Question

write an absolute value equation that has 3 and 11 as its solutions. for a fishing tournament, the minimum length of a snook is 32 inches. the maximum length is 36 inches. write and solve an absolute value equation that has these minimum and maximum lengths as its solutions.

Explanation:

Step1: Find the mid - point for the first case

The mid - point between 3 and 11 is $\frac{3 + 11}{2}=\frac{14}{2}=7$. The distance from 7 to 3 (or 11) is $|7 - 3|=4$. The general form of an absolute - value equation with solutions $x_1$ and $x_2$ is $|x - m|=d$, where $m$ is the mid - point and $d$ is the distance from the mid - point to either solution. So the equation is $|x - 7| = 4$.

Step2: Find the mid - point for the second case

The mid - point between 32 and 36 is $\frac{32+36}{2}=\frac{68}{2}=34$. The distance from 34 to 32 (or 36) is $|34 - 32| = 2$. The absolute - value equation is $|x - 34|=2$.

Step3: Solve the second equation

Case 1: $x−34 = 2$, then $x=34 + 2=36$.
Case 2: $x−34=-2$, then $x=34-2 = 32$.

Answer:

The absolute - value equation with solutions 3 and 11 is $|x - 7| = 4$. The absolute - value equation with solutions 32 and 36 is $|x - 34|=2$.