Solve the compound inequality. Give the answer in interval notation.
-$0 \leq \frac { 2 x + 3 } { 4 }

Question

Solve the compound inequality. Give the answer in interval notation.
-$0 \leq \frac { 2 x + 3 } { 4 } < 3$
A) $\left[ - \frac { 3 } { 2 } , \frac { 9 } { 2 } \right)$
B) $\left[ - \frac { 3 } { 2 } , \frac { 9 } { 2 } \right]$
C) $\left( - \frac { 3 } { 2 } , \frac { 9 } { 2 } \right]$
D) $\left( - \frac { 3 } { 2 } , \frac { 9 } { 2 } \right)$

Quizplus · Accepted Answer

To solve the compound inequality $0 \leq \frac { 2 x + 3 } { 4 } < 3$, first multiply all parts by 4 to eliminate the denominator: $0 \leq 2x + 3 < 12$. Then, subtract 3 from all parts: $-3 \leq 2x < 9$. Finally, divide all parts by 2: $-\frac{3}{2} \leq x < \frac{9}{2}$. This corresponds to the interval notation $\left[ - \frac { 3 } { 2 } , \frac { 9 } { 2 } \right)$.

Solve the Compound Inequality 0≤2x+34<30 \leq \frac { 2 x + 3 } { 4 } < 30≤42x+3​<3

Solve the Compound Inequality $0 \leq \frac { 2 x + 3 } { 4 } < 3$