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$$dfrac{|x−2|+|8−4x|}{|x−2|+3} < 2$$

How do we find integer values of $x$? I'm unsure whether I actually have to cover all cases such as $x>2$ or $x<2$ since there are more than one absolute values. What should I do?

Regards

Hint

$$|8-4x|=4|x-2|$$therefore the following inequality $$dfrac{|x−2|+|8−4x|}{|x−2|+3} < 2$$reduces to $$dfrac{5|x−2|}{|x−2|+3} < 2$$or equivalently $$5|x-2|<2|x-2|+6$$

Big hint: $|8 - 4x| = 4|x-2|$.

So let $a = |x-2|$

Find the range for $frac {a + 4a}{a+3} < 2$ and $a ge 0$.

Then solve for $|x-2| = a$.

$frac {a+4a}{a + 3} < 2$

$5a< 2a + 6$

$3a < 6$

$a< 2$

$|x - 2|< 2$

$0 le x - 2 < 2$ or $-2 < x -2 < 0$ or

$0 < x < 4$.

If you want to do it case by case:

Case 1: $x- 2 ge 0$ and $8-4x ge 0$ then $x ge 2$ and $8 ge 4x$ as $2 ge x$ so $x = 2$.

$frac {|x-2| + |8-4x|}{|x-2| + 3} < 2$

$frac {0 + 0}{0+3} < 2$. Yep. No contradiction. $x = 2$.

Case 2: $x - 2 ge 0$ and $8 -4x < 0$ then $x ge 2$ and $8 < 4x$ so $x > 2$.

$frac {|x-2| + |8-4x|}{|x-2|+3} = frac {x-2 + 4x -8}{x-2+3}=$

$frac {5x -10}{x+1} < 2$ so (as $x + 1 > 0$)

$5x -10 < 2x + 2$ so $3x < 12$ or $x < 4$ so $2< x < 4$

Case 3: $x -2 < 0$ and $8 - 4x ge 0$ so $x < 2$ and $8 ge 4x$ so $x le 2$ so $x < 2$.

$frac {|x-2| + |8-4x|}{|x-2| + 3} = frac {2-x + 8-4x}{2-x+3}=$

$frac {10-5x}{5-x}<2$ and as $x< 2$ and $5-x > 5-2 =3 > 0$ so

$10-5x < 10 - 2x$ so $0 < 3x$ so $x> 0$ and $x < 2$ so $0 < x < 2$.

Case 4: $x-2 < 0$ and $8-4x < 0$ then $x < 2$ and $8 < 4x$ so $x > 2$ so $x < 2$ and $x > 2$ which is a contradiction.

Combining the three results:

$0 < x < 4$.

But that was way too much work and there was no excuse to not recognize that $|8-4x| = 4|x-2|$.