Suppose the inequality were abs(4-x)+15 >14 instead of abs(4 -x)+ 15 >21. How would the solution change? Explain.?

Because the absolute value function always returns a positive value, the solution changes from being some of the real numbers (x< -2; x> 10) to being all the real numbers $\left(x \in \mathbb{R}\right)$

Explanation:

It looks like we're starting with the equation

$\left\mid 4 - x \right\mid + 15 > 21$

We can subtract 15 from both sides and get:

$\left\mid 4 - x \right\mid + 15 \textcolor{red}{- 15} > 21 \textcolor{red}{- 15}$

$\left\mid 4 - x \right\mid > 6$

at which point we can solve for $x$ and see that we can have x< -2; x> 10

So now let's look at

$\left\mid 4 - x \right\mid + 15 > 14$

and do the same with subtracting 15:

$\left\mid 4 - x \right\mid + 15 \textcolor{red}{- 15} > 14 \textcolor{red}{- 15}$

$\left\mid 4 - x \right\mid > - 1$

Because the absolute value sign will always return a value that is positive, there is no value of $x$ we can put into this inequality that will make $\left\mid 4 - x \right\mid < 0$, let alone $- 1$. And so the solution here is the set of all real numbers, which can be written $x \in \mathbb{R}$