To solve for \( c \) in the inequalities:
- \(\frac{2c - 11}{3} \geq 3\)
- \(\frac{3c - 2}{4} < 4\)
we'll solve each inequality separately.
For the first inequality:
\[ \frac{2c - 11}{3} \geq 3 \]
Multiply both sides by 3 (to eliminate the fraction):
\[ 2c - 11 \geq 9 \]
Add 11 to both sides:
\[ 2c \geq 20 \]
Divide by 2:
\[ c \geq 10 \]
For the second inequality:
\[ \frac{3c - 2}{4} < 4 \]
Multiply both sides by 4:
\[ 3c - 2 < 16 \]
Add 2 to both sides:
\[ 3c < 18 \]
Divide by 3:
\[ c < 6 \]
Combine the results:
We have two inequalities:
- \( c \geq 10 \)
- \( c < 6 \)
When combined as a compound inequality, since these two inequalities do not overlap (i.e., there are no values of \( c \) that satisfy both conditions at the same time), we write the solution in terms of \( c \):
Final Answer:
\( c \geq 10 \) or \( c < 6 \)
As a compound inequality with integers, the solution can be represented as:
\[ c \in (-\infty, 6) \cup [10, \infty) \]
In terms of integer solutions explicitly:
\[ c < 6 \quad \text{or} \quad c \geq 10 \]
This is the complete solution for \( c \).