Question
Solve the following systems of linear inequations graphically: $2\text{x}+3\text{y}\leq6,\text{x}+4\text{y}\leq4,\text{x}\geq0,\text{y}\geq0$
✓
Answer
We have,
$2\text{x}+3\text{y}\leq6,\text{x}+4\text{y}\leq4,\text{x}\geq0,\text{y}\geq0$
Converting the given inequation into equations, the inequations reduce to 2x + 3y = 6,
x + 4y = 4, x = 0 and y = 0.
Region represented by $2\text{x}+3\text{y}\leq6$
Putting x = 0 inequation 2x + 3y = 6
we get $\text{y}=\frac{6}{3}=2$
Putting y = 0 in the equation 2x + 3y = 6,
we get $\text{x}=\frac{6}{2}=3.$
$\therefore$ This line 2x + 3y = 6 meets the coordinate axes at (0, 2) and (3, 0). Draw a thick line joining these points. We find that (0, 0) satisfies inequation 2x + 3y 36.
Region represented by $\text{x}+4\text{y}\leq4$
Putting x = 0 in x + 4y = 4
we get $\text{y}=\frac{4}{4}=1$
Putting y = 0 in x + 4y = 4
we get x = 4
$\therefore$ This line x + 4y = 4 meets the coordinate axes at (0, 1) and (4, 0). Draw a thick line joining these points.
Now putting x = 0, y = 0 in $\text{x}+4\text{y}\leq4$ we get $0\leq4$
Clearly, we find that (0, 0) satisfying in equation $\text{x}+4\text{y}\leq4$
Region represent by $\text{x}\geq0$ and $\text{y}\geq0$
Clearly $\text{x}\geq0$ and $\text{y}\geq0$ represent in first quadrant.
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