Questions and Answers Section

Example 1: Solving a Basic Inequality

Question: Solve $5x - 3 < 3x + 1$ for real x.

Solution: We collect the x terms on one side and the constants on the other. $5x - 3x < 1 + 3$ $2x < 4$ $x < 2$ The solution set is $(-\infty, 2)$.

Example 2: Reversing the Inequality Sign

Question: Solve $7 - 3x \ge 19$.

Solution: Subtract 7 from both sides: $-3x \ge 12$. Divide by -3 and reverse the inequality sign:

$$x \le \frac{12}{-3} \implies x \le -4$$

Solution set: $(-\infty, -4]$.

Example 3: Compound Inequality

Question: Solve $-5 \le \frac{5-3x}{2} \le 8$.

Solution: Multiply all parts by 2: $-10 \le 5-3x \le 16$. Subtract 5 from all parts: $-15 \le -3x \le 11$. Divide by -3 and reverse both inequality signs: $\frac{-15}{-3} \ge x \ge \frac{11}{-3}$. $5 \ge x \ge -11/3$. The solution set is $[-11/3, 5]$.

Example 4: Graphical Solution of a System

Question: Solve the system of inequalities graphically: $x+y \le 6$ and $x+y \ge 4$.

Solution:

1. Graph $x+y \le 6$: Draw the solid line $x+y=6$. Test (0,0): $0 \le 6$ is true. Shade the region containing the origin.
2. Graph $x+y \ge 4$: Draw the solid line $x+y=4$. Test (0,0): $0 \ge 4$ is false. Shade the region not containing the origin.
3. Solution Region: The solution is the area where the shaded regions overlap, which is the strip between the two parallel lines, including the lines themselves.

Example 5: Finding Corner Points

Question: Find the corner points of the feasible region for the system: $x+2y \le 8$, $2x+y \le 8$, $x \ge 0$, $y \ge 0$.

Solution: The region is a bounded quadrilateral in the first quadrant. The vertices are the intersections of the boundary lines:

• (0,0): Intersection of $x=0$ and $y=0$.
• (4,0): Intersection of $2x+y=8$ and $y=0$.
• (0,4): Intersection of $x+2y=8$ and $x=0$.
• To find the last point, solve $x+2y=8$ and $2x+y=8$ simultaneously. This gives the point $(8/3, 8/3)$. The corner points are (0,0), (4,0), (0,4), and (8/3, 8/3).

Example 6: Quadratic Inequality

Question: Solve $x^2 - x - 6 < 0$.

Solution: First, find the roots of $x^2-x-6=0$. Factoring gives $(x-3)(x+2)=0$. The critical points are $x=3$ and $x=-2$. We place these on a number line and test the intervals $(-\infty, -2)$, $(-2, 3)$, and $(3, \infty)$. The expression is a simple upward-opening parabola, so it is negative between its roots. The solution is $(-2, 3)$.

Example 7: Rational Inequality (Wavy Curve Method)

Question: Solve $\frac{x-2}{x+5} > 0$.

Solution: The critical points are where the numerator or denominator is zero: $x=2$ and $x=-5$. We place these on a number line.

• Interval $(-\infty, -5)$: Test $x=-6$. $\frac{-8}{-1} > 0$ (True).
• Interval $(-5, 2)$: Test $x=0$. $\frac{-2}{5} > 0$ (False).
• Interval $(2, \infty)$: Test $x=3$. $\frac{1}{8} > 0$ (True). The solution is $(-\infty, -5) \cup (2, \infty)$.

Example 8: Modulus Inequality (Less Than)

Question: Solve $|3x-2| \le 1/2$.

Solution: Use the property $|u| \le a \iff -a \le u \le a$. $-1/2 \le 3x-2 \le 1/2$. Add 2 to all parts: $3/2 \le 3x \le 5/2$. Divide by 3: $1/2 \le x \le 5/6$. The solution is $[1/2, 5/6]$.

Example 9: Modulus Inequality (Greater Than)

Question: Solve $|4-x| > 1$.

Solution: Use the property $|u| > a \iff u > a \text{ or } u < -a$. $4-x > 1$ OR $4-x < -1$. $3 > x$ OR $5 < x$. So, $x < 3$ or $x > 5$. The solution is $(-\infty, 3) \cup (5, \infty)$.

Example 10: Word Problem

Question: A manufacturer has 600 litres of a 12% solution of acid. How many litres of a 30% acid solution must be added to it so that the acid content in the resulting mixture will be more than 15% but less than 18%?

Solution: Let x be the litres of 30% solution added. Total mixture = $600+x$. Total acid = $0.12(600) + 0.30(x) = 72+0.3x$. The concentration is $\frac{72+0.3x}{600+x}$. The inequality is $0.15 < \frac{72+0.3x}{600+x} < 0.18$. Solving the left inequality gives $x>120$. Solving the right inequality gives $x<300$. The solution is $120 < x < 300$.

Example 11: Inequality with Three Factors

Question: Solve $(x-1)(x-2)(x-3) \ge 0$.

Solution: Critical points are 1, 2, 3.

• For $x>3$, all factors are positive, so the product is positive. (Solution)
• For $2<x<3$, $(x-3)$ is negative, so the product is negative.
• For $1<x<2$, $(x-2)$ and $(x-3)$ are negative, so the product is positive. (Solution)
• For $x<1$, all three factors are negative, so the product is negative. Since the inequality is $\ge$, we include the endpoints. The solution is $[1, 2] \cup [3, \infty)$.

Example 12: Inequality involving $x^2$

Question: Solve $x^2+4 > 0$.

Solution: For any real number x, $x^2$ is always greater than or equal to 0. Therefore, $x^2+4$ is always greater than or equal to 4. The inequality $x^2+4 > 0$ is true for all real numbers. The solution is $\mathbb{R}$.

Example 13: Graphical Solution (Unbounded)

Question: Solve the system $x-2y \le 3$, $3x+4y \ge 12$, $x \ge 0, y \ge 1$.

Solution:

1. Graph all four boundary lines.
2. Shade the correct region for each.
3. The common overlapping region is an unbounded feasible region starting from the first and second quadrants, bounded below by the lines and $y=1$.

Example 14: Double Modulus Inequality

Question: Solve $|x-1| + |x-2| \ge 4$.

Solution: Critical points are 1 and 2.

• Case 1: $x<1$. $-(x-1)-(x-2) \ge 4 \implies -2x+3 \ge 4 \implies -2x \ge 1 \implies x \le -1/2$. Solution for this case is $(-\infty, -1/2]$.
• Case 2: $1 \le x < 2$. $(x-1)-(x-2) \ge 4 \implies 1 \ge 4$ (False). No solution in this case.
• Case 3: $x \ge 2$. $(x-1)+(x-2) \ge 4 \implies 2x-3 \ge 4 \implies 2x \ge 7 \implies x \ge 7/2$. Solution for this case is $[7/2, \infty)$. The final solution is the union: $(-\infty, -1/2] \cup [7/2, \infty)$.

Example 15: Solving $\frac{1}{x} < 1$

Question: Solve the inequality $\frac{1}{x} < 1$.

Solution: We cannot simply multiply by x. We must bring all terms to one side and use the wavy curve method. $\frac{1}{x} - 1 < 0 \implies \frac{1-x}{x} < 0$. Critical points are $x=0$ and $x=1$. We test the intervals.

• Interval $(-\infty, 0)$: Test $x=-1$. $\frac{2}{-1} < 0$ (True).
• Interval $(0, 1)$: Test $x=0.5$. $\frac{0.5}{0.5} < 0$ (False).
• Interval $(1, \infty)$: Test $x=2$. $\frac{-1}{2} < 0$ (True). The solution is $(-\infty, 0) \cup (1, \infty)$.