Solve Compound Inequalities - Or Statements - Overview
Solve Inequalities Using Addition and Subtraction - Example 1
Solve Inequalities Using Addition and Subtraction - Example 2
Solve Inequalities Using Addition and Subtraction - Example 3
Solve Inequalities Using Addition and Subtraction - Example 4
Syracuse University
Solve Compound Inequalities - Or Statements - Example 4


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Video Transcript

So in this problem, we're being asked to solve the giving compound inequality. And then we're gonna graph the solution set on the number line. So remember to solve the compound inequality that for statements that include the word or we're going to solve both inequality separately. So let's start with the first one you have X minus five is less than negative three. So the solve for X, we're gonna add five to both sides of our inequality. Well, negative three plus five is positive too. So we're left with X is less than two Now. We'll solve our second inequality. We have two X is greater than negative 10. So to solve for X, we're going to divide both sides of our inequality by two. So we have X is greater than well, negative 10 divided by two is negative. Five perfect. So now we have solved our compound inequality. We have X is less than negative. Two or X is greater than negative five. So the next thing we need to dio is graft this on the number line. So let's set up our number line. Remember, we're gonna put our key values here, so we have to a negative five, but because negative five smaller that needs to go first. And remember for graphing compound inequalities that involve the words. Or we can simply just graft both inequality separately, just on the same number line. So let's start with our first one. It says that X is less than two. Well, that means to is not a part of the solution set. So that means we have an open circle it too. And values of X that our lesson to would be to the left. So we're going to shade the entire number line to the left off to because all of these values are less than two. Therefore, they should be included in the solution set. Okay, great. Now we want a graph. The X is greater than negative five. So typically, we would start by putting an open circle of negative five because negative five would not be a solution to this. However, take a look at the shaded area we already have. Negative five is included in this first inequality because negative five is less than two. So we're not gonna actually put a circle. There were just gonna start our shading here because, really, if we had an open circle, it would have filled up. Now, where are where are our values of X that are greater than negative? Five? Well, it would be to the right of negative five. So that means we have to shade all the way past two because we're talking about any value of X that's greater than negative five. Well, let's think about it. What's gonna happen to that open circle It too. It's now going to close because two does make the second inequality. True, because two is greater than negative five. We'll take a look here. The entire number line is shaded. Well, what does that mean? Well, what that means is our solution set is all real numbers, meaning that no matter what value for X, you plug in to our compound and equality for either of those inequalities, it will make a true statement and think about it. If X is less than two, that includes all of your negative values, including zero, or if X is greater than negative five, that's gonna include all your positive values. So we've now said we're gonna include all of our negative values, all of our positive values and zero well, that is all of our real numbers, which is why that is the solution to this compound in the quality. So, like I said, you might not recognize that this is a special case at first. But the graph compound inequality statements that include the word or if you just grab both inequality separately, you will be able to pick out the special case is much simpler.