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If we have an absolute equation like this one: $$|x-1|+|x+1|=4$$ do I read it as $$|x-1|+|x+1|=y$$ or is it actually: $$|x-1|+|x+1|-4=y$$ The graphs are different so I'm confused which one is the right path to follow if I want to graph the function.

Also, if we have a normale equation like $$3x-2=4$$ do I still need to move the 4 over to the left side so it looks like a function $$3x-6=y$$ so I can graph it, or does it stay on the right side and is ignored when I am graphing my function $$3x-2=y$$?

TheSlavSQuat
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  • In general, if you are faced by an equation like $$\text{left hand side}=\text{right hand side}$$ where both sides are depending on $x$ the solutions of that equation are the zeroes of the function $$y=\text{left hand side}-\text{right hand side}.$$ – Michael Hoppe Apr 20 '20 at 14:02
  • @MichaelHoppe thanks for the quick answer – TheSlavSQuat Apr 20 '20 at 14:29

2 Answers2

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Either way works. If we work with a simple example first to demonstrate the idea such as $3x-2=4$.

If we set $y=3x-2$ we can draw a graph of this and we want to see on the graph when $y=4$ (or equivalently, when the line intersects the line $y=4$). Alternatively we could rearrange our equation to say $3x-6=0$. Then we can set $y=3x-6$ and graph it. Now we want to see when $y$ is equal to $0$ on our graph (also called the roots). So both ways work, they will give you different graphs but when you look for the appropriate $y$-value they will give you the same answers.


If we now look at the original equation $|x-1|+|x+1|=4$. We are again faced with the choice of what to set $y$ to. And in fact we can set it to whatever we like although most things won't be very helpful.

If we set $y=|x-1|+|x+1|$ and draw the graph, then we are looking for points on the graph where $y=4$. If instead we choose to set $y=|x-1|+|x+1|-4$ we are looking for points where $y=0$. Both will give you the same answers for $x$. (The answers for $y$ will be different but it doesn't matter because we made up $y$, it's not in the question.)

Joz
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If you only have one variable you can‘t graph it, because the graph (the line) represents the infintely many solutions for the equation.

Exmaple: $2x+4y=3$ reform it: $y={3-2x \over 4}$, which you can graph.

If you only have one variable there are probably finitely many solutions (values for x that satisfy the equation). If x is bigger than 0, the equation simplifies to:

$(x-1)+(x+1)=4$

$2x = 4$

$x = 2$

So $x=2$ satisfies the equation and is a solution to it.

If x is smaller than 0, the equation reads:

$-(x-1)-(x+1)=4$

$-2x = 4$

$x = -2$

So the other solution is $x=-2$.

That‘s how you can find the values for x, so that the equation with absolute values is true.

You can‘t graph it, because there are only two solutions and the axes would stand for two variables.

Saha
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  • Your logic in saying if $x>0$ the equation simplifies to $(x-1)+(x+1)=4$ isn't quite right. For example if $x=\frac{1}{2}$ then the original equation gives $|-\frac{1}{2}|+|\frac{3}{2}|$ which is 2, but yours gives $(-\frac{1}{2})+(\frac{3}{2})$ which is 1. You actually need for $x \geq 1$ to make your simplification. There's a similar problem with the $x<0$ assumption too. – Joz Apr 20 '20 at 14:49
  • Yeah, I know, but because it‘s right for this example with only integers, I didn‘t want to confuse with writing even more. – Saha Apr 21 '20 at 22:15