Solve: $\frac { x - 9 } { x + 2 }

Name: Solve: $\frac { x - 9 } { x + 2 } < 0$ A) $( - 2,9 )$ B) $( - \infty , - 2 ) \cup ( 9 , \infty )$ C) $( - \infty , 9 )$ D) $( 9 , \infty )$
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Question

Solve: $\frac { x - 9 } { x + 2 } < 0$
A) $( - 2,9 )$
B) $( - \infty , - 2 ) \cup ( 9 , \infty )$
C) $( - \infty , 9 )$
D) $( 9 , \infty )$

Quizplus · Accepted Answer

The inequality $\frac{x - 9}{x + 2} < 0$ implies that the numerator and denominator have opposite signs. The critical points are $x = 9$ and $x = -2$, where the expression changes sign. Testing intervals around these points, we find that the expression is negative for $x$ in the interval $(-2, 9)$, which matches option A.

Solve x−9x+2<0\frac { x - 9 } { x + 2 } < 0x+2x−9​<0

Solve $\frac { x - 9 } { x + 2 } < 0$