next, we want to group all of the π‘₯’s on the same side. And if we add 20π‘₯ to each side, we’ll have a positive number of π‘₯’s on the left-hand side of our inequality. Adding 20π‘₯ to each side of the inequality, then, gives 25π‘₯ is less than 51. Finally, dividing both sides of our inequality by 25 gives π‘₯ is less than 51 over 25. And we’ll keep this answer in fractional form.

So we’ve solved the first part, and now let’s consider how we can solve the second part. First, we can multiply both sides by two to eliminate the fractions, giving negative four π‘₯ plus 10 is less than π‘₯ plus three. We can then subtract three from each side giving negative four π‘₯ plus seven is less than π‘₯ and then add four π‘₯ to each side of the inequality, giving seven is less than five π‘₯. Finally, we divide by five, giving seven over five is less than π‘₯ or π‘₯ is greater than seven over five.

So, we’ve solved the two parts of our inequality separately. But actually, we can combine these answers. We can express this as a double-sided inequality, π‘₯ is greater than seven over five and less than 51 over 25. The question, though, asked us to give our answer an interval notation. We can write this then as the interval with endpoints of seven over five and 51 over 25. And as each of the inequalities are strict inequalities, meaning neither of the endpoints are included in the interval, this means that our interval will be open at both ends. So we have our answer to the problem. The solution set of this inequality is the open interval from seven over five to 51 over 25.

Now, although the steps involved in solving our two separate inequalities were different here, it is actually possible to solve this inequality by keeping it all together, although this won’t always be the case. As the denominators involved in the fractions are 10 and two and as two is a factor of 10, we can eliminate each of the denominators by multiplying each part of our inequality by 10. Which takes us to five π‘₯ minus one is less than negative 20π‘₯ plus 50, which is less than five π‘₯ plus 15. In that final part, the factor of two in the numerator has canceled with the factor of two in the denominator. But we still have a factor of five to multiply π‘₯ plus three by.

Now, we can subtract 50 from each part of the inequality to give five π‘₯ minus 51 is less than negative 20π‘₯, which is less than five π‘₯ minus 35. But at this stage, we notice that we still have π‘₯ terms involved in each part of our inequality. We need to group them all together in the center. And in this instance, because the number of π‘₯’s on the far left of the inequality, that’s five π‘₯, is the same as the number of π‘₯’s on the far right of our inequality, we can do this by subtracting five π‘₯ from each part. However, had the number of π‘₯’s not been the same on the two sides of the inequality, we wouldn’t have been able to do this. And we’d need to have used our first method.

We now have negative 51 is less the negative 25π‘₯, which is less than negative 35. Finally, we can divide each part of our inequality by negative 25. But we must be really careful here because we should remember that when we multiply or divide an inequality by a negative number, we have to reverse the direction of the inequality sign. So each of our less than signs must become greater than signs. We therefore have negative 51 over negative 25 is greater than π‘₯, which is greater than negative 35 over negative 25. In each of our fractions, the negatives in the numerators and denominators will cancel out. And the fraction 35 over 25 can be simplified to seven over five.

Writing our inequality the other way round so that the smallest number is on the left, we have seven over five is less than π‘₯ is less than 51 over 25, which is the same as the solution we found using our previous method. So we can convert our answer to interval notation in exactly the same way.

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