Absolute value equations
Absolute Inequalities 2 Absolute Inequalities 2
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- We have the absolute value of 2r minus 3 and 1/4 is less
- than 2 and 1/2, and we want to solve for r.
- So right from the get go we have to deal with this
- absolute value.
- And just as a bit of a review, if I were to say that the
- absolute value of x is less than, well let's just say,
- less than 2 and 1/2, that means that the distance from x
- to 0 is less than 2 and 1/2.
- That means that x would have to be less than 2 and 1/2, and
- x would have to be greater than negative 2 and 1/2.
- And just think about it for a second.
- If I were to draw it on a number line right here, that
- is 0, that is 2 and 1/2, and that is negative 2 and 1/2.
- These two numbers are exactly 2 and 1/2 away from 0, because
- both of their absolute values is 2 and 1/2.
- Now, if we want all of the numbers whose absolute value
- is less than 2 and 1/2, or that are less than 2 and 1/2
- away from 0, it would be all of the numbers in between.
- And that's exactly what these two statements are saying. x
- has to be less than 2 and 1/2, and it has to be greater than
- negative 2 and 1/2.
- If this absolute value were the other way, that the
- absolute value of x has to be greater than 2 and 1/2, then
- it would be the numbers outside of this, and
- it would be an or.
- But we're dealing with the less than situation right
- there, so let's just do what we were able to figure out
- when it was just an x.
- The distance from this thing to 0 has to be less than 2 and
- 1/2, so we can write that 2r minus 3 and 1/4 has to be less
- than 2 and 1/2 and 2r minus 3 and 1/4 has to be greater than
- negative 2 and 1/2.
- Same exact reasoning here.
- Let me draw a line so we don't get confused.
- Same exact reasoning here.
- This quantity right here has to be between
- negative 2 and 1/2.
- It has to be greater than negative 2
- and 1/2 right there.
- And it has to be less than 2 and 1/2, so
- that's all I wrote there.
- So let's solve each of these independently.
- Well, this first went over here, you've learned before
- that I don't like improper fractions, and I don't like
- fractions in general.
- So let's make all of these fractions.
- Sorry, I don't like mixed numbers.
- I want them to be improper fractions.
- So let's turn all of these into improper fractions.
- So if I were to rewrite it, we get 2r minus 3 and 1/4 is the
- same thing as 3 times 4 is 12, plus 1 is 13.
- 2r minus 13/4 is less than-- 2 times 2 is 4, plus 1 is five--
- is less than 5/2.
- So that's the first equation.
- And then the second question-- and do the same thing here--
- we have 2r minus 13 over 4 has to be a greater
- than negative 5/2.
- All right, now let's solve each of these independently.
- To get rid of the fractions, the easiest thing to do is to
- multiply both sides of this equation by 4.
- That'll eliminate all of the fractions, so let's do that.
- Let's multiply-- let me scroll to the left a little bit--
- let's multiply both sides of this equation by 4.
- And what do we get?
- 4 times 2r is 8r, 4 times negative 13 over 4 is negative
- 13, is less than-- and I multiplied by a positive
- number so I didn't have to worry about swapping the
- inequality-- is less than 5/2 times 4 is 10, right?
- You get a 2 and a 1, it's 10.
- So you get 8r minus 13 is less than 10.
- Now we can add 13 to both sides of this equation so that
- we get rid of it on the left-hand side.
- Add 13 to both sides and we get 8r-- these guys cancel
- out-- is less than 23, and then we divide
- both sides by 8.
- And once again, we didn't have to worry about the inequality
- because we're dividing by a positive number.
- And we get r is less than 23 over 8.
- Or, if you want to write that as a mixed number, r is less
- than-- what is that-- 2 and 7/8.
- So that's one condition, but we still have to worry about
- this other condition.
- There was an and right here.
- Let's worry about it.
- So our other condition tells us 2r minus 13 over 4 has to
- be greater than negative 5/2.
- Let's multiply both sides of this equation by 4.
- So 4 times 2r is 8r.
- 4 times negative 13 over 4 is negative 13, is greater than
- negative 5/2 times 4 is negative 10.
- Now we add 13 to both sides of this equation.
- The left-hand side-- these guys cancel out, you're just
- left with 8r-- is greater than negative 10 plus 13 is 3.
- Or divide both sides of this by 8, and you're left with r
- has to be greater than 3/8.
- So our two conditions, r has to be less than 2 and 7/8 and
- greater than 3/8.
- Or we can just write it like this: r is greater than 3/8,
- so it's greater than-- maybe I should say 3/8 is less than r,
- which is less than 2 and 7/8.
- So if we were to plot the solution on the number line--
- which I'm about to do, so that's my number line-- this
- is 0 right here, maybe this is 1, 2, and 3.
- We have 2 and 7/8.
- We have to be less than 2 and 7/8.
- Let's say that this is 2 and 7/8 right there.
- And we have to be greater than 3/8.
- Say that is 3/8, so 3/8 will be some place
- right around there.
- And everything in between is a valid solution.
- And we could try it out.
- Let's try out something that, based on what I just drew,
- should be a valid solution.
- 1 should be a valid solution.
- Let's try it out here.
- 2 times 1 minus 3 and 1/4, what is that?
- That's 2 minus 3 and 1/4.
- And so what is that?
- 2 minus 3 and 1/4 is-- well, 3 and 1/4 minus 2 is 1 and 1/4,
- so this will be negative 1 and 1/4.
- But we're taking the absolute value of it, so we take the
- absolute value of it, which is equal to 1 and 1/4, which is
- indeed less than 2 and 1/2.
- Now let's try another number.
- Let's try 0.
- Based on this, 0 should not work.
- So what happens if we put 0 here?
- You get 2 times 0, which is 0, minus 3 and 1/4.
- If you take the absolute value of negative 3 and 1/4, you'll
- get positive 3 and 1/4, which won't work.
- 3 and 1/4 is greater than 2 and 1/2, so that's
- true, that works out.
- And same thing for 3.
- 2 times 3 is 6, minus 3 and 1/4 is 2 and 3/4.
- Take the absolute value, it's 2 and 3/4, still bigger than 2
- and 1/2, so it won't work.
- So at least the points that we tried out seem to validate
- this solution that we got.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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