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 $180$ hours availability in a week. Manufacturing one bag of grade I fertilizer requires $4$ hours in the plant. Manufacturing one bag of grade II fertilizer requires $10$ hours in the plant. Express this using linear inequalities.

On solving the inequalities $5x + y $\displaystyle \le$ 100$\displaystyle ,$x + y $\displaystyle \le$ 60$\displaystyle ,$x $\displaystyle \ge$ 0$\displaystyle ,$y $\displaystyle \ge$ 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 $11$\displaystyle people$x$\displaystyle and$y$ can be related by the inequality.

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

On solving the inequalities; we get $6x + y $\displaystyle \ge$ 18$\displaystyle ,$x + 4y $\displaystyle \ge$ 12$\displaystyle \ge$ …" style="color:#cc0000">\displaystyle , $2x + y $\displaystyle \ge$ 10

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

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