***
So we want to solve |x + 2| + |2 – x| = 8.
This is tricky because there’s three cases to consider:
x is less than -2, x is bigger than 2, or x is between -2 and 2.
Choosing different x changes the way in which we can remove the absolute value bars.
When x is less than -2 (so x < -2), then we have x + 2 < 0,
so |x + 2| = -(x + 2).
Also, when x is less than -2, then x is less than 2.
Knowing |2 – x| = |x – 2| and x < 2 means that |x – 2| = -(x – 2).
So when x is less than -2, we want to solve
-(x + 2) – (x – 2) = 8. Solving this gives x = -4.
We can check that |-4 + 2| + |2 – (-4)| = 8.
___
Similarly, when x is bigger than 2, then x is bigger than -2 (so x > -2).
So we have x + 2 > 0, and so |x + 2| = (x + 2).
Also, when x is bigger than 2,
knowing |2 – x| = |x – 2| and x > 2 means that |x – 2| = (x – 2).
So when x is bigger than 2, we want to solve
(x + 2) + (x – 2) = 8. Solving this gives x = 4.
We can check that |4 + 2| + |2 – 4| = 8.
__
Finally, when x is between 2 and -2,
we want to solve
(x + 2) + (2 – x) = 8.
But this reduces to 4 = 8, so
there is no x between 2 and -2 which makes the expression 8.
So only for x = -4 and 4 do we have |x + 2| + |2 – x| is equal to 8.
__
Now we’re pretty much done.
Draw out a number line and mark -4 and 4.
Now there are three intervals we just broke the number line into:
- infinity to -4, -4 to 4, and 4 to infinity.
Now choose easy “test points” in each interval, like
-100000 in the first, 0 in the second, and 100000 in the third.
If we plug x = -100000 into |x + 2| + |2 – x|, we see that it’s way bigger than 8.
If we plug x = 100000 into |x + 2| + |2 – x|, we
also see that it’s way bigger than 8.
If we plug x = 0 into 8, we see than the expression is less than 8.
Therefore, our expression must be less than 8 over the entire interval -4 < x < 4,
since that’s the interval that contains x = 0.