CBSE Class 11 Mathematics Linear Inequalities Assignment Set 02

VERY SHORT ANSWER TYPE QUESTIONS (1 MARK)

Question 1. Solve \(5x < 24\) when \(x \in N\)
Answer: Given inequality is \(5x < 24\)
Dividing both sides by 5:
\(x < \frac{24}{5}\)
\(x < 4.8\)
Since \(x \in N\) (set of natural numbers), the natural numbers less than 4.8 are 1, 2, 3, and 4.
Solution set = \(\{1, 2, 3, 4\}\)
In simple words: We found numbers belonging to natural numbers that are strictly less than 4.8.

Exam Tip: Always pay close attention to the given replacement set (\(N, Z,\) or \(R\)) before writing the final solution set.

Question 2. Solve \(3x < 11\) when \(x \in Z\)
Answer: Given inequality is \(3x < 11\)
Dividing both sides by 3:
\(x < \frac{11}{3}\)
\(x < 3.67\)
Since \(x \in Z\) (set of integers), the integers less than 3.67 are ..., \(-3, -2, -1, 0, 1, 2, 3\).
Solution set = \(\{\dots, -3, -2, -1, 0, 1, 2, 3\}\)
In simple words: All integers less than 3.67 form our solution set.

Exam Tip: Use ellipsis (\(\dots\)) to indicate an infinite set of integers extending in one direction.

Question 3. Solve \(3 - 2x < 9\) when \(x \in R\)
Answer: Given inequality is \(3 - 2x < 9\)
Subtracting 3 from both sides:
\(-2x < 9 - 3\)
\(-2x < 6\)
Dividing both sides by \(-2\) (and reversing the inequality sign):
\(x > \frac{6}{-2}\)
\(x > -3\)
In simple words: Isolate \(x\) by moving terms, and remember to flip the inequality sign when dividing by a negative number.

Exam Tip: Multiplying or dividing an inequality by a negative number reverses the direction of the inequality sign.

Question 4. Show the graph of the solution of \(2x - 3 > x - 5\) on number line.
Answer: Given inequality is \(2x - 3 > x - 5\)
Subtracting \(x\) from both sides:
\(x - 3 > -5\)
Adding 3 to both sides:
\(x > -5 + 3\)
\(x > -2\)
To represent on a number line, draw a directed line, mark an open circle at \(-2\) (since \(-2\) is not included), and shade the line to the right towards positive infinity.

In simple words: Simplify the inequality to \(x > -2\) and draw an open circle at \(-2\) pointing right.

Exam Tip: Use an open circle for strict inequalities (\(<\) or \(>\)) and a solid filled circle for slack inequalities (\(\le\) or \(\ge\)).

Question 5. Solve \(5x - 8 \ge 8\) graphically
Answer: Given inequality is \(5x - 8 \ge 8\)
Adding 8 to both sides:
\(5x \ge 16\)
Dividing both sides by 5:
\(x \ge \frac{16}{5}\)
\(x \ge 3.2\)
Graphical representation on a number line involves drawing a solid circle at \(3.2\) and shading the region to the right towards infinity.

In simple words: Solve for \(x\) to get \(x \ge 3.2\), then plot a solid dot at 3.2 and shade to the right.

Exam Tip: A solid dot indicates that the boundary point is included in the solution set.

Question 6. Solve \(\frac{1}{x - 2} \le 0\)
Answer: Given inequality is \(\frac{1}{x - 2} \le 0\)
For the fraction to be less than or equal to zero, since the numerator is positive (1), the denominator must be strictly less than zero (since denominator cannot be zero).
\(x - 2 < 0\)
\(x < 2\)
Also, \(x \neq 2\) is automatically satisfied since \(x < 2\).
Solution in interval form: \(x \in (-\infty, 2)\)
In simple words: A fraction with a positive numerator is negative only when its denominator is negative.

Exam Tip: Never cross-multiply an inequality with an unknown variable in the denominator unless you are certain of its sign.

Question 7. Solve \(0 < \frac{-x}{3} < 1\)
Answer: Given inequality is \(0 < \frac{-x}{3} < 1\)
Multiply all parts by 3:
\(0 < -x < 3\)
Multiply all parts by \(-1\) (and reverse all inequality signs):
\(0 > x > -3\)
Which can be written as:
\(-3 < x < 0\)
Solution in interval form: \((-3, 0)\)
In simple words: Multiply across by 3 and then by \(-1\), remembering to flip inequality directions.

Exam Tip: When multiplying a compound inequality by a negative number, all inequality signs must be reversed.

Write the solution in the form of intervals for \(x \in R\). for Questions 8 to 10

Question 8. \(\frac{2}{x - 3} < 0\)
Answer: Given inequality is \(\frac{2}{x - 3} < 0\)
Since the numerator 2 is positive, the fraction is less than zero if and only if the denominator is negative:
\(x - 3 < 0\)
\(x < 3\)
Expressed in interval notation: \((-\infty, 3)\)
In simple words: The numerator is positive, so the denominator must be negative, giving \(x < 3\).

Exam Tip: Inequalities involving strict less than zero with a positive constant numerator simplify to denominator \(< 0\).

Question 9. \(-3 \le -3x + 2 < 4\)
Answer: Given compound inequality is \(-3 \le -3x + 2 < 4\)
Subtract 2 from all parts:
\(-3 - 2 \le -3x < 4 - 2\)
\(-5 \le -3x < 2\)
Divide all parts by \(-3\) (and reverse all inequality signs):
\(\frac{-5}{-3} \ge x > \frac{2}{-3}\)
\(\frac{5}{3} \ge x > -\frac{2}{3}\)
Rewriting in standard order from smaller to larger:
\(-\frac{2}{3} < x \le \frac{5}{3}\)
Solution in interval form: \((-\frac{2}{3}, \frac{5}{3}]\)
In simple words: Isolate \(x\) in the middle by subtracting 2 and dividing by \(-3\), then flip the signs.

Exam Tip: Pay careful attention to parentheses versus square brackets in interval notation (\([\dots]\) for inclusive, \((\dots)\) for strict).

Question 10. \(3 + 2x > -4 - 3x\)
Answer: Given inequality is \(3 + 2x > -4 - 3x\)
Add \(3x\) to both sides:
\(3 + 5x > -4\)
Subtract 3 from both sides:
\(5x > -4 - 3\)
\(5x > -7\)
Divide both sides by 5:
\(x > -\frac{7}{5}\)
Solution in interval form: \((-\frac{7}{5}, \infty)\)
In simple words: Collect all \(x\) terms on one side and constants on the other, then solve for \(x\).

Exam Tip: Always express interval solutions from left to right on the real number line (smaller value to larger value).

Question 11. Draw the graph of the solution set of \(x + y \ge 4\).
Answer: 1. First, draw the line corresponding to the equation \(x + y = 4\).
2. Find intercept points: when \(x = 0\), \(y = 4\) \((0, 4)\); when \(y = 0\), \(x = 4\) \((4, 0)\).
3. Draw a solid line passing through \((0, 4)\) and \((4, 0)\) because the inequality includes '\(\ge\)'.
4. Test the origin \((0, 0)\): \(0 + 0 \ge 4\) (False). Therefore, shade the half-plane that does not contain the origin (the upper-right half-plane).

In simple words: Draw the solid boundary line \(x + y = 4\) and shade the region away from the origin.

Exam Tip: Always test the point \((0, 0)\) to quickly determine which side of the line to shade, provided the line does not pass through the origin.

Question 12. Draw the graph of the solution set of \(x \le y\)
Answer: 1. Consider the boundary line \(y = x\).
2. Draw a solid line passing through the origin \((0, 0)\), \((1, 1)\), and \((2, 2)\).
3. Since the line passes through the origin, choose a test point not on the line, for example \((1, 0)\).
4. Substitute \((1, 0)\) into \(x \le y \implies 1 \le 0\) (False).
5. Therefore, shade the region that does not contain \((1, 0)\), which is the upper-left half-plane containing \((0, 1)\).

In simple words: Plot the line \(y = x\) and shade the region where \(y\) values are greater than or equal to \(x\).

Exam Tip: When a boundary line passes through \((0, 0)\), always pick an easily identifiable test point like \((1, 0)\) or \((0, 1)\) to determine the correct half-plane.

SHORT ANSWER TYPE QUESTIONS (4 MARKS)

Solve the inequalities for real \(x\)

Question 13. \(\frac{2x - 3}{4} + 9 \ge 3 + \frac{4x}{3}\)
Answer: Given inequality is:
\(\frac{2x - 3}{4} + 9 \ge 3 + \frac{4x}{3}\)
Subtracting 3 from both sides and simplifying the constants:
\(\frac{2x - 3}{4} + 6 \ge \frac{4x}{3}\)
Taking LCM on the left side:
\(\frac{2x - 3 + 24}{4} \ge \frac{4x}{3}\)
\(\frac{2x + 21}{4} \ge \frac{4x}{3}\)
Cross-multiplying (since 12 is positive):
\(3(2x + 21) \ge 4(4x)\)
\(6x + 63 \ge 16x\)
Rearranging terms:
\(63 \ge 16x - 6x\)
\(63 \ge 10x\)
\(10x \le 63\)
\(x \le \frac{63}{10}\)
Solution in interval form: \((-\infty, \frac{63}{10}]\)
In simple words: Simplify the numbers, take the LCM to eliminate denominators, and isolate \(x\).

Exam Tip: Eliminating fractions early by multiplying the entire inequality by the LCM of denominators prevents calculation errors.

Question 14. \(\frac{2x + 3}{4} - 3 < \frac{x - 4}{3} - 2\)
Answer: Given inequality is:
\(\frac{2x + 3}{4} - 3 < \frac{x - 4}{3} - 2\)
Adding 3 to both sides:
\(\frac{2x + 3}{4} < \frac{x - 4}{3} - 2 + 3\)
\(\frac{2x + 3}{4} < \frac{x - 4}{3} + 1\)
Simplify the right side by taking LCM:
\(\frac{2x + 3}{4} < \frac{x - 4 + 3}{3}\)
\(\frac{2x + 3}{4} < \frac{x - 1}{3}\)
Cross-multiplying by 12:
\(3(2x + 3) < 4(x - 1)\)
\(6x + 9 < 4x - 4\)
Subtracting \(4x\) from both sides:
\(2x + 9 < -4\)
Subtracting 9 from both sides:
\(2x < -13\)
\(x < -\frac{13}{2}\)
Solution in interval form: \((-\infty, -\frac{13}{2})\)
In simple words: Simplify constant terms, combine fractions using LCM, and solve for \(x\).

Exam Tip: Always be careful with negative signs when expanding brackets in numerators during cross-multiplication.

Question 15. \(-5 \le \frac{2 - 3x}{4} \le 9\)
Answer: Given compound inequality is:
\(-5 \le \frac{2 - 3x}{4} \le 9\)
Multiply all parts by 4:
\(-20 \le 2 - 3x \le 36\)
Subtract 2 from all parts:
\(-20 - 2 \le -3x \le 36 - 2\)
\(-22 \le -3x \le 34\)
Divide all parts by \(-3\) and reverse all inequality signs:
\(\frac{-22}{-3} \ge x \ge \frac{34}{-3}\)
\(\frac{22}{3} \ge x \ge -\frac{34}{3}\)
Rewriting in standard ascending order:
\(-\frac{34}{3} \le x \le \frac{22}{3}\)
Solution in interval form: \([-\frac{34}{3}, \frac{22}{3}]\)
In simple words: Clear the denominator, isolate the middle term by subtracting 2, then divide by \(-3\) while flipping the inequality symbols.

Exam Tip: Remember to reverse BOTH inequality symbols when dividing a compound inequality by a negative number.

Question 16. \(|x - 2| \ge 5\)
Answer: Given absolute inequality is \(|x - 2| \ge 5\)
By absolute value properties (\(|X| \ge a \implies X \le -a\) or \(X \ge a\)):
\(x - 2 \le -5 \quad \text{or} \quad x - 2 \ge 5\)
Solving the first inequality:
\(x \le -5 + 2 \implies x \le -3\)
Solving the second inequality:
\(x \ge 5 + 2 \implies x \ge 7\)
Combining both solutions using union:
\(x \in (-\infty, -3] \cup [7, \infty)\)
In simple words: Split the absolute inequality into two separate cases connected by 'or'.

Exam Tip: Remember that inequality with 'greater than or equal to' splits into an 'or' compound statement (\(\le -a\) or \(\ge a\)).

Question 17. \(|4 - x| + 1 < 3\)
Answer: Given inequality is \(|4 - x| + 1 < 3\)
Subtracting 1 from both sides:
\(|4 - x| < 2\)
Using absolute value property (\(|X| < a \implies -a < X < a\)):
\(-2 < 4 - x < 2\)
Subtracting 4 from all parts:
\(-2 - 4 < -x < 2 - 4\)
\(-6 < -x < -2\)
Multiplying all parts by \(-1\) and reversing the inequality signs:
\(6 > x > 2\)
Rewriting in standard ascending order:
\(2 < x < 6\)
Solution in interval form: \((2, 6)\)
In simple words: Isolate the absolute value term, convert it into a compound inequality, and solve for \(x\).

Exam Tip: Inequalities with 'less than' translate into a single combined 'and' inequality (\(-a < X < a\)).

Question 18. \(\frac{3}{x - 2} < 1\)
Answer: Given inequality is \(\frac{3}{x - 2} < 1\)
Subtract 1 from both sides:
\(\frac{3}{x - 2} - 1 < 0\)
\(\frac{3 - (x - 2)}{x - 2} < 0\)
\(\frac{3 - x + 2}{x - 2} < 0\)
\(\frac{5 - x}{x - 2} < 0\)
Multiply numerator and denominator by \(-1\) (and flip sign):
\(\frac{x - 5}{x - 2} > 0\)
Using the wavy curve method (critical points are 2 and 5):
\(x < 2\) or \(x > 5\)
Solution in interval form: \((-\infty, 2) \cup (5, \infty)\)
In simple words: Bring 1 to the left side, simplify into a single rational expression, and use critical points.

Exam Tip: Never cross-multiply algebraic denominators when they can be positive or negative; instead, bring all terms to one side.

Question 19. \(\frac{x - 1}{x - 5} > \frac{1}{2}\)
Answer: Given inequality is \(\frac{x - 1}{x - 5} > \frac{1}{2}\)
Subtract \(\frac{1}{2}\) from both sides:
\(\frac{x - 1}{x - 5} - \frac{1}{2} > 0\)
\(\frac{2(x - 1) - (x - 5)}{2(x - 5)} > 0\)
\(\frac{2x - 2 - x + 5}{2(x - 5)} > 0\)
\(\frac{x + 3}{2(x - 5)} > 0\)
Critical points are \(x = -3\) and \(x = 5\).
Analyzing intervals using sign scheme: \(x < -3\) or \(x > 5\)
Solution in interval form: \((-\infty, -3) \cup (5, \infty)\)
In simple words: Transpose terms to make the right side zero, simplify into a single fraction, and find intervals where the expression is positive.

Exam Tip: Always exclude values that make any denominator zero (\(x \neq 5\) here).

Question 20. \(\frac{x + 3}{x - 2} > 0\)
Answer: Given inequality is \(\frac{x + 3}{x - 2} > 0\)
The critical points are \(-3\) and \(2\).
For the quotient to be positive, both numerator and denominator must have the same sign:
Case 1: Both positive
\(x + 3 > 0\) and \(x - 2 > 0 \implies x > -3\) and \(x > 2 \implies x > 2$
Case 2: Both negative
\(x + 3 < 0\) and \(x - 2 < 0 \implies x < -3\) and \(x < 2 \implies x < -3$
Combining both cases: \(x < -3\) or \(x > 2\)
Solution in interval form: \((-\infty, -3) \cup (2, \infty)\)
In simple words: A fraction is positive when numerator and denominator have identical signs (both positive or both negative).

Exam Tip: The sign of a rational function changes only across its critical points (roots of numerator and denominator).

Question 21. \(x + 2 \le 5\), \(3x - 4 > -2 + x\)
Answer: Solve each inequality separately.
First inequality:
\(x + 2 \le 5 \implies x \le 3\)
Second inequality:
\(3x - 4 > -2 + x\)
\(3x - x > -2 + 4\)
\(2x > 2 \implies x > 1\)
Combining both solutions: \(1 < x \le 3\)
Solution in interval form: \((1, 3]\)
In simple words: Solve each inequality independently, then find the overlapping intersection region.

Exam Tip: A system of simultaneous inequalities requires the intersection (common values) of all individual solution sets.

Question 22. \(3x - 7 > 2(x - 6)\), \(6 - x > 11 - 2x\)
Answer: Solve each inequality separately.
First inequality:
\(3x - 7 > 2x - 12\)
\(3x - 2x > -12 + 7\)
\(x > -5\)
Second inequality:
\(6 - x > 11 - 2x\)
\(-x + 2x > 11 - 6\)
\(x > 5\)
Intersection of \(x > -5\) and \(x > 5\) is \(x > 5\)
Solution in interval form: \((5, \infty)\)
In simple words: Solve both linear inequalities and take the common range satisfying both.

Exam Tip: When two inequalities overlap, the stricter condition (\(x > 5\) over \(x > -5\)) forms the final solution.

Question 23. The water acidity in a pool is considered normal when the average PH reading of three daily measurements is between 7.2 and 7.8. If the first two PH readings are 7.48 and 7.85, find the range of PH value for the third reading that will result in the acidity level being normal.
Answer: Let the third PH reading be \(x\).
The three readings are 7.48, 7.85, and \(x\).
The average is given by:
\(\text{Average} = \frac{7.48 + 7.85 + x}{3}\)
Given that the average is between 7.2 and 7.8:
\(7.2 < \frac{7.48 + 7.85 + x}{3} < 7.8\)
\(7.2 < \frac{15.33 + x}{3} < 7.8\)
Multiply all parts by 3:
\(21.6 < 15.33 + x < 23.4\)
Subtract 15.33 from all parts:
\(21.6 - 15.33 < x < 23.4 - 15.33\)
\(6.27 < x < 8.07\)
In simple words: Set up an inequality for the average of the three readings and solve for the unknown third reading \(x\).

Exam Tip: Word problems on averages translate directly into compound inequalities involving the sum divided by the number of observations.

Question 24. While drilling a hole in the earth, it was found that the temperature (\(T\) °C) at \(x\) km below the surface of the earth was given by \(T = 30 + 25(x - 3)\), when \(3 \le x \le 15\). Between which depths will the temperature be between 200°C and 300°C?
Answer: Given temperature formula is \(T = 30 + 25(x - 3)\).
We need to find the depths \(x\) for which \(200 < T < 300\):
\(200 < 30 + 25(x - 3) < 300\)
Subtract 30 from all parts:
\(170 < 25(x - 3) < 270\)
Divide all parts by 25:
\(\frac{170}{25} < x - 3 < \frac{270}{25}\)
\(6.8 < x - 3 < 10.8\)
Add 3 to all parts:
\(6.8 + 3 < x < 10.8 + 3\)
\(9.8 < x < 13.8\)
In simple words: Substitute the temperature formula into the given range and solve for depth \(x\).

Exam Tip: In applied problems, always verify that your calculated solution interval lies within the domain specified in the question (\(3 \le x \le 15\)).

Solve the following systems of inequalities graphically : (Questions 25, 26)

Question 25. \(x + y > 6\), \(2x - y > 0\)
Answer: 1. Consider boundary line 1: \(x + y = 6\). Intercepts are \((0, 6)\) and \((6, 0)\). Since the inequality is strict ('\(>\)'), draw a dashed line.
2. Consider boundary line 2: \(2x - y = 0 \implies y = 2x\). Passes through \((0, 0)\), \((1, 2)\), \((2, 4)\). Draw a dashed line.
3. Test origin \((0, 0)\) for \(x + y > 6 \implies 0 > 6\) (False). Shade region away from origin.
4. Test point \((1, 0)\) for \(2x - y > 0 \implies 2(1) - 0 > 0\) (True). Shade the region containing \((1, 0)\).
5. The common overlapping region of both inequalities represents the solution set.

In simple words: Draw dashed boundary lines for both strict inequalities and shade their overlapping half-planes.

Exam Tip: Use dashed lines for strict inequalities (\(<\) or \(>\)) to indicate that points on the boundary line are excluded.

Question 26. \(3x + 4y \le 60\), \(x + 3y \le 30\), \(x \ge 0\), \(y \ge 0\)
Answer: 1. Line 1: \(3x + 4y = 60\). Intercepts: \((0, 15)\) and \((20, 0)\). Solid line.
2. Line 2: \(x + 3y = 30\). Intercepts: \((0, 10)\) and \((30, 0)\). Solid line.
3. Non-negativity constraints \(x \ge 0\) and \(y \ge 0\) restrict the graph to the first quadrant.
4. Testing \((0, 0)\) for both inequalities gives true statements, so shade towards the origin for both.
5. The feasible region is bounded by the axes and the intersection of the two half-planes in the first quadrant.

In simple words: Plot the boundary lines, restrict to the first quadrant, and shade the common region towards the origin.

Exam Tip: Constraints \(x \ge 0\) and \(y \ge 0\) always confine the graphical solution to the first quadrant.

LONG ANSWER TYPE QUESTIONS (6 MARKS)

Solve the system of inequalities for real \(x\)

Question 27. \(\frac{5x}{4} + \frac{3x}{8} > \frac{39}{8}\) and \(\frac{2x - 1}{12} - \frac{x - 1}{3} < \frac{3x + 1}{4}\)
Answer: Solve each inequality separately.
First inequality:
\(\frac{5x}{4} + \frac{3x}{8} > \frac{39}{8}\)
Multiply through by 8:
\(10x + 3x > 39\)
\(13x > 39 \implies x > 3\)

Second inequality:
\(\frac{2x - 1}{12} - \frac{x - 1}{3} < \frac{3x + 1}{4}\)
Multiply through by 12 (LCM):
\((2x - 1) - 4(x - 1) < 3(3x + 1)\)
\(2x - 1 - 4x + 4 < 9x + 3\)
\(-2x + 3 < 9x + 3\)
Subtracting 3 from both sides:
\(-2x < 9x\)
Add \(2x\) to both sides:
\(0 < 11x \implies x > 0\)
Combining both solutions (\(x > 3\) and \(x > 0\)):
The intersection is \(x > 3\).
Solution in interval form: \((3, \infty)\)
In simple words: Solve each linear inequality independently, then find the common overlap.

Exam Tip: When solving system inequalities, always check the final intersection on a number line to avoid missing overlapping bounds.

Solve the following system of inequalities graphically (Questions 28 to 30)

Question 28. \(3x + 2y \le 24\), \(x + 2y \le 16\), \(x + y \le 10\), \(x \ge 0\), \(y \ge 0\)
Answer: 1. Line 1: \(3x + 2y = 24\) (Intercepts: \((0, 12)\), \((8, 0)\)).
2. Line 2: \(x + 2y = 16\) (Intercepts: \((0, 8)\), \((16, 0)\)).
3. Line 3: \(x + y = 10\) (Intercepts: \((0, 10)\), \((10, 0)\)).
4. Non-negativity constraints \(x \ge 0, y \ge 0\) restrict region to the first quadrant.
5. Test origin \((0, 0)\) for all inequalities (all true).
6. Shade the common region satisfying all constraints simultaneously in the first quadrant.

In simple words: Plot all boundary lines and identify the polygon region satisfying all conditions together.

Exam Tip: Corner points of the feasible region can be found by solving pairs of intersecting boundary equations simultaneously.

Question 29. \(2x + y \ge 4\), \(x + y \le 3\), \(2x - 3y \le 6\)
Answer: 1. Line 1: \(2x + y = 4\) (Intercepts: \((0, 4)\), \((2, 0)\)). Test \((0, 0)\): \(0 \ge 4\) (False, shade away from origin).
2. Line 2: \(x + y = 3\) (Intercepts: \((0, 3)\), \((3, 0)\)). Test \((0, 0)\): \(0 \le 3\) (True, shade towards origin).
3. Line 3: \(2x - 3y = 6\) (Intercepts: \((0, -2)\), \((3, 0)\)). Test \((0, 0)\): \(0 \le 6\) (True, shade towards origin).
4. The intersection of these three half-planes forms the common solution region.

In simple words: Draw each boundary line, test the origin or a test point for each, and shade the overlapping intersection.

Exam Tip: Double-check test points for inequalities involving greater-than versus less-than to ensure accurate half-plane shading.

Question 30. \(x + 2y \le 2000\), \(x + y \le 1500\), \(y \le 600\), \(x \ge 0\), \(y \ge 0\)
Answer: 1. Line 1: \(x + 2y = 2000\) (Intercepts: \((0, 1000)\), \((2000, 0)\)).
2. Line 2: \(x + y = 1500\) (Intercepts: \((0, 1500)\), \((1500, 0)\)).
3. Line 3: \(y = 600\) (Horizontal line parallel to X-axis at \(y = 600\)).
4. Constraints \(x \ge 0, y \ge 0\) restrict the graph to the first quadrant.
5. All inequalities satisfy the origin test (\(0 \le\) constant), so shade towards the origin.
6. The feasible region is the polygon bounded by these lines in the first quadrant.

In simple words: Plot the scaled boundary lines and determine the bounded polygonal region in the first quadrant.

Exam Tip: For large numbers, choose appropriate scale units on your axes (e.g., 1 unit = 100 or 500 units) to fit the graph neatly.