Answer
a) $| x-430 | = 20$, minimum weight = $410$ grams, maximum weight = $450$ grams
b) The weight is not acceptable.
Work Step by Step
a) The general absolute value equation for minimum and maximum is $$| x - h | = d,\tag{1}$$ where $h$ is the halfway point and $d$ is the distance of the halfway point from either $min$ or $max$ value.
Here, in the question, $h = 430$ and $d=20$.
Thus, equation $(1)$ can be written $$| x - 430| = 20.\tag{2}$$ We solve equation $(2)$.
i) Equating $x- 430$ to $20$ $$x-430 = 20.$$ Add $430$ on both sides $$x = 450.$$
ii) Equating $x- 430$ to $-20$ $$x-430 = -20$$ Add $430$ on both sides $$x = 410.$$
Hence, the minimum weight is $410$ grams and the maximum weight is $450$ grams.
b) The weight of the soccer ball after reduction due to wear and tear is $$423 - 16=407\text{ grams}.$$ From part a), the minimum acceptable weight is $410$ grams. Since $407$ is less than $410$, this weight is not acceptable.
