When 4 times a number is subtracted from 5, the absolute value of...

Question

When 4 times a number is subtracted from 5, the absolute value of the difference is at most 13. Use interval notation to express the set of all numbers that satisfy this condition.

Accepted Answer

Hello everyone in this video. We're going to be looking at this practice problem where we want to find the solution set for this equation. That is described as the absolute value function of the sum of two times a number And three is at most six. So the first thing that we're going to want to do is to convert these words into an equation that we can solve the solution set for. So we already know that there's going to be an absolute value. And inside the absolute value there's going to be a sum of two times a number. So a number is an ambiguous part. So we'll say a number is X. And that's what we're trying to find. The solution set for the number which I've defined as X. So say two x plus three because inside the absolute value is the sum of two times the number, two times X. And three. So plus three. And the final part is that it is at most six, which means that six is the maximum value. So the absolute value of two X plus three has to be less than or equal to six. So now we have this absolute value inequality and we can sulfur it if we think about what's inside the absolute value, recall that the absolute value makes any negative number positive. So two X plus three could truly be positive. So I can write that as two X plus three is less than or equal to six but two X plus three could also be negative. So I could have negative two x plus three less than or equal to six. So in order to find the solution set for this equation I need to consider both of these inequalities. So I'll start with if two X plus three is positive and in order to find the solutions that I need to solve for the number which I've defined as X. So I need to solve for X. And I'll start by subtracting three. So I get two, X is less than or equal to positive three. Then I'll divide by two. So I get X is less than or equal to 3/2. So that solves my first inequality. And if I look at the second inequality you can see that I have this negative so all Divide by - on both sides in order to get rid of it. And because I'm dividing a negative across an inequality it changes directions from less than or equal to two greater than or equal to. Now I have two X plus three on the left hand side and negative six on the right hand side. So in order to solve for X or subtract three. So that leaves me with two, X is greater than or equal to negative nine and all divided by two. So it leaves me with X is greater than or equal to negative nine halfs. So now I have my second inequality solved. But in order to change this into interval notation, I'm going to draw a number line to visualize this better. So I'll say that this point here is negative 9/2, I will say that this is zero And this point here is 3/2. So this first interval is saying that X is less than or equal to negative two. So because it has that equal to, I know that I'm going to draw a bracket and X is less than 3/2, which means I need to draw a line going to the left and my second interval is saying X is greater than or equal to negative nine halfs. So because X is greater than negative nine halfs, I know that that is a lower limit. And again because there's an equal to there's a bracket And now X is greater than negative 9/2, which means I can go to the right. So my solution set for this function is where these two inequalities overlap because that's where the values for X are true for both of these inequalities. So if you look at where they overlap, you can see that it's in this section between negative nine halves and positive three halves. So that becomes my solution set. And if I write that in interval notation, my lower boundary is negative nine halfs all the way up to Positive 3/2 and again with the brackets because there's an equal to at both of those boundaries. So this becomes my final interval for the solution set of this problem and you can see that that aligns with answer choice B. So hopefully this video helped and see you next time.

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When 4 times a number is subtracted from 5, the absolute value of the difference is at most 13. Use interval notation to express the set of all numbers that satisfy this condition.

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