Inequality involving absolute value: $|x-1|>2-x$

Question

How does one go about solving the following inequality?

$$|x-1|>2-x$$

Answer 1

-if $x - 1 \geq 0$ ($x \geq 1$), then $\vert x - 1 \vert = x - 1$, implies $x - 1 > 2- x$ which gives $x > \frac{3}{2}$.

-if $x - 1 \leq 0$ ($x \leq 1$), then $\vert x - 1 \vert = 1 - x$, implies $1 - x > 2- x$, which admits no solution.

Answer 2

Hint:

For an algebraic solution you can divide the inequality in two systems: $$ \begin{cases} x-1\ge 0\\ x-1>2-x \end{cases} \quad \land \quad \begin{cases} x-1< 0\\ 1-x>2-x \end{cases} $$

Answer 3

: Given inequality : $|x-1|>2-x$ \, there can obtain two possible solutions of any absolute value inequality, because the expression inside the absolute value sign can be positive and negative. Thus, there are two cases, (1) $x-1>2-x$ and (2) $x-1>-(2-x)$

Case-1: $x-1>2-x$

$\implies$ $x-1+x>2$

$\implies$ $2x>2+1$

$\implies$ $2x>3$

$\implies$ $x>3/2$

Case-2: $x-1>-(2-x)$

$\implies$ $x-1>-2+x$

$\implies$ $x-1-x>-2$

$\implies$ $0>-2+1$

$\implies$ $0>-1$

It means there is no solution in Case-2.

Thus, there is only positive value possible inside the absolute value sign for this given inequality and only one solution that is $x>3/2.$

Answer 4

There is a technique for solving inequalities called the "split point method", which doesn't require so much "parity chasing." The method is: 1. Solve the corresponding equation and find all discontinuities. These are the "split points." 2. Plot the split points on a number line. This cuts the line into a number of intervals. Each of these intervals is either part of the solution set or not. 3. Test each interval by choosing a nice point in the interior of each interval. If the one point makes the inequality true, then the whole interval does. Otherwise not. Test each split point separately.

So solve $\mid x-1 \mid = 2-x$. Either $x-1 = 2-x$ and so $x=3/2$ or $-(x-1) = 2-x$ which has no solution. There are no disconinutites, so there is only one split point. Plot $x=3/2$ on the number line. To test the interval $(-\infty,3/2)$ plug $0$ into the inequality for $x$ to get $\mid 0-1 \mid > 2-0$. This is false, so that interval is not part of the solution set. To test the interval $(3/2,\infty)$ plug $2$ in for $x$ to get $\mid 2-1 \mid >2-2$ which is true. So that interval is part of the solution set. The value $x=3/2$ makes the inequality false. So the solution set is $(3/2,\infty)$.

The reason this works is that you've found all the points where the graph of the right hand side minus the left hand side can cross the $x$-axis.