A Twist for Absolute Value

We use the simple fact that $|y|\geq 0$ for real $y$ .

Thus $\large{|x-2|\geq 0 \Rightarrow 2-x\geq 0 \Rightarrow x\leq 2}$

For the equation $|y|=-y$ to be satisfied, we need to have $y\le 0$ .

Here, we have $y=x-2 \le 0$

Subject to this condition, the maximum value is $\boxed{x=2}$ .

There are 2 answers...0,2

for this math,

modulus of (x-2) =2-x

or, x-2 =2- x

or, 2x = 4

or, x=2

$(x-2)^2=2-x$

$x^2-4x+4=2-x$

$x^2-3x+2=0$

$(x-2)(x-1)=0$

$x=2$

$x=1$

$answer:x=2$

solving or graphing the equations will get you x is atmost 2 :D

|x-2| = 2-x :

If x-2 >or= 0 then x >or= 2 (and |x-2| = x-2) thus x-2 = 2-x thus 2x = 4 and x = 2 is a solution

If x-2 < 0 then x < 2 (and |x-2| = -(x-2) = 2-x) thus 2-x = 2-x thus any x< 2 is a solution

Thus x <or= 2 is the solution

Thus x = 2 is the greatest solution.