2 < x < 4
Short explanation
(in case you just needed a little help):
|x - 3| > 1 implies 4 things
(assuming x-3 is positive):
1. x - 3 > 1, so x > 4
2. 3 - x < 1, so x > 2
(One of these is true)
(Assuming x-3 is negative):
3. x-3 < 1, so x< 4
4. 3-x > 1, so x <2
(One of these is true)
The only inequalities from each pair that make sense here are
x>2 and x <4.
2 < x < 4.
LONGER EXPLANATION
(in case your teacher didn’t teach the subject well):
|x - 3| > 1 implies
(assuming x-3 is positive):
1. x - 3 > 1
2. 3 - x < 1
Why? Because the absolute value takes any negative or positive value and spits out the positive version of it. If x-3 is positive and |x - 3| > 1,
then x-3 > 1.
Also if x-3 is positive, 3-x must be negative so 3-x < 1 since all negative numbers are less than 0 which is less than 1.
Now we just work out values of x from the inequalities we just discovered.
1. x-3 > 1:
x > 4 (Add 3 to both sides)
2. 3-x < 1:
-x < -2 (subtract 3 from both sides)
x > 2 (multiplying both sides by (-1) will also flip the inequality).
That was all if x - 3 was positive. But if x-3 is negative then:
3. x-3 < 1, so x < 4
4. 3-x > 1. so x < 2
So we have 2 pairs of inequalities but only 1 inequality in each pair can be true:
1. x>2
2. x>4
3. x<4
4. x<2
The only inequalities from each pair that make sense here are #1 and #3
x>2 and x <4.
#2 doesn’t make sense because if x>4 (first pair) then that contradicts both inequalities in the second pair x<4 and x<2, and we know one of them has to be true.
#2 doesn’t work, so #1 must be true: x>2. Now we need to see which inequality is true in the second pair: x<4 or x < 2.
If x>2 then that contradicts what we found with #1. x can’t be less than 2 at the same time. So #4 can’t be true and it must be the case that #3 is true so x<4. So we found a pair that works and have a solution!
x<4 and x>2
So 2 < x < 4.
