A fertilizer company produces two types of fertilizers called grade I and grade II. Each of these types is processed through a critical chemical plant unit. The plant has maximum of $\displaystyle 180$ hours availability in a week. Manufacturing one bag of grade I fertilizer requires $\displaystyle 4$ hours in the plant. Manufacturing one bag of grade II fertilizer requires $\displaystyle 10$ hours in the plant. Express this using linear inequalities.

On solving the inequalities $\displaystyle 5x + y \le 100$, $\displaystyle x + y \le 60$, $\displaystyle x \ge 0$, $\displaystyle y \ge 0$, we get the following solution:

An employer recruits experienced $\displaystyle (x)$ and fresh workmen $\displaystyle (y)$ for his under the condition that he cannot employ more than $\displaystyle 11$ people $\displaystyle x$ and $\displaystyle y$ can be related by the inequality.

The solution set of the equations $\displaystyle x + 2 > 0$ and $\displaystyle 2x - 6 > 0$ is

On solving the inequalities; we get $\displaystyle 6x + y \ge 18$, $\displaystyle x + 4y \ge 12$, $\displaystyle 2x + y \ge 10$

Solve for $\displaystyle x$ of the inequalities $\displaystyle 2 \le \frac{3x - 2}{5} \le 4$ where $\displaystyle x \in N$

The common region in the graph of the inequalities $\displaystyle x + y \le 4$, $\displaystyle x - y \le 4$, $\displaystyle x \ge 2$ is