Solving Compound Inequalities
Compound inequalities are two or more inequalities joined by the word “and” or “or.” These first two compound inequalities are “and” compound inequalities, even though you don’t actually see the word “and.” They could be written as 19 is greater than or equal to negative 3x, plus 4, and negative 3x plus 4 is greater than or equal to 10, but often we see them written as one just like this.
This compound inequality is an “or” compound inequality, and you see it right there. This is really the only way you can write it; you can’t write an “or” compound inequality like you can these “and” compound inequalities, and they mean two very different things.
For an “and” compound inequality our solutions have to satisfy both inequalities, but for an “or” compound inequality our solutions only have to satisfy one inequality or the other one. The solutions don’t have to satisfy both. Let’s solve our first “and” compound inequality.
We’re going to solve it like we do regular inequalities or equations, meaning that we’re solving for x by undoing the operations around x. The first thing I need to do is subtract 4, and if we subtract 4 from the middle, we have to subtract 4 from our other two parts. 19 minus 4 is 15, is greater than or equal to negative 3x, is greater than or equal to 6.
Then we divide by negative 3, which means we have to divide our other two parts by negative 3. Since we’re dividing by a negative number, we must reverse our inequality sign from a “greater than or equal to” to a “less than or equal to.” 15 divided by negative 3 is negative 5, is less than or equal to x, which is less than or equal to negative 2.
We can show this answer graphically on our number line, (negative 5, negative 4, negative 3, negative 2, negative 1) so our solutions are the numbers that are greater than or equal to negative 5, and less than or equal to negative 2 at the same time. We can graph this a couple of different ways. I could graph these separately first, so I’ll show you that first.
This says that negative 5 is less than or equal to x, meaning that x is greater than or equal to, (so it’s all the numbers over here) “or equal to” means we fill in this circle. These are all the numbers that negative 5 is less than or equal to (all of this). Then we would graph x is less than or equal to negative 2.
We again need a closed circle on negative 2 because of the “or equal to,” and then we would shade all the numbers less than or equal to negative 2, (which is all of those numbers in that direction) but if you’ll remember the solutions to an “and” compound inequality have to satisfy both inequalities.
Which means the only part of the number line that should be shaded is where our inequalities overlap, so that’s between negative 5 and negative 2, (so you could go back and erase these other parts, they are not part of the solution) it’s just that segment of your number line between negative 5 and negative 2.
Looking at the way it’s written this really makes sense, your answers (where you’re shading) is between those two numbers, so that’s where we shaded. Let’s try another one. We want to solve for x, so the first thing we need to do is subtract 3, which means we’re subtracting 3 from all three parts of our inequality.
Now we work from left to right. 9 is greater than or equal to x divided by 2, which is greater than negative 1. Then, we multiply all three parts by 2. 2 times 9 is 18, is greater than or equal to x, which is greater than negative 2. True statement. 18 is greater than or equal to x, and that is greater than negative 2, but generally in math, we like to express things from least to greatest, so it’s usually a good idea to reverse this entire compound inequality.
Starting with negative 2 (which now has a reverse sign) is less than (notice the small point pointing at the negative 2, it still is) x, (open mouth to the x, still has the open mouth to the x) is less than or equal to (so that’s this sign here, still pointing at the x) 18. Now we can graph. It’s a pretty big spread, so on this one you may want to use a scale you might count by 5s (so negative 5, 0. 5, 10, 15, 20, so we can get all our numbers on there).
Again, your solutions are between these numbers, so we’re going to be shading a segment of the number line. You can graph the two inequalities separately first if you like, and then find where they overlap and shade that, or what I like to do is look at where my circles go, and what kind of circles they are and shade between my numbers.
Negative 2 is about here and since it’s only “less than” x, that means we should be using an open circle on negative 2, saying negative 2 is not an answer. Then x is less than or equal to 18, (so 18 would be about here) and on this we should be using a closed circle, since 18 is an answer, and then all of our solutions are between those two numbers.
To check this, pick any number in the shaded region and substitute it back in to your original compound inequality and it will make a true statement (same thing with the last one we just solved). That’s how you solve an “and” compound inequality, but what about an “or?”
Well an “or” is really just solving the same kind of inequality you would if that wasn’t there, but that you just graph both of them on the same number line. I’m going to solve this one first by adding 6. I have negative 4x is less than negative 12, then dividing by negative 4, which means we reverse our sign, so it is now greater than, we have x is greater than 3.
Bring down your “or,” and then solve the other inequality. Add 8 to both sides. We have 2x is less than negative 4. Divide both sides by 2, and we have x is less the negative 2. Now we would just graph both of our inequalities on the same number line (negative 3, negative 2, negative 1, 0, 1, 2, 3, 4, and we can throw one more number on there, 5).
I will graph this one first, x is greater than 3. Well, if it’s just “greater than,” that means that 3 is not a solution, so use an open circle on 3 to indicate that 3 is not a solution, and then shade the numbers that are greater than 3, which would be 4, 5, 6, etc. Then we graph x is less than negative 2.
Again, since it’s just a “less than” not a “less than or equal to,” that means negative 2 is not a solution, just the numbers that are less than negative 2 are, like negative 3, negative 4, negative 5, etc. This is a graphical representation of all the solutions for this compound inequality: 4, 5, 6, 3 and 1/2, negative 2 and 4 tenths, all those values that are shaded, but remember with an “or,” when you pick one of these numbers, it will only work in one or the other inequality—it won’t work in both—that’s the key with “or”s. Your solutions for your “and” compound inequalities have to work, (or make your whole inequality true) but in an “or” compound inequality your shaded solutions will only work in one of your inequalities, not in both.