Can you explain why $x>\frac 23$ is not a solution to this inequality

Question

$|2x + 2| + |x - 1| > 3$

why can't $x>\frac 23$ be a part of the solution?

Thanks for your help!

Answer 1

You take the “highest common area” if you're solving simultaneous inequalities. If you split an inequality into cases, you have to take the union of the areas.

A solution of the problem can go like this: subdivide the real line in the intervals $(-\infty,-1)$, $[-1,1]$ and $(1,\infty)$; they are so chosen because the expressions in the absolute values change sign at $-1$ and $1$ respectively.

Interval $(-\infty,-1)$: since $2x+2<0$ and $x-1<0$, the inequality becomes $$ \begin{cases} x<-1\\[4px] -2x-2-x+1>3 \end{cases} $$ that is, $x<-4/3$.

Interval $[-1,1]$: since $2x+2\ge0$ and $x-1<0$, the inequality becomes $$ \begin{cases} -1\le x\le 1\\[4px] 2x+2-x+1>3 \end{cases} $$ that is, $0<x\le 1$.

Interval $(1,\infty)$: since $2x+2>0$ and $x-1>0$, the inequality becomes $$ \begin{cases} x>1\\[4px] 2x+2+x-1>3 \end{cases} $$ that is, $x>1$.

This is probably the case that's bothering you: the second inequality has indeed the solution set $x>2/3$, but we must match it with the previous condition that $x>1$.

Finally, we put together the solution sets: $$ (-\infty,-4/3)\cup(0,1]\cup(1,\infty)= (-\infty,-4/3)\cup(0,\infty) $$ or, in other terms, the solutions are $$ x<-\frac{4}{3}\quad\text{or}\quad x>0 $$

Answer 2

Suppose $x>2/3$, then if $x \ge 1 $, then $2x+2 + x -1 = 3x + 1 \ge 3 + 1 > 3$. If $2/3 < x < 1$, then $2x+2 - x +1 = x + 3 > 3$.

This is, however, not a complete solution (probably why it is not a solution!?). Consider $-1 \le x \le 2/3$, then $2x+2 - x +1 = x + 3 >3 $. This is only true if $x>0$.

If $x<-1$, then $-2x-2 - x +1 = -3x - 1 > 3.$ This holds if $x < -4/3$.

Thus the complete solution is $x \in (-\infty,-4/3) \cup (0,\infty)$.