Mathematics

Inequalities

Study of inequalities: 1st and 2nd degree, higher degree polynomials, rational expressions, and absolute value inequalities. Systems of linear inequalities with one and two variables (feasible region).

They are inequalities that can take the following 4 forms: >> (greater than), ≥\ge (greater than or equal to), << (less than), ≤\le (less than or equal to).
They are solved by isolating the variable, just like in equations. The result is expressed as an interval, representing the set of values that satisfy the inequality. Ex.:
2(x+3)≤4x+10⇒2x−4x≤10−62(x+3) \le 4x + 10 \quad \Rightarrow \quad 2x - 4x \le 10 - 6
−2x≤4-2x \le 4
x≥4−2⇒x≥−2⇒x∈[−2,∞)x \ge \dfrac{4}{-2} \quad \Rightarrow \quad x \ge -2 \quad \Rightarrow \quad x \in [-2,\infty)
Note that when isolating x, because its coefficient is negative, the direction of the inequality sign is flipped. The interval is closed at x=-2 because the inequality contains an or equal to.
In general, to solve any inequality, follow these steps:
  1. Simplify the inequality and compare it to 0. If it is rational, do not eliminate the denominators.
  2. Find the roots of the numerator and denominator. Plot these roots on the real number line, considering:
    • If the inequality contains an or equal to, use closed circles.
    • If the inequality does not contain an or equal to, or if they are roots of the denominator, use open circles.
  3. Evaluate the sign of the algebraic expression in the resulting intervals on the real number line.
  4. The solution is the set of intervals that satisfy the inequality.
1. Simplified inequality: x2−x−2>0x^2 - x - 2 > 0
2. Roots of the polynomial: x=1±1+82⇒x=−1  ;x=2\displaystyle x = \dfrac{1 \pm \sqrt{1+8}}{2} \Rightarrow x = -1 \;; \quad x = 2
Roots on the real number line:
Roots -1 and 2 on real line
3. Sign of x2−x−2x^2 - x - 2 in the intervals:
Sign in intervals
4. The result is the positive intervals. In this case, they are open because the inequality does not contain an or equal to.
x∈(−∞,−1)∪(2,∞)x \in (-\infty, -1) \cup (2, \infty)
  • Simplified inequality: (x−3)xx+2≥0\displaystyle \dfrac{(x-3)x}{x+2} \ge 0
  • Roots of the numerator and denominator: x=−2  ;x=0  ;x=3x = -2 \;; \quad x = 0 \;; \quad x = 3
Roots on the real number line:
Roots -2, 0 and 3
  • Sign of (x−3)xx+2\displaystyle \dfrac{(x-3)x}{x+2} in the intervals:
Signs
  • Result, positive intervals:
x∈(−2,0]∪[3,∞)x \in (-2, 0] \cup [3, \infty)
  • Solve the inequalities independently.
  • The overall solution is the intersection of the individual solution intervals of each inequality. Example:
{2(x−2)≤4xSol: x∈[−2,∞)x2−x−2>0Sol: x∈(−∞,−1)∪(2,∞)\begin{cases} 2(x-2) \le 4x & \text{Sol: } x \in [-2, \infty) \\ x^2 - x - 2 > 0 & \text{Sol: } x \in (-\infty, -1) \cup (2, \infty) \end{cases}
Graph the solutions on the real number line to visualize the intersection:
Intersection on real line
The solution is the region where both lines overlap:
x∈[−2,−1)∪(2,∞)x \in [-2, -1) \cup (2, \infty)
  • Once the inequality is simplified, graph the corresponding function. If the inequality includes an or equal to, draw a solid line; otherwise, draw a dashed line. This divides the plane into two regions.
  • Test any point from each region to determine which area satisfies the inequality. The solution region is shaded.

2x+3y≥62x + 3y \ge 6

Graph 2x+3y=62x+3y=6 with a solid line

y>x2−4y > x^2 - 4

Graph y=x2−4y=x^2-4 with a dashed line

Plane region line
Plane region parabola
Graph each inequality independently. The solution is the region where they overlap (the feasible region).

{x≤4y≤32x+3y≥6\begin{cases} x \le 4 \\ y \le 3 \\ 2x + 3y \ge 6 \end{cases}
Feasible region
  • << or ≤\le: Split the inequality into two. The solution is the intersection of both. Example:
∣2x−3∣≤3−x⇒{2x−3≤3−x⇒x≤22x−3≥−(3−x)⇒x≥0Sol: x∈[0,2]|2x-3| \le 3-x \quad \Rightarrow \quad \begin{cases} 2x-3 \le 3-x & \Rightarrow \quad x \le 2 \\ 2x-3 \ge -(3-x) & \Rightarrow \quad x \ge 0 \end{cases} \qquad \text{Sol: } x \in [0, 2]
  • >> or ≥\ge: Split the inequality into two. The solution is the union of both. Ex.:
∣x+1∣>3⇒{x+1>3⇒x>2x+1<−3⇒x<−4Sol: x∈(−∞,−4)∪(2,∞)|x+1| > 3 \quad \Rightarrow \quad \begin{cases} x+1 > 3 & \Rightarrow \quad x > 2 \\ x+1 < -3 & \Rightarrow \quad x < -4 \end{cases} \qquad \text{Sol: } x \in (-\infty, -4) \cup (2, \infty)