same. especially when he brought up "not all real solutions"

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.

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

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.

Yes, you would be right.

If |x - a| is distance between x and a, then what is |x + a|? What is just |x|?

@@user-qc9cd5iz3l 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

@@user-qc9cd5iz3l 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.

Thank you very much for this, I was trying to think of a way to do it algebraically.

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

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 didn't use the definition it's just the method we learned at school

Since when do red, blue and green bananas exist?

alcohol is for algebra. cocaine is saved for number theory. look man, i don't make the rules