Solve each inequality. Give the solution set in interval notation...

Question

Solve each inequality. Give the solution set in interval notation. | (2/3)x + 1/2 | ≤ 1/6

Accepted Answer

Hello everyone in this video. We're going to be looking at this practice problem where we want to express the solution set for the absolute value inequality, absolute value of 3/5 X plus three halfs less than or equal to 3/10. So in this case because we're working with an absolute value, we want to think about what the absolute value is doing to be values inside of it. Which in this case would be 3/5 X plus three halfs. So recall that an absolute value means that the components inside the absolute value could be positive. So I could have positive 3/5 X plus three halfs less than or equal to 3/ but they could also be negative. So I could have negative 3/5 X plus three halfs less than or equal to 3/10 with the negative on the outside because recall that it would apply to the whole term inside of the absolute value. So in order to find the solution set, I'm trying to find the values of X that make both of these inequalities true. So I need to solve for both of these scenarios. So I'm going to start with if the number inside the absolute value was positive And I need to isolate X. So I'll subtract 3/2 from both sides. So it leaves me with 3/5 X less than or equal to 3/ -3/2. Well I have two fractions on the right hand side. So in order to combine them I need to find a common denominator. So I can get three halves to get a denominator of 10 by multiplying by five Over five. So this becomes 15/10. So I'll rewrite 3/2 as 15/10. And now I have 3/5 x less than or equal to 3/10 -15/10 is negative 10th. And in order to get rid of the 3/5, I'll multiply by 5/3. Can see the three cancels out and the five cancels out. But what I do to one side, I need to do to the other. So this becomes X is less than or equal to negative 12 times five is negative 60 and 10 times three is 30 I can write this as X is less than or equal to negative two. So now I've found the or I simplified and isolated X. Or if the number was positive and I can deal with if the number was negative and you can see that if the number was negative, I have this negative on the outside So I can divide by -1. And if I divide a negative across the inequality, the inequality changes direction. So instead of being less than I will now have greater than or equal to. So I have 3/5 X plus three halves greater than or equal to negative three tents. And once again I need to isolate X. I'll subtract three halves from both sides. So it leaves me with 3/5 X greater than or equal to negative 3/10 minus three halfs. And recall that we said that three houses equal to 15 tents. So in order to combine the fractions I'm gonna go ahead and plug in 15 tents for 3/2. So now I have 3/5 X greater than or equal to negative 3/ minus 15 10th is negative tents. And once again to completely isolate X. I need to multiply by 5/3. So that completely cancels the 3/5 on the left hand side. And that leaves me with X is greater than or equal to negative 18 times five, which is negative 90 divided by 10 times three, which is 30. So X is greater than or equal to negative three. So now I've solved both of the possible scenarios and I need to turn these inequalities into an interval. So in order to simplify writing this in an interval notation, I'm going to go ahead and visualize it on a number line. I'll say that this is negative three and this is negative two. So my first interval Is X is less than or equal to -2, which means that -2 is a number boundary and X is less than or equal to. So I'll draw a line going to the left and my second interval X is greater than or equal to negative three is seeing that negative three is the lower boundary. So I'll draw again a bracket there because it's an equal to but now X is going to the right because X is greater than negative three. So I'll draw an arrow going or a line going to the right. And my solution set is where these two inequalities overlap. So you can see that that portion is here in between negative three and negative two. So if I write that in interval notation, I have that lower boundary at negative three And the upper boundary -2. So my interval for the solution set is Brockett -3 2 -2 Bracket. And if we look at the answer choices, you can see that C also has that same interval. So hopefully this video helped and see you next time.

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Solve each inequality. Give the solution set in interval notation. | (2/3)x + 1/2 | ≤ 1/6

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