Solve the inequality. Write the solution set in interval notation.
-$$\frac { 2 x } { x + 7 }

Question

Solve the inequality. Write the solution set in interval notation.
-$$\frac { 2 x } { x + 7 } < x$$
A) $( - 7 , - 5 ) \cup ( 0 , \infty )$
B) $( - \infty , 5 ) \cup ( 7 , \infty )$
C) $( - \infty , - 7 ) \cup ( - 5,0 )$
D) $( - \infty , - 7 ) \cup ( 0 , \infty )$

Quizplus · Accepted Answer

To solve the inequality $\frac{2x}{x+7} < x$, first bring all terms to one side: $\frac{2x}{x+7} - x < 0$. Simplifying, we get $\frac{2x - x(x+7)}{x+7} < 0$, which simplifies to $\frac{-x^2 + 2x + 7x}{x+7} < 0$, and further to $\frac{-x^2 + 9x}{x+7} < 0$. Factoring the numerator gives $\frac{-x(x - 9)}{x+7} < 0$. The critical points are $x = 0$, $x = 9$, and $x = -7$. Testing intervals around these points, we find that the solution set is $x \in (-7, 0) \cup (9, \infty)$, but due to a mistake in simplification, the correct approach should have been to simplify directly to $\frac{2x - x(x+7)}{x+7} < 0$ correctly as $\frac{2x - x^2 - 7x}{x+7} < 0$, leading to $\frac{-x^2 - 5x}{x+7} < 0$. Solving this correctly would involve finding the roots of the numerator and analyzing the sign changes around the roots and the undefined point $x = -7$. The correct interval considering the nature of the inequality and the critical points would be $(-7, -5) \cup (0, \infty)$, acknowledging an error in the initial step-by-step simplification.

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Solve the Inequality $\frac { 2 x } { x + 7 } < x$