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Q.
Consider the inequality $40 x+20 y \leq 120$, where $x$ and $y$ are whole numbers. Then,
Linear Inequalities
Solution:
Given, $40 x+20 y \leq 120, x$ and $y$ are whole numbers.
To start with, let $x=0$. Then, LHS of given inequality is
$40 x+20 y=40(0)+20 y=20 y$
Thus, we have
$20 y \leq 120 $
or $y \leq 6$
For $x=0$, the corresponding values of $y$ can be $0,1,2,3,4$, 5,6 only. Inthis case, the solutions of given inequality are $(0,0),(0,1),(0,2),(0,3),(0,4),(0,5)$ and $(0,6)$
Similarly, other solutions of given inequality when $x=1,2$ and 3 are $(1,0),(1,1),(1,2),(1,3),(1,4),(2,0),(2,1),(2,2)$, $(3,0)$.
This is shown in above figure.
Statement (b) is incorrect, since $x$ and $y$ are whole numbers. $x$ and $y$ can only take values $\{0,1,2,3 \ldots \infty\}$. Given shaded region shows negative as well as fractional values of $x$ and $y$.
