Solve the rational inequality. Express the solution in interval notation.
-$$\frac { x + 15 } { x + 9 }

Question

Solve the rational inequality. Express the solution in interval notation.
-$$\frac { x + 15 } { x + 9 } < 3$$
A) $( - \infty , - 9 ) \cup ( 9 , \infty )$
B) $( - 9 , - 6 )$
C) $( - \infty , - 9 ) \cup ( - 6 , \infty )$
D) $( - \infty , - 6 ) \cup ( 9 , \infty )$

Quizplus · Accepted Answer

First, solve the inequality $\frac{x + 15}{x + 9} < 3$. Subtract 3 from both sides and get a common denominator: $\frac{x + 15}{x + 9} - \frac{3(x + 9)}{x + 9} < 0$, simplifying to $\frac{x + 15 - 3x - 27}{x + 9} < 0$, which simplifies further to $\frac{-2x - 12}{x + 9} < 0$. Factoring out -2 gives $\frac{-2(x + 6)}{x + 9} < 0$. The critical points are x = -6 and x = -9. Testing intervals around these points shows the inequality is satisfied for $x < -9$ and $-6 < x < -9$, but due to the nature of the inequality (strict inequality), x cannot equal -9. Thus, the solution in interval notation is $(-\infty, -9) \cup (-6, \infty)$.

Solve the Rational Inequality x+15x+9<3\frac { x + 15 } { x + 9 } < 3x+9x+15​<3

Solve the Rational Inequality $\frac { x + 15 } { x + 9 } < 3$