Conceptualizing this sum of "absolute sums" as a combined distance makes a lot of sense! It's much more satisfying, and easier, than breaking the equation up in three cases and doing the algebra (as I did before watching the video). Very neat!
I've been there before, plastic eraser holder and the piece of felt that is attached with a sticky tape/glue to it over time just comes off. That's why I bring my own eraser to classrooms, I can't trust any other teacher and their sloppy ways of ruining it and more importantly... not replacing it when it's broken.
i thought you meant that the solutions aren't all real numbers so i was just patiently waiting for a complex solution 😂 my brain wasn't braining edit: tHaNkS fOr 1 LiKe!1!1!1!1!1!1!1!1!1!1!1!1!1!1!1!1!1
@@windowsxpmemesandstufflol oh i forgot absolute value is the modulus of the complex number, well that way there must be complex solutions tho, in fact infinitely many
At the beginning of the video I was thinking like the solutions were complex or something like this, but after I think he kinda "opened" my mind and I remained shocked about this way to understand better absolute value
When you said “it’s not all real numbers”, I thought you meant complex solutions as well. Thinking about the standard norm in the complex plane, I saw that all solutions form an ellipse around the foci -1 and 5 (flat line since the sum of distances is 6). Looking at the comments, I’m not the only one to have noticed this, but still thought it was really neat.
This reminds me of the property of an ellipse. If the right-hand side of the equation is something more than 6, the solution forms an ellipse with its focus at -1 and 5 respectively, where the vertical axis is the imaginary number axis.
I check wit intervals, the interval (-∞,-1) both absoluts values their negative inside and gives x=-1 but -1 its not on (-∞,-1) (Actually you can include -1 since 0=-0 but well I realized later) Then [-1,5) And you get the negative inside from x-5 but the positive from x+1 You get -x+5+x+1=6 that give us 6=6 so, al the values between [-1,5) its an answer (in this part was when I realized that 0=-0) And the las interval was [5,∞) That make the both positives than give us x=5 So [-1,5)U[5]=[-1,5] I didn't check if there are complex solutions
Solution: The edge cases are x < 5 and x < -1 Case 1: x ≥ 5 x - 5 + x + 1 = 6 2x - 4 = 6 |+4 2x = 10 |:2 x = 5 Since x = 5 and x ≥ 5 works, it is a valid solution, but ONLY x = 5. Case 2: -1 ≤ x < 5 -(x - 5) + x + 1 = 6 -x + 5 + x + 1 = 6 6 = 6 So everything inside -1 ≤ x < 5 is a valid solution Case 3: x < -1 -(x - 5) + -(x + 1) = 6 -x + 5 - x - 1 = 6 -2x + 4 = 6 |-4 -2x = 2 |:-2 x = -1 Since x = -1 and x < -1 is a contradiction, x can not be below -1 Putting the valid results of case 1 and 2 together, we end up with -1 ≤ x ≤ 5 or x = [-1, 5]
Apply triangle inequality to LHS => |2x-4| \leq 6 => -6 \leq 2x -4 \leq 6 => -2 \leq 2x \leq 10 => -1 \leq x \leq 5. To verify these are the only solutions, by way of contradiction suppose x ot \in [-1,5] and |x-5| + |x+1| =6. Then either x>5 or x5 => |x-5|>0 and |x+1|>6 =>adding these 2 we get |x-5| + |x+1|>6 implying |x-5| + |x+1| =\=6, a contradiction. Case 2: x x+10 =>x-56 => adding these 2 we get |x-5| + |x+1|>6 implying |x-5| + |x+1| =\=6, a contradiction.
@@PranitSuman Kind of need to memorize some of the concepts. The why is the more important thing because the result can be generated from the why. For example, the absolute value is just the distance from 0. It's why the function was made. The addition is just moving the function left and right, which is what happens in every function when there is an addition to a particular variable. If you add it outside the function, then you move the graph up and down. Multiplication of the variable just shrinks or extends the graph.
if you want complex solutions, it's an ellipse and the equation is already in the canonical form. if you're some kind of real number degenerate, it's an extremely simple piecewise function
It seems to me that it was easier to explain through a graph, if you know what a graph looks like, these are functions, and then the problem is solved orally!
You look at the cases for the expressions inside the absolute values. Either they both are negative, both are positive, or one is positive, and the other is negative. In the case of both negative, x - 5 < 0 and x + 1 0 x - 5 + x + 1 = 6 => 2x - 4 = 6 => 2x = 10 => x = 5. x - 5 >= 0 and x +1 > 0 with x = 5, valid solution. In the case of one is positive and the other is negative, x -5 < 0 and x + 1 > 0 (Accidentally wrote the wrong symbol, but I corrected it below.) -x + 5 + x + 1 = 6 6 = 6, valid for all x in the interval x < 5 and x > -1 Put it all together -1
In the general case one of the intervals might have produced an equation where there were two constants that were unequal. That just eliminates that interval producing any solution at all or the number isn't in the range, and therefore, it equally is not a solution.
You can also do the steps mentioned in the video till 3:29 then prove why all answers belongs between -1 and 5 by intuition. Basically the gap between -1 and 5 is 6, so if x is somewhere between -1 and 5, it would be akin to breaking the line from -1 to 5 in two parts; from -1 to x and x to 5 so the total distance of both would obviously be 6. Since the distance from -1 to 5 is 6, the distance from -1 to an x greater than 5 would naturally be greater than 6, and since the distance from 5 to -1 is 6, the distance from 5 to a number less than -1 would be greater than six Technically it’s the same method as the video, but we’re eliminating ranges instead of proving the same via induction. You can try plugging 5.01 and 4.99 in the equation and look at the equation made as a way to see why mathematically the numbers work out as you move the numbers up and down. Maybe the confusion is happening because the integer value that were chosen doesn’t make it very clear why it would be the same for all numbers in that range.
What makes you think this is not a mathematical way of working it out? It's a geometric way of solving an algebraic problem. If solving a geometric problem with algebra is a mathematical approach, then why going the other way shouldn't be? You have two points on a line distance 6 apart and you are asked to find all points such that the sum of distances from this point to two given points is also 6. This is only possible if you are between the two points. This can be proven in a number of ways, but more often than not it's an axiom in geometry that for any three points A, B and P, where d(A,B) denotes the distance between A and B d(A,P)+d(B,P)=d(A,B) iff P is between A and B. It's on par with the triangle inequality, giving a condition for equality. Depending on your choice of axioms it can be a theorem, but there's nothing not mathematical in this approach IMO.
if you don't hate yourself, you render the absolute values into piecewise functions. if you do hate yourself (equivalently, if you love algebra), you treat the absolute values as square roots
If you plot the sum of the two absolute sums you'll get a graph that looks like the silhouette of a truncated cone, or a V but instead of the point in the bottom you have a line, or this ⊔ but instead of right angles they are greater than 90°. Equating this with a specific value (as in |x-5| + |x+1| = 6) means looking for where this graph intersects with a line parallel to the x-axis. What you see very neatly from this graph is that this lower line is a segment of the line y=6. That's why you get infinitely many solutions, all numbers between -1 and 5. If you "pull up" the line parallel to the x-axis, you'll get two points intersecting with that graph, in equal distances from -1 and 5. For the value 7, as in your example, it will be at an additional distance of 0.5 in both directions, i.e. -1-0.5 (i.e. -1.5) on the left, and 5+0.5 (i.e. 5.5) on the right. If however you "pull down" the line parallel to the x-axis below y=6, then this line will not intersect with the graph and there are no solutions at all (e.g. no solutions for |x-5| + |x+1| = 3). Like I said, if you plot the graphs, you'll easily see all of this. It's a lot more straightforward to see than me describing the plot.
In this case you can split your solution into several cases, which solves the module and lets you turn this into trivial linear equation. In your example cases are 1) -inf < x ≤ -1; 2) -1 ≤ x ≤ 5 and 3) 5 ≤ x < +inf. After you solve each case make sure the solution satisfies initial condition. First case yields solution x = -1.5, second case has no solutions and third case yields solution x = 5.5. Solutions for cases 1) and 3) satisfy their own initial conditions, so final answer is x = -1.5 and x = 5.5.
Still works, but now there will only be two answers: 5.5 and -1.5. The idea is still the same, the distance between -1 and 5 is 6 and for every number between those two the summed distances cover the interval from -1 to 5, however, every number outside [-1,5] covers [-1,5] only once and then covers the interval from the number to the closest end of [-1,5] twice and this double covering should be 7-6=1, so it's two numbers (in the case of real numbers) distance 0.5 from -1 and 5 outside [-1,5], giving solutions -1.5 and 5.5. It also works for complex numbers as well, but there are infinitely many solutions, all lying on an ellipse with foci -1 and 5 by the definition of an ellipse. That also shows that for any d smaller than 6 |x-5|+|x+1|=d there are no solutions real or complex. As for the sum of three absolute values of this form or scaled absolute values... It becomes quite a mess really quickly. And that is not taking into account expressions not of the form x-a inside the absolute values. At this point doing something other than considering cases doesn't really work.
|x-5| + |x+1| = 6 First, we find the roots x-5 = 0 x+1=0 x = 5 x = -1 Then we draw the signs table -inf -1 5 +inf x-5 - - 0 + x+1 - 0 + + If x € ]-inf;-1[ : |x-5| + |x+1| = 6 (-x + 5) + (-x - 1) = 6 -x + 5 - x - 1 = 6 -2x + 4 = 6 x = -1 unacceptable If x € ]-1;5[: |x-5| + |x+1| = 6 -x + 5 + x + 1 = 6 0x = 0 all possible x € ]-1;5[ If x € ]5;+inf[ |x-5| + |x+1| = 6 x - 5 + x + 1 = 6 2x = 10 x = 5 unacceptable If x = -1 6 = 6 so x = -1 is a solution If x = 5 6 = 6 so x = 5 is a solution x € [-1;5]
Can someone solve this for me There are 3 kinds of bananas red, green, and blue. For every 4500 red bananas sold you get 1500 dollars, for every 3000 green bananas sold you get 1000 dollars, and for every 2500 blue bananas sold you get 2500 dollars. Now, for every 100k bananas sold you'll gain 1% more money(multiplicative). If I want to sell 50 million bananas of each kind, which banana color gives me more money?
Conceptualizing this sum of "absolute sums" as a combined distance makes a lot of sense!
It's much more satisfying, and easier, than breaking the equation up in three cases and doing the algebra (as I did before watching the video).
Very neat!
So [-1, 5]
What a fun solution!
2:59 nice eraser
I've been there before, plastic eraser holder and the piece of felt that is attached with a sticky tape/glue to it over time just comes off. That's why I bring my own eraser to classrooms, I can't trust any other teacher and their sloppy ways of ruining it and more importantly... not replacing it when it's broken.
I'm not going to lie, as soon as you said "the distance between" I thought you were going to start invoking complex numbers as part of the solution.
same. especially when he brought up "not all real solutions"
i thought you meant that the solutions aren't all real numbers so i was just patiently waiting for a complex solution 😂
my brain wasn't braining
edit: tHaNkS fOr 1 LiKe!1!1!1!1!1!1!1!1!1!1!1!1!1!1!1!1!1
how would there be complex solutions without a x^2💀
@@Yilmaz4I think the x terms are in absolute values
@@windowsxpmemesandstufflol oh i forgot absolute value is the modulus of the complex number, well that way there must be complex solutions tho, in fact infinitely many
Plot the complex number on the y-axis of the number line. It is obvious that any point in complex plane off the number line will total greater than 6
same
At the beginning of the video I was thinking like the solutions were complex or something like this, but after I think he kinda "opened" my mind and I remained shocked about this way to understand better absolute value
When you said “it’s not all real numbers”, I thought you meant complex solutions as well. Thinking about the standard norm in the complex plane, I saw that all solutions form an ellipse around the foci -1 and 5 (flat line since the sum of distances is 6). Looking at the comments, I’m not the only one to have noticed this, but still thought it was really neat.
When you can't understand the problem so he lets you visualise it. Best math teacher❤.
I really love your way of teaching.
You can also make use of the triangle inequality to get the same result.
OMG I FINALLY UNDERSTOOD WHAT I WASNT ABLE TO FOR LIKE A WEEK TYSMM
Great video. Also how many expo markers do you go through a day?
This reminds me of the property of an ellipse. If the right-hand side of the equation is something more than 6, the solution forms an ellipse with its focus at -1 and 5 respectively, where the vertical axis is the imaginary number axis.
I check wit intervals, the interval (-∞,-1) both absoluts values their negative inside and gives x=-1 but -1 its not on (-∞,-1) (Actually you can include -1 since 0=-0 but well I realized later)
Then [-1,5)
And you get the negative inside from x-5 but the positive from x+1
You get -x+5+x+1=6 that give us
6=6 so, al the values between [-1,5) its an answer (in this part was when I realized that 0=-0)
And the las interval was [5,∞)
That make the both positives than give us
x=5
So [-1,5)U[5]=[-1,5]
I didn't check if there are complex solutions
Solution:
The edge cases are x < 5 and x < -1
Case 1: x ≥ 5
x - 5 + x + 1 = 6
2x - 4 = 6 |+4
2x = 10 |:2
x = 5
Since x = 5 and x ≥ 5 works, it is a valid solution, but ONLY x = 5.
Case 2: -1 ≤ x < 5
-(x - 5) + x + 1 = 6
-x + 5 + x + 1 = 6
6 = 6
So everything inside -1 ≤ x < 5 is a valid solution
Case 3: x < -1
-(x - 5) + -(x + 1) = 6
-x + 5 - x - 1 = 6
-2x + 4 = 6 |-4
-2x = 2 |:-2
x = -1
Since x = -1 and x < -1 is a contradiction, x can not be below -1
Putting the valid results of case 1 and 2 together, we end up with -1 ≤ x ≤ 5 or x = [-1, 5]
Thank you very much for this, I was trying to think of a way to do it algebraically.
Apply triangle inequality to LHS => |2x-4| \leq 6 => -6 \leq 2x -4 \leq 6 => -2 \leq 2x \leq 10 => -1 \leq x \leq 5.
To verify these are the only solutions, by way of contradiction suppose x
ot \in [-1,5] and |x-5| + |x+1| =6. Then either x>5 or x5
=> |x-5|>0 and |x+1|>6
=>adding these 2 we get |x-5| + |x+1|>6 implying |x-5| + |x+1| =\=6, a contradiction.
Case 2: x x+10
=>x-56
=> adding these 2 we get |x-5| + |x+1|>6 implying |x-5| + |x+1| =\=6, a contradiction.
Can we say the solution of
|x-5|+|x+1|>6 is R-[-1,5]
And
|x-5|+|x+1|
Yes, you would be right.
If |x - a| is distance between x and a, then what is |x + a|? What is just |x|?
@@PranitSuman He showed it was the distance between x and -a. |x| is the distance between x and 0.
I am not understanding maths after grade 10 I am just 'memorising' maths
@@PranitSuman Kind of need to memorize some of the concepts. The why is the more important thing because the result can be generated from the why.
For example, the absolute value is just the distance from 0. It's why the function was made. The addition is just moving the function left and right, which is what happens in every function when there is an addition to a particular variable. If you add it outside the function, then you move the graph up and down. Multiplication of the variable just shrinks or extends the graph.
Can you move the absolute value to the 8? Like from |x-2|=8 to x-2=|8|
Please can you do more questions related to the modulus function I would really appreciate it!
There is discussion in the comments forva solution in the complex plane. Would you do a follow-up and address that, please.
I like how if the question was not"x" and rather it was "z" , everyone would know it has complex solutions.
if you want complex solutions, it's an ellipse and the equation is already in the canonical form. if you're some kind of real number degenerate, it's an extremely simple piecewise function
So this only works when the two "fixed points" are exactly 6 apart?
It seems to me that it was easier to explain through a graph, if you know what a graph looks like, these are functions, and then the problem is solved orally!
It's easier to just make a rough graph of it, range of it's solutions will be pretty clear that way.❤
Can you please make a video on Lagrange interpolation please please please
How do we work this out mathematically, as opposed to drawing a number line and making a few guesses?
You look at the cases for the expressions inside the absolute values. Either they both are negative, both are positive, or one is positive, and the other is negative.
In the case of both negative, x - 5 < 0 and x + 1 0
x - 5 + x + 1 = 6 => 2x - 4 = 6 => 2x = 10 => x = 5. x - 5 >= 0 and x +1 > 0 with x = 5, valid solution.
In the case of one is positive and the other is negative,
x -5 < 0 and x + 1 > 0 (Accidentally wrote the wrong symbol, but I corrected it below.)
-x + 5 + x + 1 = 6
6 = 6, valid for all x in the interval x < 5 and x > -1
Put it all together -1
In the general case one of the intervals might have produced an equation where there were two constants that were unequal. That just eliminates that interval producing any solution at all or the number isn't in the range, and therefore, it equally is not a solution.
You can also do the steps mentioned in the video till 3:29 then prove why all answers belongs between -1 and 5 by intuition. Basically the gap between -1 and 5 is 6, so if x is somewhere between -1 and 5, it would be akin to breaking the line from -1 to 5 in two parts; from -1 to x and x to 5 so the total distance of both would obviously be 6. Since the distance from -1 to 5 is 6, the distance from -1 to an x greater than 5 would naturally be greater than 6, and since the distance from 5 to -1 is 6, the distance from 5 to a number less than -1 would be greater than six
Technically it’s the same method as the video, but we’re eliminating ranges instead of proving the same via induction. You can try plugging 5.01 and 4.99 in the equation and look at the equation made as a way to see why mathematically the numbers work out as you move the numbers up and down. Maybe the confusion is happening because the integer value that were chosen doesn’t make it very clear why it would be the same for all numbers in that range.
What makes you think this is not a mathematical way of working it out? It's a geometric way of solving an algebraic problem. If solving a geometric problem with algebra is a mathematical approach, then why going the other way shouldn't be? You have two points on a line distance 6 apart and you are asked to find all points such that the sum of distances from this point to two given points is also 6. This is only possible if you are between the two points. This can be proven in a number of ways, but more often than not it's an axiom in geometry that for any three points A, B and P, where d(A,B) denotes the distance between A and B d(A,P)+d(B,P)=d(A,B) iff P is between A and B. It's on par with the triangle inequality, giving a condition for equality. Depending on your choice of axioms it can be a theorem, but there's nothing not mathematical in this approach IMO.
if you don't hate yourself, you render the absolute values into piecewise functions. if you do hate yourself (equivalently, if you love algebra), you treat the absolute values as square roots
Feeling proud as I did it in my head and I got it correct 😎😎
What if the distance between the two numbers is not equal to the sum of the absolute values, though, like if you had |x-5| + |x+1| =7?
You'd be taking on the outside of the two limits, so like x = 5.5 would be a solution
I think -2 and 6
If you plot the sum of the two absolute sums you'll get a graph that looks like the silhouette of a truncated cone, or a V but instead of the point in the bottom you have a line, or this ⊔ but instead of right angles they are greater than 90°.
Equating this with a specific value (as in |x-5| + |x+1| = 6) means looking for where this graph intersects with a line parallel to the x-axis.
What you see very neatly from this graph is that this lower line is a segment of the line y=6. That's why you get infinitely many solutions, all numbers between -1 and 5. If you "pull up" the line parallel to the x-axis, you'll get two points intersecting with that graph, in equal distances from -1 and 5. For the value 7, as in your example, it will be at an additional distance of 0.5 in both directions, i.e. -1-0.5 (i.e. -1.5) on the left, and 5+0.5 (i.e. 5.5) on the right.
If however you "pull down" the line parallel to the x-axis below y=6, then this line will not intersect with the graph and there are no solutions at all (e.g. no solutions for |x-5| + |x+1| = 3).
Like I said, if you plot the graphs, you'll easily see all of this. It's a lot more straightforward to see than me describing the plot.
In this case you can split your solution into several cases, which solves the module and lets you turn this into trivial linear equation. In your example cases are 1) -inf < x ≤ -1; 2) -1 ≤ x ≤ 5 and 3) 5 ≤ x < +inf. After you solve each case make sure the solution satisfies initial condition. First case yields solution x = -1.5, second case has no solutions and third case yields solution x = 5.5. Solutions for cases 1) and 3) satisfy their own initial conditions, so final answer is x = -1.5 and x = 5.5.
Still works, but now there will only be two answers: 5.5 and -1.5. The idea is still the same, the distance between -1 and 5 is 6 and for every number between those two the summed distances cover the interval from -1 to 5, however, every number outside [-1,5] covers [-1,5] only once and then covers the interval from the number to the closest end of [-1,5] twice and this double covering should be 7-6=1, so it's two numbers (in the case of real numbers) distance 0.5 from -1 and 5 outside [-1,5], giving solutions -1.5 and 5.5.
It also works for complex numbers as well, but there are infinitely many solutions, all lying on an ellipse with foci -1 and 5 by the definition of an ellipse.
That also shows that for any d smaller than 6 |x-5|+|x+1|=d there are no solutions real or complex.
As for the sum of three absolute values of this form or scaled absolute values... It becomes quite a mess really quickly. And that is not taking into account expressions not of the form x-a inside the absolute values. At this point doing something other than considering cases doesn't really work.
bluepenredpen
Where are the non real solutions mentioned a few times?
He's not talking about non-real solution, he mentions that there are infinite solutions, but that the solution isn't "all real numbers".
@@MadaraUchihaSecondRikudo ah, the phrasing sounded like he meant some of the solutions weren't real numbers
I would like to draw this function graph😂
I always see the thumbnail of your videos and think I know the correct answer, but realize I don’t when I watch the video.
|x-5| + |x+1| = 6
First, we find the roots
x-5 = 0 x+1=0
x = 5 x = -1
Then we draw the signs table
-inf -1 5 +inf
x-5 - - 0 +
x+1 - 0 + +
If x € ]-inf;-1[ :
|x-5| + |x+1| = 6
(-x + 5) + (-x - 1) = 6
-x + 5 - x - 1 = 6
-2x + 4 = 6
x = -1 unacceptable
If x € ]-1;5[:
|x-5| + |x+1| = 6
-x + 5 + x + 1 = 6
0x = 0 all possible
x € ]-1;5[
If x € ]5;+inf[
|x-5| + |x+1| = 6
x - 5 + x + 1 = 6
2x = 10
x = 5 unacceptable
If x = -1
6 = 6 so x = -1 is a solution
If x = 5
6 = 6 so x = 5 is a solution
x € [-1;5]
I didn't use the definition it's just the method we learned at school
I didn't understand what do you mean
Is 0 not an answer as well
Do you have a new camera?
this is a wondeful informative video
Yeah anything below -1 does not work.
wtf thats crazy
Alright, we got 2 answers, x = -1 and x = 5, that's it, done. 😁😁
*bprp: JOB'S NOT FINISHED!!!!* ❌❌
use graphic go brrrrrrr
Can someone solve this for me
There are 3 kinds of bananas red, green, and blue. For every 4500 red bananas sold you get 1500 dollars, for every 3000 green bananas sold you get 1000 dollars, and for every 2500 blue bananas sold you get 2500 dollars.
Now, for every 100k bananas sold you'll gain 1% more money(multiplicative). If I want to sell 50 million bananas of each kind, which banana color gives me more money?
Since when do red, blue and green bananas exist?
Nvm i think -2 does not work
Dear god im pretty sure every number works
yo
Noone else noticed the siren being picked up as background noise?
Does it contain cocaine?
alcohol is for algebra. cocaine is saved for number theory. look man, i don't make the rules
First 😊