Question
Solve the following systems of linear inequations graphically: $2\text{x}+3\text{y}\leq6,3\text{x}+2\text{y}\leq6,\text{x}\geq0,\text{y}\geq0$
✓
Answer
We have,
$2\text{x}+3\text{y}\leq6,3\text{x}+2\text{y}\leq6,\text{x}\geq0,\text{y}\geq0$
Converting the given inequation into equations, the inequations reduce to 2x + 3y = 6,
3x + 2y = 6, 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 $\text{x}=\frac{6}{3}=3.$
$\therefore$ 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 3x + 2y 56:
Putting x = 0 in the equation
3x + 2y = 6, we get $\text{y}=\frac{6}{2}=3$
Putting y = 0 in the equation
3x +2y = 6, we get $\text{x}=\frac{6}{2}=2$
$\therefore$ This line 3x + 2y = 6 meets the coordinate axes at (0, 3) and (2, 0). Draw a thick line joining these points. we find that (0, 0) satisfies inequation $2\text{x}+3\text{y}\leq6.$
Region represented by $\text{x}>0$ and $\text{y}\geq0$
Clearly $\text{x}>0$ and $\text{y}\geq0$ represent the first quadrant.
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