Question Video: Finding the Solution of Inequalities Using Interval Notation Mathematics

Find all the values of 𝑥 that satisfy 11 < 7𝑥 + 4 ≤ 32. Give your answer in interval form.

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Video Transcript

Find all the values of 𝑥 that satisfy seven 𝑥 plus four is greater than 11 and less than or equal to 32. Give your answer in interval form.

This is what’s known as a double-sided or compound inequality. In general, a compound inequality contains at least two inequalities separated by either the word “or” or the word “and”. Here, we have an expression, seven 𝑥 plus four, and the inequality tells us that the value of this expression is greater than 11 and it’s less than or equal to 32. There’re two approaches we can take to solving a double-sided inequality.

The first approach is to treat the two parts of the inequality separately. So we have one inequality telling us that seven 𝑥 plus four is greater than 11 and another telling us that seven 𝑥 plus four is less than or equal to 32. We then solve each inequality. For the inequality on the left, the first step is just subtract four from each side, giving seven is less than seven 𝑥. We can then divide each side of this inequality by seven to give one is less than 𝑥 or 𝑥 is greater than one. So, we’ve solved our first inequality.

To solve the second, we also subtract four from each side, giving seven 𝑥 is less than or equal to 28 and then divide both sides by seven to give 𝑥 is less than or equal to four. We’ve found then that the value of 𝑥 is greater than one and less than or equal to four. So we must make sure to put these two parts of the solution back together again at the end. We can write our solution as the double-sided inequality 𝑥 is greater than one and less than or equal to four. As an interval then, this would have endpoints of one and four.

And now we need to consider the type of brackets or parentheses for each end. At the lower end, the sign is a strict inequality; 𝑥 is strictly greater than one. So the value one is not included in the solution set. And so, our interval is open at the lower end. However, at the upper end, we have a weak inequality, 𝑥 is less than or equal to four. So the value of four is included in the solution set, and our interval is closed at its upper end. And so, we have our answer to the problem. The solution set for this inequality is the interval from one to four, open at its lower end and closed at its upper end.

Now, notice that the steps involved in solving these two separate inequalities were exactly the thing. In each case, we subtracted four first and then divided by seven. So, in fact, there was no need for us to treat the two parts of this inequality separately. The second and probably more efficient approach, then, is to keep all three parts of the inequality together. In this case, we solve in exactly the same way but we must make sure that we perform the same operation to all three parts of the inequality. We begin by subtracting four from each part. 11 minus four is seven. Seven 𝑥 plus four minus four is seven 𝑥. And 32 minus four is 28. So we now have the statement seven 𝑥 is greater than seven and less than or equal to 28.

We then divide each part of our inequality by seven, giving the statement 𝑥 is greater than one and less than or equal to four, which we notice is identical to the solution we had using our first method. We would then express this in interval notation in exactly the same way. This method is certainly quicker, but we must make sure that we treat all three parts of the inequality identically. So if we’re subtracting four, we must make sure we do it from every part. And if we’re dividing by some number, in this case seven, again we need to do it to every part.

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