Find the solution set for the nonlinear inequality $\frac { 2 x + 1 } { x - 4 } \leq 5$ .
A) $( - \infty , - 4 ) \cup ( 7 , \infty )$
B) $( - \infty , - 7 ) \cup [ 4 , \infty )$
C) $( - \infty , 4 ) \cup [ 7 , \infty )$
D) $( - \infty , 4 ]$
E) $( - \infty , 7 )$

Question

Quizplus · Accepted Answer

To solve the inequality $\frac { 2 x + 1 } { x - 4 } \leq 5$, first bring all terms to one side: $\frac { 2 x + 1 } { x - 4 } - 5 \leq 0$, which simplifies to $\frac { 2 x + 1 - 5(x - 4) } { x - 4 } \leq 0$, or $\frac { 2 x + 1 - 5x + 20 } { x - 4 } \leq 0$, leading to $\frac { -3x + 21 } { x - 4 } \leq 0$. This simplifies further to $\frac { 3(7 - x) } { x - 4 } \leq 0$. The critical points are x = 7 and x = 4. Testing intervals around these points shows that the solution set is $( - \infty , 4 ) \cup [ 7 , \infty )$.

Find the Solution Set for the Nonlinear Inequality 2x+1x−4≤5\frac { 2 x + 1 } { x - 4 } \leq 5x−42x+1​≤5

Find the Solution Set for the Nonlinear Inequality $\frac { 2 x + 1 } { x - 4 } \leq 5$