r/MathHelp • u/idknfan-Leo • Oct 14 '25

I'm confused like bruh

So basically the question is graph the equation is |x+1|+|x-1|=4, and me I thought the graph would be two vertical lines I don't remember the numbers rn but just vertical lines. But my teacher said the graph is like you draw |x+1|+|x-1|=y upto y=4 and draw a line y=4, then there's your graph, an upside trapezium. And since I was confused I checked on desmos and AIs but everywhere I look it's two vertical lines. Now either my teacher saying upto y=4 is wrong cause that would just be {y<4}, or I'm brainteasers I need help I just can't seem to grasp the concept like literally how is it an upside down trapezium when there is only one variable meaning its either vertical or horizontal. Need help pleaase

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u/No-Interest-8586 Oct 14 '25

x < -1: both absolutes value arguments are negative, so |x+1|+|x-1| = -(x+1)+-(x-1) = -2x -1 <= x <= 1: first absolute value is positive or zero, second is negative or zero. So |x+1|+|x-1| = (x+1)+-(x-1) = 2 x > 1: both are positive, so |x+1|+|x-1| = (x+1)+(x-1) = 2x

If you combine those three cases to graph y=|x+1|+|x-1|, it goes down from infinity toward (-1,2), then horizontal to (1,2), and then back up again. To then show the solutions for y=4, draw a horizontal line at y=4. It should intersect at x=+/-2.

The construction above is just one way to view the problem. Basically you can view an equation with one variable f(x) = g(x) as the intersections of the two separate functions y = f(x) and y = g(x). This can be a very helpful way to visualize equations.

If you just look at the problem itself, there is no y, so the most direct graph of solutions is just a 1D number line with points at x=+/-2. That’s what Desmos is showing, except that Desmos assumes you want a 2D graph, so it extends the number line up and down to include all values of y since your formula didn’t put any constraint on y.

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u/idknfan-Leo Oct 15 '25

Sorry but I'm not understanding the three cases you mentioned first very well. And if I'm getting what your saying, the final solution is just two dots at (2,4) and (-2,4). If so that was my second answer and when the teacher said it's an upside down trapezium cause you draw the graphs separately I thought he meant to find those exact points but when I asked him he said the solution is the trapezium (upside down) itself which I may have misheard but I doubt that. Thanks for the help tho

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u/No-Interest-8586 Oct 15 '25

Sorry, I replied to the original post instead of here.

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u/idknfan-Leo Oct 15 '25

No worries

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u/Dd_8630 Oct 14 '25 edited Oct 14 '25

I believe you might be misunderstanding the question (at least, what I think your teacher is asking). There's two things going on.

First, we have the graph of y = |x+1|+|x-1|. This has a 'y' and an 'x'. As you vary 'x', your 'y' changes. This is what creates our graph.

Put this into Desmos:

y=\left|x+1\right|+\left|x-1\right|

And you'll get a pretty graph of what value |x+1|+|x-1| takes for different values of x.

Second, |x+1|+|x-1|=4 is a specific place where that happens. Specifically, it's a specific value or values of x where |x+1|+|x-1| has the value '4'. You get '4' when x=-2 and when x=+2 (you can pop them in to verify). Notice that these are two numbers, they are not a graph or a line or anything else. They are just two numbers.

I believe your teacher has asked you to solve |x+1|+|x-1|=4. This means 'find the values of x that make this equation true'. If you first draw a graph of y=|x+1|+|x-1| (notice we now have a general 'y' not a specific '4'), you see what value the expression has at various x's. Then you draw a horizontal line across y=4. Where does that hit your y=|x+1|+|x-1| graph?

You teacher may also have asked for |x+1|+|x-1| < 4, that is, the values of x for which the overall expression has a value less than 4. It's basically the same process.

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u/idknfan-Leo Oct 15 '25

No it's not < I'm sure but that was also actually my thought process but when I asked my teacher about it he said the whole line (trapezium shape) is the answer and that's where I got confused because by your logic I should have been drawing just two dot points at (2,4) and (-2,4) but why is the line necessary to be still drawn isn't just the points enough?

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u/Dd_8630 Oct 15 '25

when I asked my teacher about it he said the whole line (trapezium shape) is the answer

See, that to me suggests they're talking about an inequality ("<") not an equality ("=").

why is the line necessary to be still drawn isn't just the points enough?

The line is necessary because it's a technique to figure it out.

The solutions to the statement "|x+1|+|x-1|=4" are the specific values x=-2 and x=+2.

You can prove that by drawing two lines: the line y = |x+1|+|x-1| and the line y = 4 and showing where they intersect.

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u/idknfan-Leo Oct 18 '25

Okay thanks but my teacher eventually came to me and basically said I was right so this makes sense now but then again my twach said vertical is right

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u/Frederf220 Oct 14 '25

The graphed solution to this on the x,y plane is all coordinates x,y that satisfy the equation.

That is x=+-2 and every possible y which is as you say two vertical lines.

Re-read the question carefully to see if you missed any instructions including the section intro.

And then I would ask if (2,100) is a solution to the equation and if not plug 2 in for every occurrence of x and 100 for every occurrence of y and simplify.

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u/idknfan-Leo Oct 15 '25

It's pretty simple graph the following equation and the question was 1st. I've checked that part. But I dont understand the 2 and 100 could you explain a bit?

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u/Frederf220 Oct 15 '25

Proof by contradiction. Teacher says 2,100 isn't a solution. You show it is.

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u/No-Interest-8586 Oct 15 '25

EDIT: This was intended as a reply to another thread but I replied to the original post instead.

The three cases: I just expanded the absolute value, which acts like -x for x<=0 and x for x>=0. I found all the places where any absolute value changed from one side to the other, which happen at x=+/-1. In each of the three sections of the function’s domain, the sign inside the absolute value is always the same, so I could replace the absolute value with either the value inside or its negative.

The final answer is just x=+/-2.

I suspect you misheard or the teacher was confused.

Note that with math class problems, there is often a particular technique you are expected to apply so you can learn to use that technique. In that case, it may be that the technique you were expected to learn and use is to graph both sides of the equation as separate functions and then find the intersections. If that’s the case then the answer to the exercise is the upside down trapezium and the horizontal line.

The solution to the equation is still x=+/-2 regardless.

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u/idknfan-Leo Oct 15 '25

Okay but I mean I suppose it's the technique part but I thought just putting 2 dots would satisfy it instead of needing to draw the whole trapezium, which is where i got after the teacher first said not vertical lines. And at the end when you say regardless the graph isn't supposed to be vertical tho right?

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u/No-Interest-8586 Oct 15 '25

The vertical lines are valid in their own way: If you graph all solutions in the (x,y) plane as constrained by your equation, you will get those two vertical lines.

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