Solve the inequality: $\frac { x } { 8 } + \frac { x } { 3 } \geq 17$
A) $x \geq \frac { 8 } { 3 }$
B) $x \geq \frac { 408 } { 11 }$
C) $x \geq \frac { 17 } { 11 }$
D) $x \geq 204$
E) $x \geq 93.5$

Question

Quizplus · Accepted Answer

First, find a common denominator for the fractions, which is 24. The inequality becomes $\frac{3x}{24} + \frac{8x}{24} \geq 17$, simplifying to $\frac{11x}{24} \geq 17$. Multiplying both sides by 24/11 gives $x \geq \frac{408}{11}$.

Solve the Inequality x8+x3≥17\frac { x } { 8 } + \frac { x } { 3 } \geq 178x​+3x​≥17

Solve the Inequality $\frac { x } { 8 } + \frac { x } { 3 } \geq 17$