Solve: $\frac { 8 } { x - 3 } \leq 3$
A) $( - \infty , 3 ) \cup \left[ \frac { 17 } { 3 } , \infty \right)$
B) $\left[ \frac { 17 } { 3 } , \infty \right)$
C) $\left( - \infty , \frac { 17 } { 3 } \right]$
D) $\left( 3 , \frac { 17 } { 3 } \right]$

Question

Quizplus · Accepted Answer

To solve the inequality $\frac { 8 } { x - 3 } \leq 3$, first bring all terms to one side: $\frac { 8 } { x - 3 } - 3 \leq 0$. Simplifying, we get $\frac { 8 - 3(x - 3) } { x - 3 } \leq 0$, which simplifies to $\frac { 8 - 3x + 9 } { x - 3 } \leq 0$, or $\frac { 17 - 3x } { x - 3 } \leq 0$. This inequality is satisfied when $x < \frac { 17 } { 3 }$ but not equal to 3, because the denominator cannot be zero. Thus, the solution includes all values less than 3 and values greater than or equal to $\frac { 17 } { 3 }$, excluding 3 itself.

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