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Solve Linear Inequalities

Solve Linear Inequalities

Author: Colleen Atakpu

This lesson demonstrate how to solve linear inequalities.

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Today we're going to talk about solving inequalities. Solving an inequality is very similar to solving an equation except that your solution is going to be a range of values. So we'll start by reviewing the steps for solving an equation, and then what we do some examples solving inequalities.

So let's review the steps for solving an equation, and then compare it to the steps for solving an inequality. So if I want to solve this equation, I'm trying to isolate x. So I'm going to cancel out or undo my subtracting operation and my multiplying operation.

So I'm going to start by adding 8 to both sides to cancel out the subtracting operation, leaving me with 7x. And then 6 plus 8 is equal to 14. To cancel out my multiplying operation, I'm going to divide by 7. And I need to do it on both sides. 7's will cancel here, and I've isolated my x.

14 divided by 7 will give me 2. So x equals 2. So in an equation, 2 is the solution. And if we were to substitute 2 back into our original equation, it's going to give us a true statement. So 7 times 2 minus 8 should equal 6. 14 minus 8 is equal to 6.

So x equals 2 is my solution. We can see that that works. So if we were to solve an inequality you do it in the same way, following the same steps. We just use the inequality symbol instead of the equal sign.

So again, I'm going to start by adding 8 on both sides. This will give me 7x is less than 14. These cancel out. I'm going to divide by 7 on both sides. These will cancel.

So now I have x is less than 2. So now my solution for an inequality is a range of values. x can be anything less than, but not exactly equal to 2. So we can see how that works if we pick a value for x that less than 2 and substituting it back into our original inequality.

So I'm just going to go ahead and pick 1. So I would we have 7 times 1 minus 8 should be less than 6. 7 times 1 is 7 minus 8, and 7 minus 8 gives me a negative 1, which indeed is less than 6.

So we could pick any value that's, again, less than but not equal to 2. And it would work, if we substituted it back to our original inequality. Let's do some more examples.

So here's my first example. I'm going to solve this inequality and then show how to graph the solution on a number line. So again, I'm going to start by solving just as I would have regular equation. I'm going to subtract 6 from both sides to cancel out this adding operation.

Now I've got negative 3x is less than or equal to 18 minus 6, which is 12. Then I'm going to divide both sides by negative 3. This will cancel out my multiplying operation, and I'll be left with x and negative 4.

Now when you are dividing by a negative number, or if you multiply by a negative number, your inequality symbol is going to switch direction. So instead of being less than or equal to, since I divided by a negative number, it's going to change to greater than or equal to.

So again, whenever you divide or multiply by a negative number when you're solving an inequality, your symbol is going to switch direction. And that would also be the case if this had just been less than, this would switch to be greater than.

So now that I have my solution x is greater than negative 4, I can plot that on my number line. So I'm going to start at negative 4. And I know that I need to use a filled in circle because my solution set includes the value, the exact value of negative 4-- so a filled in circle.

And x can be anything greater than that. So I have my solution going towards the direction of numbers that are bigger than negative 4. So let's show that this solution does work with our original inequality.

So I have negative 3x plus 6 is less than or equal to 18. So if I pick a number that's within this range, then it should work when I substitute it back into my x. So let's just pick 0.

So negative 3 times 0 plus 6 should be less than or equal to 18. Negative 3 times 0 is just 0. And we see that 0 plus 6, which is just 6, is indeed less than 18. So 0 is in my solution set.

So let's take a number that's not in my solution set, and we should see that it does not work with our inequality. So I'm going to pick something to the left of negative 4. Let's do a negative 5, substitute that in to my original inequality for x.

And it should not be less than 18. This should not be a true statement. So negative 3 times negative 5 should be positive-- or is positive 15 plus 6-- 15 plus 6 is 21.

And we see that 21 is not less than or equal to 18, so this is not true. So we can see that negative 5, which is to the left of our solution set, is not in the solution set.

So for this example, we have something called a compound inequality. It's a compound inequality because it has more than 1 inequality symbol. So specifically, this compound inequality is telling us that the value of 2x plus 5 falls within the range although 11 to 15.

So let's see how to solve this, and then we'll graph the solution on the number line. So I'm going to again solve it as if I'm solving a regular equation or inequality by subtracting 5 to cancel out the adding. But anything I do in between my inequality symbols, I need to do on the other side of both inequality symbols.

So I'm going to subtract 5 from my 15 and from my 11. So here they will cancel, and we'll be left with 2x. 15 minus 5 is 10, and 11 minus 5 gives me 6.

I'm going to divide by 2 to cancel out my multiplying operation. And again, I need to divide by 2 on the other side of both inequality symbols. So I'm left with x in the middle. 10 divided by 2 is 5. And 6 divided by 2 gives me 3.

So that tells me that my solution for x can fall between the ranges including 3 and 5, and then anything in between. So when we graph that on a number line, we're going to use open circles-- or, sorry, closed circles over the 3 and the 5. And we want to show that it can also be anything between 3 and 5.

So let's see if that solution works by picking values within our range and outside of our range, and substituting it back into our original compound inequality. First let's pick something within our range. So if I substitute 4 back into my inequality, 2 times 4 plus 5 should be in between 11 and 15.

So 2 times 4 gives me 8 plus 5. 8 plus 5 gives me 13. So we can see that yes, 13 does fall within the range of 11 to 15. Let's pick something outside of our range.

So let's do negative 2. So I've got it 2 times negative 2 plus 5. And since it's outside of our range, this should not be between the values of 11 and 15. So let's simplify. 2 times negative 2 will give me a negative 4. Negative 4 plus 5 gives me 1.

And I can see that 1 does not fall within the range of 11 to 15. So we can see that, no, 1 is not a solution to our set because it does not satisfy our compound inequality.

Let's go over our key points from today. Solving inequalities is similar to solving equations, except that we use an inequality symbol instead of an equal sign. And we saw that when you're solving an inequality, and you multiply or divide by a negative number, your inequality symbol is going to switch directions.

And then when you're solving a compound inequality, any operation done between the inequality symbols must be done on the other side of both inequality symbols. So I hope that these key points and examples helped you understand a little bit more about solving inequalities. Keep practicing and using your notes, and soon you'll be a pro. Thanks for watching.