|x-4|<1
the answer my textbook gives are 3 and 5
but that is equal to one can someone explain this please?
To solve this inequality you have to consider two cases by using the definition of the absolut value of a number:
(1) suppose $\displaystyle x-4\geq 0$ therefore the inequality to solve is $\displaystyle x-4<1$
(2) suppose $\displaystyle x-4<0$ therefore the inequality to solve is $\displaystyle -(x-4)<1$
If you summarize the solutions of (1) and (2) then you'll get the solution of the original inequality.
I agree the method Archie Meade provided is much easier, but to use the way I proposed:
(1)$\displaystyle x-4<1 \Leftrightarrow x<5$
(2)$\displaystyle -(x-4)<1 \Leftrightarrow x-4>-1 \Leftrightarrow x>3$
Therefore the solution is $\displaystyle 3<x<5$.
Perhaps you were confused by the way the answer was written? Is the answer in the textbook written with those exact words, "3 and 5", or does the text use notation like $\displaystyle (3, 5)$?
If it is the latter, you should be aware that this notation represents an interval of real numbers. The first number, 3, represents the left endpoint of the interval, and the second number, 5, represents the right endpoint of the interval. $\displaystyle (3, 5)$ denotes the set of numbers strictly between 3 and 5, that is the set of $\displaystyle x$ such that $\displaystyle 3<x<5$. This is called an open interval.
A similar notation, using brackets instead of parentheses, is used to indicate intervals where the endpoints are included. $\displaystyle [3, 5]$ would mean the set of numbers between 3 and 5, including 3 and 5 themselves, or $\displaystyle 3\le x\le 5$. This is called a closed interval.
It is also possible to have half-open intervals like $\displaystyle [3, 5)$. This would translate to $\displaystyle 3\le x<5$.
The correct answer to your problem would be, in interval notation, $\displaystyle (3, 5)$, as others have demonstrated.