5.1 Q-5

Question Statement

Graph the solution region of the following system of linear inequalities by shading.

$$\begin{aligned} (i) \quad & 3x - 4y \leq 12, \quad 3x + 2y \geq 3, \quad x + 2y \leq 9 (ii) \quad & 3x - 4y \leq 12, \quad x + 2y \leq 6, \quad x + y \geq 1 (iii) \quad & 2x + y \leq 4, \quad 2x - 3y \geq 12, \quad x + 2y \leq 6 (iv) \quad & 2x + y \leq 10, \quad x + y \leq 7, \quad -2x + y \leq 4 (v) \quad & 2x + 3y \leq 18, \quad 2x + 2y \leq 10, \quad -2x + y \leq 2 \end{aligned}$$

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Background and Explanation

This question involves graphing the solution region of a system of linear inequalities. To do this:

1. Convert each inequality to an equation.
2. Plot the boundary lines corresponding to these equations.
3. Identify the feasible region where all inequalities hold true, and shade that area.

Key concepts involved:

• Linear Inequalities: Represented by boundary lines, either solid (if the inequality includes equal to) or dashed (if it doesn’t).
• Feasible Region: The area where all the inequalities are satisfied.

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Solution

(i) Graphing the inequalities:

1. First Inequality: $3x - 4y \leq 12$

   • Convert to equation: $3x - 4y = 12$
   • Find intercepts:
     • Let $x = 0$: $0 - 4y = 12$ ⟹ $y = -3$. Point: $(0, -3)$.
     • Let $y = 0$: $3x = 12$ ⟹ $x = 4$. Point: $(4, 0)$.

2. Second Inequality: $3x + 2y \geq 3$

   • Convert to equation: $3x + 2y = 3$
   • Find intercepts:
     • Let $x = 0$: $2y = 3$ ⟹ $y = \frac{3}{2}$. Point: $(0, \frac{3}{2})$.
     • Let $y = 0$: $3x = 3$ ⟹ $x = 1$. Point: $(1, 0)$.

3. Third Inequality: $x + 2y \leq 9$

   • Convert to equation: $x + 2y = 9$
   • Find intercepts:
     • Let $x = 0$: $2y = 9$ ⟹ $y = \frac{9}{2} = 4.5$. Point: $(0, 4.5)$.
     • Let $y = 0$: $x = 9$. Point: $(9, 0)$.

Plot the boundary lines for all three equations. The feasible region is where all inequalities are satisfied. The points that lie inside this region (where all inequalities hold true) should be shaded.

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(ii) Graphing the inequalities:

1. First Inequality: $3x - 4y \leq 12$

   • Convert to equation: $3x - 4y = 12$
   • Find intercepts:
     • Let $x = 0$: $0 - 4y = 12$ ⟹ $y = -3$. Point: $(0, -3)$.
     • Let $y = 0$: $3x = 12$ ⟹ $x = 4$. Point: $(4, 0)$.

2. Second Inequality: $x + 2y \leq 6$

   • Convert to equation: $x + 2y = 6$
   • Find intercepts:
     • Let $x = 0$: $2y = 6$ ⟹ $y = 3$. Point: $(0, 3)$.
     • Let $y = 0$: $x = 6$. Point: $(6, 0)$.

3. Third Inequality: $x + y \geq 1$

   • Convert to equation: $x + y = 1$
   • Find intercepts:
     • Let $x = 0$: $y = 1$. Point: $(0, 1)$.
     • Let $y = 0$: $x = 1$. Point: $(1, 0)$.

Plot the boundary lines and shade the feasible region where all inequalities are satisfied.

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(iii) Graphing the inequalities:

1. First Inequality: $2x + y \leq 4$

   • Convert to equation: $2x + y = 4$
   • Find intercepts:
     • Let $x = 0$: $y = 4$. Point: $(0, 4)$.
     • Let $y = 0$: $2x = 4$ ⟹ $x = 2$. Point: $(2, 0)$.

2. Second Inequality: $2x - 3y \geq 12$

   • Convert to equation: $2x - 3y = 12$
   • Find intercepts:
     • Let $x = 0$: $-3y = 12$ ⟹ $y = -4$. Point: $(0, -4)$.
     • Let $y = 0$: $2x = 12$ ⟹ $x = 6$. Point: $(6, 0)$.

3. Third Inequality: $x + 2y \leq 6$

   • Convert to equation: $x + 2y = 6$
   • Find intercepts:
     • Let $x = 0$: $2y = 6$ ⟹ $y = 3$. Point: $(0, 3)$.
     • Let $y = 0$: $x = 6$. Point: $(6, 0)$.

Plot all the boundary lines and shade the region that satisfies all inequalities.

Rest are done in similar manner

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Key Formulas or Methods Used

• To graph linear inequalities, we convert them into linear equations.
• Identify the intercepts by substituting $x = 0$ and $y = 0$ into the equation.
• Use the shading method to indicate the solution region based on the inequality signs (less than, greater than).

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Summary of Steps

1. Convert each inequality to an equation.
2. Find the intercepts by substituting $x = 0$ and $y = 0$.
3. Plot the boundary lines for each inequality.
4. Identify the feasible region where all inequalities are satisfied.
5. Shade the region of the graph where all inequalities hold true.

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