Questions: Use the graphical value method to solve the rational inequality. Write the solution set in interval notation. (x-7)/(x+8) ≥ 3

Solution Steps

To solve the rational inequality \(\frac{x-7}{x+8} \geq 3\), we first rearrange it to form a single rational expression by subtracting 3 from both sides. This gives us \(\frac{x-7}{x+8} - 3 \geq 0\). We then find a common denominator and simplify the expression. Next, we determine the critical points by setting the numerator and denominator equal to zero separately. Finally, we test intervals around these critical points to determine where the inequality holds true and express the solution in interval notation.

We start with the inequality: \[ \frac{x-7}{x+8} \geq 3 \] Subtract 3 from both sides to form a single rational expression: \[ \frac{x-7}{x+8} - 3 \geq 0 \]

Combine the terms over a common denominator: \[ \frac{x-7 - 3(x+8)}{x+8} \geq 0 \] Simplify the numerator: \[ \frac{x-7 - 3x - 24}{x+8} \geq 0 \] \[ \frac{-2x - 31}{x+8} \geq 0 \]

Determine the critical points by setting the numerator and denominator equal to zero separately.

1. Numerator: \(-2x - 31 = 0\) \[ x = -\frac{31}{2} = -15.5 \]

2. Denominator: \(x + 8 = 0\) \[ x = -8 \]

The critical points are \(x = -15.5\) and \(x = -8\).

Test the intervals determined by the critical points: \((-\infty, -15.5)\), \((-15.5, -8)\), and \((-8, \infty)\).

• For \(x \in (-\infty, -15.5)\), choose \(x = -16\): \[ \frac{-2(-16) - 31}{-16 + 8} = \frac{32 - 31}{-8} = \frac{1}{-8} < 0 \]

• For \(x \in (-15.5, -8)\), choose \(x = -10\): \[ \frac{-2(-10) - 31}{-10 + 8} = \frac{20 - 31}{-2} = \frac{-11}{-2} = 5.5 > 0 \]

• For \(x \in (-8, \infty)\), choose \(x = 0\): \[ \frac{-2(0) - 31}{0 + 8} = \frac{-31}{8} < 0 \]

The inequality \(\frac{-2x - 31}{x+8} \geq 0\) holds true in the interval \((-15.5, -8)\).

Final Answer