In Exercises 95–102, use interval notation to represent all value...

Question

In Exercises 95–102, use interval notation to represent all values of x satisfying the given conditions.

y1 = (2/3)(6x - 9) + 4, y2 = 5x + 1, and y1 > y2

Accepted Answer

Hello. Everyone in this video we're going to be looking at this practice problem where we want to find all the values of X that satisfy the condition. Z one minus two is less than or equal to Z two. We want to write that in interval notation and we're also given that Z one is equal to X plus one and Z two is equal to X divided by two plus four. So in this case you can see that the inequality Z one minus two less than or equal to Z two tells us nothing about the range of X values that will satisfy this condition. So that's where we need to take into consideration one. Z 1 and Z two are equal to, so Z one is equal to X plus one and Z two is equal to X divided by two plus four. So in this case if we plug in the values of Z one and Z two into our inequality, we're going to form an inequality that will result in a range of X values and that range of X values satisfies the conditions. E one minus two less than or equal to Z two because that's what we're plugging in. So if I do that I have Z one minus two comes X plus one minus two less than or equal to Z two. So X divided by two plus four. And you can see here that I haven't changed the inequality, I've only substituted in one, Z one and Z two are equal to. So if I solve this inequality for X, it's going to give me a range of X values that satisfy the conditions. E one minus two less than or equal to Z two, which is what we want. So I'm going to solve this inequality for X. Trying to get X all the X values on one side. So this left hand side X plus two minus two becomes X minus one less than or equal to X divided by two plus four. Fine subtract four. I have X minus five less than or equal to X, divided by two. Multiply both sides by two. I have two X minus 10 less than or equal to X. And then I can move the two X over. So I'm off with negative 10 less than or equal to negative X. And I'm going to divide by a negative. So my inequality switches signs And I'm left with greater than or equal to X. And if I rearrange that Then I have X is less than or equal to 10. And now you can see that we have this inequality that is telling us X has to be less than or equal to 10. So if I had a number line and say this was negative five, this was 05 and 10. This is saying that any value below 10 will satisfy the condition C1 -2 less than or equal to Z two. I noticed how I drew a bracket here because there is an equal to. So now I'm going to convert this inequality into interval notation, where I have my upper bound of 10 with the bracket because it's equal to or less than or equal to. And there is no lower limit for the X. So I can write negative infinity as my lower limit. And this becomes the interval notation for the range of values that satisfy the condition. And you can see that this aligns with answer choice A. So hopefully this video helped and see you next time.

In Exercises 95–102, use interval notation to represent all values of x satisfying the given conditions. y1 = (2/3)(6x - 9) + 4, y2 = 5x + 1, and y1 > y2

Key Concepts

Interval Notation

Inequalities

Linear Functions

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