# modulus equation

• Dec 23rd 2009, 02:23 AM
renlok
modulus equation
i have the equation $\displaystyle \mid x - 1 \mid = 1 - x$
i opened up the modulus to get $\displaystyle x - 1 = 1 - x$ & $\displaystyle 1 - x = 1 - x$

and the answer is supposed to be the inequality $\displaystyle x \le 1$ but ive no clue how to get to this. (Headbang)
• Dec 23rd 2009, 02:28 AM
red_dog
First of all, the right member must be nonnegative, as the left member is.

That means $\displaystyle 1-x\geq 0\Rightarrow x\leq 1$.

In this case the equation is $\displaystyle 1-x=1-x$ which is true for all $\displaystyle x\leq 1$
• Dec 23rd 2009, 02:49 AM
HallsofIvy
Quote:

Originally Posted by renlok
i have the equation $\displaystyle \mid x - 1 \mid = 1 - x$
i opened up the modulus to get $\displaystyle x - 1 = 1 - x$

Provided $\displaystyle x-1\ge 0$.
x- 1= 1- x reduces to 2x= 2 and then x= 1. Yes, that is satisifies x- 1= 0 and so is a valid solution.

Quote:

& $\displaystyle 1 - x = 1 - x$
Provided x- 1< 0.
1-x= 1- x reduces to 0= 0 which is true for all x but we still have "x- 1< 0". Every x such that x- 1< 0, which says x< 1, is a solution. Putting that together with the first case says that all x such that $\displaystyle x\le 1$ is a solution.

Quote:

and the answer is supposed to be the inequality $\displaystyle x \le 1$ but ive no clue how to get to this. (Headbang)