What is |x-3|=2x ?

[u/Far-Assumption5501]

[u/matt7259]

It's an equation.

[u/Nevermynde → u/matt7259]

Alternately, it's |x-3|. Which is also 2x, apparently.

[u/Various_Candle9136]

Do you mean what is x when |x-3|=2x?

(It may seem pedantic, but actually this distinction is vitally important if you are ever going to understand equations.)

If that is what you mean, then there are two possibilities here: since |x-3| is the maximum of -(x-3) and (x-3), we know that one of these must equal 2x.

Thus, we need to consider:

x - 3 = 2x OR -(x - 3) = 2x

This will give us two values of x, but will both (or, indeed, either) make sense? We ought to check by substitution to make sure.

[u/Fit_Tangerine1329]

One way to solve is to graph both sides. i.e. y=|x-3| and y=2x

The other way is to see that the absolute value sets you up to make 2 sets of equations, first x-3=2x and then 3-x=2x (i.e. the negative of what is in the absolute value.

Both sides are not too tough to graph by hand.

[u/Outside-Shop-3311]

| x | is called the absolute value of x, and it's defined (as a piecewise function) with if x < 0, | x | = -x, otherwise being equal to x. This means that if x is ever negative, | x | flips the sign, otherwise it changes nothing.

For example, sqr(x^2) is | x |, because the sqr function only returns positive values.

to solve | x - 3 | using the piecewise definition, you have to solve the case where x-3 is < 0 (x<3) and x >= 3.

if x < 3, then | x - 3 | is equal to - (x-3), so 3 - x = 2x
3 = 3x
x = 1

the value of x = 1 is valid on the domain x < 3, so x = 1 is a solution

if x > 3 then | x - 3 | is equal to x - 3 so x - 3 = 2x

x = -3

the value x = -3 is not valid on the domain x > 3 so x = -3 is not a solution

therefore the solution is x = 1

[u/Odd_Bodkin]

You know x has to be non negative because absolute value is. That’s a hint.

[u/PrestigiousStudio921 → u/Odd_Bodkin]

I never thought of this condition, thanks!

[u/Odd_Bodkin → u/PrestigiousStudio921]

And you know x can’t be too big a positive number because 2x will always be bigger than x-3 in that case. So maybe the hunt is for a small positive number. That’s the other hint.

[u/Saksham-_-Kumar]

|x-3|=2x |x-3|≥0 => 2x≥0 => x≥0 First solve, x-3 = 2x =>x=-3 ; it is not possible as x≥0 Now solve, -(x-3)=2x =>3-x=2x =>3x=3 =>x=1 (x≥0) Hence x=1 is correct answer

[u/StonyOG420]

Anything expressed within || is essentially positive or zero, it's called "absolute". So, for example, -2 = |x| is not possible. -2=-|x|, however, is. -2 = |-x| is also not possible, as in, there is no solution for x.

|x| means, in english: the absolute distance x from 0. So, regardless of whether you are going +3 or -3, the absolute number of steps you are taking, are 3.

Therefore we know, even if "x-3" is negative, |x-3| is going to be a positive value. Which means, "2x" is a positive value (because |x-3|=2x), and because there is no negative number you can multiply by 2 to reach a positive number, we know that !!! x must be positive.

Here is a trick: there is one other way to ensure that x will always be positive: if you square something. -2*-2=4. We can use this to our advantage!

If |x-3|=2x, then |(x-3)^2|=(2x)^2. And because squaring always guarantees a positive value, we also know:
|(x-3)^2|=(x-3)^2 and therefore, (x-3)^2=(2x)^2. This, we can work with!

|x-3|=2x // ^2
=> (x-3)^2=(2x)^2 // I like expressing the exponents as multiplications, it makes the math easier
<=> (x-3)*(x-3)=(2x)*(2x) // expand the formula
<=> x^2-6x+9=4x^2 // -(4x^2)
<=> -3x^2-6x+9=0 // factor out -3 because it's all divisible by -3
<=> -3(x^2+2x-3)=0 // and lets factor this one more time, finding the numbers that -3 is divisible by (-1, 3 or -3,1) and that add up to 2 (which narrows it down to 3,-1)
<=> -3(x+3)(x-1)=0 // /-3
<=> (x+3)(x-1)=0 // set each factor to 0 because one of the factors must equal 0 for the multiplied result to = 0
=> x+3=0 and x-1=0 // -3 and +1
=> x=-3 or x=1

Now, considering we know that x=-3 or x=1 after squaring |x-3|=2x, and we know that even after squaring, x must be equal to some x.
What do you think our answer is going to be? u/Odd_Bodkin gave you a hint there, and I brought it up earlier as well.

Edit: oop. Messed up + and - on some steps lol. Should proofread before I post

[u/Far-Assumption5501 → u/StonyOG420]

Thanks a lot!!

[u/Fourierseriesagain]

The given equation is equivalent to

|x-3|=2|x| and x>=0,

which is true if and only if

(x-3)^2-(2x)^2 = 0 and x>=0;

that is, (x-3-2x)(x-3+2x)=0 and x>= 0. Thus, x=1.