Question

solve the equation: (|x + 3|-4 = 1), (2|3x - 1|=5), (8|4x - 3|=-24), (5|3x + 2|-8=-2). 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: Solve the equation \(|x + 3|-4=1\)

Isolate the absolute - value term. Add 4 to both sides of the equation:
\(|x + 3|=1 + 4\), so \(|x + 3|=5\).
Then we have two cases:
Case 1: \(x+3 = 5\), then \(x=5 - 3=2\).
Case 2: \(x + 3=-5\), then \(x=-5 - 3=-8\).

Step2: Solve the equation \(2|3x - 1|=8\)

First, divide both sides by 2: \(|3x - 1| = 4\).
We have two cases:
Case 1: \(3x-1 = 4\), then \(3x=4 + 1=5\), and \(x=\frac{5}{3}\).
Case 2: \(3x - 1=-4\), then \(3x=-4 + 1=-3\), and \(x=-1\).

Step3: Solve the equation \(8|4x - 3|=-24\)

Divide both sides by 8: \(|4x - 3|=-3\).
Since the absolute - value of any real number is non - negative (\(|a|\geq0\) for all real \(a\)), this equation has no solution.

Step4: Solve the equation \(5|3x + 2|-8=-2\)

Add 8 to both sides: \(5|3x + 2|=-2 + 8 = 6\).
Divide both sides by 5: \(|3x+2|=\frac{6}{5}\).
We have two cases:
Case 1: \(3x + 2=\frac{6}{5}\), then \(3x=\frac{6}{5}-2=\frac{6 - 10}{5}=-\frac{4}{5}\), and \(x=-\frac{4}{15}\).
Case 2: \(3x + 2=-\frac{6}{5}\), then \(3x=-\frac{6}{5}-2=-\frac{6 + 10}{5}=-\frac{16}{5}\), and \(x=-\frac{16}{15}\).

Step5: Write an absolute - value equation that has 3 and 11 as its solutions

The mid - point between 3 and 11 is \(m=\frac{3 + 11}{2}=7\).
The distance from the mid - point to either 3 or 11 is \(d=|7 - 3|=4\).
The absolute - value equation is \(|x - 7| = 4\).

Step6: Write and solve an absolute - value equation for the snook length problem

The mid - point between 32 and 36 is \(m=\frac{32+36}{2}=34\).
The distance from the mid - point to either 32 or 36 is \(d=|34 - 32| = 2\).
The absolute - value equation is \(|L - 34|=2\), where \(L\) is the length of the snook.
We have two cases:
Case 1: \(L-34 = 2\), then \(L=34 + 2=36\).
Case 2: \(L - 34=-2\), then \(L=34-2=32\).

Answer:

For \(|x + 3|-4 = 1\), \(x = 2\) or \(x=-8\); for \(2|3x - 1|=8\), \(x=\frac{5}{3}\) or \(x=-1\); for \(8|4x - 3|=-24\), no solution; for \(5|3x + 2|-8=-2\), \(x=-\frac{4}{15}\) or \(x=-\frac{16}{15}\); absolute - value equation with solutions 3 and 11 is \(|x - 7| = 4\); absolute - value equation for snook length is \(|L - 34|=2\) with solutions \(L = 32\) and \(L = 36\)