Question

solve the rational inequality and graph the solution set on a real number line. express the solution set in interval notation.
\\(\frac{x + 8}{x - 3} > 0\\)
move all terms to the left side of the inequality and then define the left side of the inequality as \\(f(x)\\). then list the intervals determined by the real zeros of \\(f\\) and the real numbers for which \\(f\\) is undefined. complete the following table.

interval	sign
\\((-8, 3)\\)	negative
\\((3, \infty)\\)	positive

(type your answers in interval notation. use ascending order.)
solve the inequality. what is the solution set? select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the solution set is \\((-\infty, -8) \cup (3, \infty)\\). (simplify your answer. type your answer in interval notation. type an exact answer, using radicals as needed. use integers or fractions for any numbers in the expression.)
b. the solution set is the empty set.
choose the correct graph below.
a. graph a
b. graph b
c. graph c
d. graph d
e. graph e
f. graph f

Explanation:

Step1: Find critical points

To solve the rational inequality \(\frac{x + 8}{x - 3}>0\), we first find the critical points by setting the numerator and denominator equal to zero.
For the numerator \(x + 8 = 0\), we get \(x=-8\).
For the denominator \(x - 3 = 0\), we get \(x = 3\).
These critical points divide the real number line into three intervals: \((-\infty,-8)\), \((-8,3)\), and \((3,\infty)\).

Step2: Test intervals

• For the interval \((-\infty,-8)\), let's choose a test point, say \(x=-9\). Substitute into \(\frac{x + 8}{x - 3}\): \(\frac{-9 + 8}{-9 - 3}=\frac{-1}{-12}=\frac{1}{12}>0\). So the sign of the expression in this interval is positive.
• For the interval \((-8,3)\), let's choose a test point, say \(x = 0\). Substitute into \(\frac{x + 8}{x - 3}\): \(\frac{0+8}{0 - 3}=\frac{8}{-3}=-\frac{8}{3}<0\). So the sign of the expression in this interval is negative.
• For the interval \((3,\infty)\), let's choose a test point, say \(x = 4\). Substitute into \(\frac{x + 8}{x - 3}\): \(\frac{4+8}{4 - 3}=\frac{12}{1}=12>0\). So the sign of the expression in this interval is positive.

Step3: Determine solution set

We want to find where \(\frac{x + 8}{x - 3}>0\). From the test intervals, the expression is positive in \((-\infty,-8)\) and \((3,\infty)\). We exclude \(x=-8\) and \(x = 3\) because at \(x=-8\) the numerator is zero (making the expression zero, not greater than zero) and at \(x = 3\) the denominator is zero (making the expression undefined).

Answer:

The solution set in interval notation is \((-\infty,-8)\cup(3,\infty)\)