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This is the equation:

$|sqrtx-1 - 2| + |sqrtx-1 - 3| = 1$

Any help would be appreciated. Thanks!

Let $a =sqrtx-1$,

$|a-2|+|a-3|=1$

Check for solutions in the different regions for $a$.

Region 1: $a<2$.
Then we have $(2-a)+(3-a)=1$, i.e. $5-2a=1$, so that $a=2$.
Region 2: $2leq aleq 3$. We have $(a-2)+(3-a)=1$. This is true for every such $a$.

Final region 3: $3<a$. We have $(a-2)+(a-3)=1$ so that $a=3$.

In summary, $2 leq sqrtx-1 leq 3$.

Thus $5 leq x leq 10$.

Let $x$ be a solution of the equation. Notice $x geq 1$, since $sqrtx-1$ has its domain as $x geq 1$.

If $x geq 10$, then each of the absolute value is just the term inside (i.e.$|sqrtx-1-2| = sqrtx-1-2$ and similarly $|sqrtx-1-3| =sqrtx-1-3$) so that the given equation in this case becomes $2sqrtx-1-5=1$. Solve for $x$ you get $x=10$. Thus we have shown that IF there is a solution that is greater than or equal to $10$ then $x$ must be $10$; we have not yet proven that $10$ is a solution. However, we can check that $x=10$ is a solution by plugging it back in.

If $10 > x geq 5$ then $|sqrtx-1-2| = sqrtx-1-2$ and $|sqrtx-1-3| =-sqrtx-1+3$ so that the given equation becomes (after simplification) $1=1$; this does not give us a new information, but we can check that any number $x$ such that $10 > x geq 5$ satisfies the original equation.

If $5 > x geq 1$ then $|sqrtx-1-2| = -sqrtx-1+2$ and $|sqrtx-1-3| = -sqrtx-1+3$, so that the given equation becomes (after simplification) $sqrtx-1=2$, implying $x=5$, contradicting our assumption that $5 > x$. Thus, there cannot be any solution in this case.

Hence, the solution are $5 leq x leq 10$.

In general, any time you see an absolute value in a given equation, it's a good idea to divide into cases according to whether each expression in an absolute value is negative or non-negative.