Video: Solving Absolute Value Inequalities Algebraically

Find algebraically the solution set of the inequality 1/|4π‘₯ βˆ’ 9| >17.

08:04

Video Transcript

Find algebraically the solution set of the inequality one over the absolute value of four π‘₯ minus nine is greater than 17.

First, let’s look closely at the inequality. Here we notice that we have an π‘₯ in the denominator. And that means there is one place where π‘₯ will not exist. We cannot have a zero in the denominator. And that means four π‘₯ minus nine cannot equal zero.

We want to first find out what π‘₯ would be to make this statement zero. So we solve for π‘₯. We add nine to both sides of the equation. On the left, it cancels out, leaving us with four π‘₯ cannot equal nine. We still need to isolate π‘₯, and we do that by dividing by four on both sides. Four divided by four is one, leaving us with π‘₯ cannot equal nine-fourths. There is no solution at π‘₯ equals nine-fourths.

But now we need to find out where there is a solution. And to do that, we need to think about what we know about absolute value. When we solve an inequality with absolute value, we deal with two cases. The positive case: one over four π‘₯ minus nine is greater than 17. And the negative case: one over negative four π‘₯ minus nine is greater than 17.

Let’s start with our positive case. Our goal is to isolate π‘₯. And to do that, we’ll need to get it out of the denominator. We multiply both sides of our inequality by four π‘₯ minus nine over one. On the left side of our equation, these two things cancel out, leaving us with one is greater than 17 times four π‘₯ minus nine.

To get rid of our parentheses, we’ll distribute our 17. 17 times four π‘₯ equals 68π‘₯. 17 times negative nine equals negative 153. Bring down our inequality symbol. We now have one greater than 68π‘₯ minus 153.

Continuing as we try to isolate π‘₯, we add 153 to the right side and the left side. Negative 153 plus positive 153 cancels each other out. They’re equal to zero. One plus 153 is 154. 154 is greater than 68π‘₯. We can isolate π‘₯ by dividing the right side by 68. To keep everything equal, we have to also divide the left by 68. 68 divided by 68 equals one. 154 over 68 is greater than π‘₯.

I noticed that both the numerator and the denominator here are even, which means we can simplify this by dividing the numerator and the denominator by two. 154 divided by two equals 77. 68 divided by two equals 34. 77 over 34 is greater than π‘₯. Or in the more common format, π‘₯ is less than 77 over 34. And this is one-half of the solution set.

We now need to look at the negative case. The first thing we wanna do is distribute this negative to both the four π‘₯ and the negative nine. If I do that, I end up with negative four π‘₯ plus nine is greater than 17.

Again, we have an π‘₯ in the denominator, and we want to get it out of the denominator. We can do that by multiplying its reciprocal on the left and the right. On the left side, negative four π‘₯ plus nine over one cancels out one over negative four π‘₯ plus nine. We’re left with just one. One is greater than 17 times negative four π‘₯ plus nine. We distribute our 17. 17 times negative four π‘₯ equals negative 68π‘₯. 17 times nine equals positive 153.

Our inequality now says one is greater than negative 68π‘₯ plus 153. We subtract 153 from both sides. Positive 153 minus 153 equals zero. One minus 153 equals negative 152, which is greater than negative 68π‘₯.

In our last step to isolate π‘₯, we divide the right side by negative 68 and we divide the left side by negative 68. And here is our key moment. When we divide by negatives and we have inequalities, we must flip the sign. We divided both sides by a negative number. Therefore, we have to flip the inequality symbol.

On the left, we have negative 152 over negative 68. A negative over a negative would give us a positive number. Negative 68 over negative 68 cancels out, leaving us with π‘₯. 152 and 68 are both divisible by four. If we divide 152 by four, we get 38. If we divide 68 by four, we get 17. 38 over 17 is less than π‘₯. Again, we might want to write this in its more common format. π‘₯ is greater than 38 over 17.

Now how do we take these three pieces of information and put them together? Looking at the inequality symbol, we see that we’re dealing with less than and greater than, but not equal to. And that means the brackets we’ll use will face outward. π‘₯ must be greater than 38 over 17 and less than 77 over 34.

What we’ve said so far, if you imagine on a number line, we’re saying that π‘₯ cannot be equal to 38 over 17, but everything between 38 over 17 and 77 over 34. However, we need to exclude nine over four. There’s a hole in this domain at nine-fourths. To exclude nine-fourths, we can say minus nine-fourths. π‘₯ falls between thirty-eight seventeenths and seventy-seven thirty-fourths, with the exception of nine-fourths.

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