Any way to solve $|x-8| = |2-x|-6$ algebraically?

Question

Everything I've tried has given me $x = 2$ (which is obviously incorrect, since $-6 \neq 6$).

The actual answer is $x \geq 8$ which I obtained by observing a graph.

Would love assistance!

Answer 1

There are really only three cases here, so you can use exhaustion pretty easily.

Case 1: assume $x<0$

If $x<0$, then $|2-x|=-(2-x)$, and $|x-8|=-(x-8)$, so

$$-(x-8)=-(2-x)-6$$

Solve for $x$ and you get $x=8$. This contradicts $x<0$, so $x<0$ cannot be true.

Case 2: assume $x=0$

If $x=0$, then $8=2-6$. This is obviously false, so $x$ is not $0$.

Case 3: assume $x>0$

If $x>0$, then there must be some smallest positive number $x$ such that the equation is true. If $|2-x|$ is less than $6$, then the right hand side of the equation will be negative, which is impossible because the left hand side is an absolute value, which will always be positive. Therefore, the smallest possible number for which the equation might be true is the solution to the equation $$|2-x|=6$$ This is $8$, so $x$ is at least $8$. Plug in $8$ for $x$ and the equation is still true so $x=8$ is a solution.

If for some positive number $a$, $8+a$ is also a solution, then, by the properties of addition, you know that every number greater than $8$ is also a solution.

As there is no such number, $a$, the only solution is $x=8$.

Answer 2

Write your equation in the form $$|x-8|=|x-2|-6$$ and consider the cases: $$x\geq 8$$ then $$x-8=x-2-6$$ $$2\le x<8$$ then we get $$-x+8=x-2-6$$ $$x<2$$ then we have $$-x+8=-x+2-6$$ The solution is given by $$8\le x<\infty$$ and $$x\in \mathbb{R}$$

Answer 3

By triangle inequality we have

$|x-8| +|6|\geq |x-2|=|2-x|$

with equality iff 6 and x-8 have equaly sign, so $x\geq 8$, but then $|x-2|=x-2$ so this equality is true for all $x\geq 8$.

Answer 4

For this equation $$|x-8| = |2-x| - 6$$ there are two modulus terms $|x-8|$ and $|2-x|$ so these two terms will behave differently at the points $x=2$ and $x=8$. Now break the terms in the following cases:

Case1:

$x>=8$ , this means $-x\leq-8$ or $2-x\leq-6$ which reduces the equation to

$$ x-8 = x-2-6$$ which reduces to an identity and this means $x\geq8$ is one of the solution for the equation.

Case 2:

$x<8$, this means $-x>-8$ or $2-x>-6$ which reduces the equation to $$8-x = x-2-6$$ if $$-6<2-x<0$$ or $$2x=16$$ when $$2<x<8$$ giving no solution from this case. Another equation is $$8-x=2-x-6$$ if $$x<2$$ giving $$8=-4$$ and $$x<2$$ again giving no solution

Conclusion: There is only one possibility that $x\geq8$. Another way of investigating this is by inspection. After proving case 1 and case 2 just put values that are greater or equal 8 in the above equation and you will find that for all $x\geq8$ the equation holds true whereas for the rest of the cases it does not hold. Try putting $x=2$ or any other value. The equation wont hold.