How do you solve and graph the inequality abs(4 – v)< 5?

Apr 20, 2015

Let's start from a graph of a function $y = | v |$.
For non-negative $v$ the absolute value of $v$ is itself. So, if $v \ge 0$, $| v | = v$ and our function is the same as $y = v$.
For negative $v$ the absolute value of $v$ is its opposite, $- v$. So, if $v < 0$, $| v | = - v$ and our function is the same as $y = - v$.
That results in this graph of $y = | v |$:
graph{|x| [-10, 10, -5, 5]}

Now let's draw a graph of $y = | 4 - v |$. Since $| 4 - v | = | v - 4 |$, we will draw a graph of $y = | v - 4 |$.
According to principles of graph transformation, a graph of $y = | v - 4 |$ is the result of a shift to the right by $4$ of a graph of a function $= | v |$. Therefore, our graph looks like this:
graph{|x-4| [-10, 10, -5, 5]}

Now, to solve $| 4 - v | < 5$, we have to find all values of $v$, where the graph lies below the horizontal line that intersects the Y-axis at point $y = 5$. Obviously, it's a segment $\left(- 1 , 9\right)$ because, while $v$ is changing from $- 1$ to $9$, the value of $| 4 - v |$ is changing from $5$ down to zero and up to $5$, always staying below the horizontal line $y = 5$.
Outside of this segment, that is if $v \le - 1$ or $y \ge 9$ the value of $| 4 - v |$ is equal or greater than $5$.

Apr 20, 2015

For those interested in purely algebraic solution, here is how to do it.
Since, by definition,
$| X | = X$ if $X \ge 0$ and
$| X | = - X$ if $X < 0$,
we will consider two cases.

Case 1. Seeking solutions that satisfy the inequality
$4 - v \ge 0$ or, equivalently, $v \le 4$.
In this case $| 4 - v | = 4 - v$ and our original inequality looks like this:
$4 - v < 5$.
Solution to this is $v > - 1$.
Combined with the condition $v \le 4$ of this case, we have a segment of values of $v$, where $| 4 - v | < 5$:
$- 1 < v \le 4$

Case 2. Seeking solutions that satisfy the inequality
$4 - v < 0$ or, equivalently, $v > 4$.
In this case $| 4 - v | = - \left(4 - v\right)$ and our original inequality looks like this:
$- \left(4 - v\right) < 5$, that is $- 4 + v < 5$.
Solution to this is $v < 9$.
Combined with the condition $v > 4$ of this case, we have another segment of values of $v$, where $| 4 - v | < 5$:
$4 < v < 9$

Now it's appropriate to combine two segments that represent the solutions of an original inequality into one segment since these segments are adjacent:
$- 1 < v \le 4$ combined with $4 < v < 9$ results in a segment $- 1 < v < 9$
As you see, we get the same solution as using a graph above (which should not be a surprise).