$\frac { x + 17 } { x + 4 }

Question

$\frac { x + 17 } { x + 4 } < 4$ A) $- 4 < x < \frac { 1 } { 3 }$ B) $x < - 4 \text { or } x > \frac { 1 } { 3 }$ C) $x < \frac { 1 } { 3 } \text { or } x > 4$ D) No solution

Quizplus · Accepted Answer

To solve the inequality $\frac{x + 17}{x + 4} < 4$, first bring all terms to one side: $\frac{x + 17}{x + 4} - 4 < 0$. Simplifying gives $\frac{x + 17 - 4(x + 4)}{x + 4} < 0$, which simplifies to $\frac{-3x + 1}{x + 4} < 0$. The critical points are found by setting the numerator and denominator equal to zero: $x = \frac{1}{3}$ and $x = -4$, respectively. Testing intervals around these points shows that the solution is $x < -4$ or $x > \frac{1}{3}$.

x+17x+4<4\frac { x + 17 } { x + 4 } < 4x+4x+17​<4

$\frac { x + 17 } { x + 4 } < 4$