To solve the compound inequality \(5 \geq \frac{4c + 11}{3} > -3\), we'll break it down into two parts and solve each part separately.
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Solving the left part: \[ 5 \geq \frac{4c + 11}{3} \] Multiply both sides by 3 to eliminate the fraction (remember that multiplying by a positive number doesn't change the inequality): \[ 15 \geq 4c + 11 \] Now, subtract 11 from both sides: \[ 15 - 11 \geq 4c \] \[ 4 \geq 4c \] Now, divide both sides by 4: \[ 1 \geq c \] or \[ c \leq 1 \]
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Solving the right part: \[ \frac{4c + 11}{3} > -3 \] Again, multiply both sides by 3: \[ 4c + 11 > -9 \] Subtract 11 from both sides: \[ 4c > -9 - 11 \] \[ 4c > -20 \] Now, divide both sides by 4: \[ c > -5 \]
Combining both results, the compound inequality is: \[ -5 < c \leq 1 \]
This can be written as: \[ c \in (-5, 1] \]
In terms of integers, the integer values for \(c\) that satisfy this inequality are: \[ c = -4, -3, -2, -1, 0, 1 \]
Putting this into a compound inequality with integers: \[ -5 < c \leq 1, \text{ where } c \in {-4, -3, -2, -1, 0, 1} \]
Thus, the final answer is: \[ -5 < c \leq 1 \] with integer solutions as stated above.