Finding Systems of Inequalities in Two Variables

This tutorial explains how to determine a system of inequalities in two variables whose solution set corresponds to a shaded region on a graph. By analyzing the boundary lines and curves, we can algebraically represent the graphical region. Each example below includes a diagram and a detailed step-by-step explanation.

Table of Contents

Method: Step-by-Step Procedure

To find the system of inequalities that describes a shaded region:

  1. Identify each boundary: List every line or curve that forms the edge of the shaded region.
  2. Find the equation of each boundary: Use two points on the boundary (for a line) or vertex/point information (for a curve).
  3. Determine the inequality sign:
    • Solid boundary → use \( \le \) or \( \ge \) (boundary included).
    • Dashed boundary → use \( < \) or \( > \) (boundary excluded).
  4. Determine the shading side: Pick a test point in the shaded region and check whether it satisfies \( y > f(x) \) or \( y < f(x) \).
  5. Combine all inequalities into a single system.

Graphical Examples with Detailed Solutions

Example 1: Triangular Region in the 3rd Quadrant

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Find the system of inequalities whose solution set is represented by the region shown below.

Triangular region in the third quadrant bounded by the x-axis, y-axis, and a diagonal line through (-4,0) and (0,-2)

Solution

The region is bounded by the \(x\)-axis, the \(y\)-axis, and a diagonal line.

  • The region is to the left of the \(y\)-axis, meaning \(x \le 0\).
  • The region is below the \(x\)-axis, meaning \(y \le 0\).
  • The diagonal line passes through \((-4,0)\) and \((0,-2)\). Its slope is \(m = \frac{-2 - 0}{0 - (-4)} = -\frac{1}{2}\), and its \(y\)-intercept is \(-2\). The equation is \(y = -\frac{1}{2}x - 2\). Since the shaded region is above this line, the inequality is \(y \ge -\frac{1}{2}x - 2\).

System of Inequalities:

\[ \begin{cases} y \le 0 \\ x \le 0 \\ y \ge -\frac{1}{2}x - 2 \end{cases} \]

Example 2: Rectangular Bounded Region

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Find the system of inequalities whose solution set is shown below.

Rectangular bounded region between vertical lines x = -3, x = -1 and horizontal lines y = -3, y = 1

Solution

The shaded region is a rectangle bounded by horizontal and vertical lines.

  • Vertically, the region lies below \(y = 1\) and above \(y = -3\). This gives: \(y \le 1\) and \(y \ge -3\).
  • Horizontally, the region lies to the left of \(x = -1\) and to the right of \(x = -3\). This gives: \(x \le -1\) and \(x \ge -3\).

System of Inequalities:

\[ \begin{cases} y \le 1 \\ x \le -1 \\ y \ge -3 \\ x \ge -3 \end{cases} \]

Example 3: Triangle Formed by Three Intersecting Lines

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Find a system of inequalities that has the region shown below as its solution set.

Triangular region formed by three intersecting lines L1, L2, and L3

Solution

The boundary lines are determined by finding the equation for each line using pairs of points on the graph:

  • Line \(L_1\): Passes through \((3,0)\) and \((0,3)\).
    Slope \(m = \frac{3 - 0}{0 - 3} = -1\).
    Equation: \(y = -x + 3\). The region is below this dashed line, so \(y < -x + 3\).
  • Line \(L_2\): Passes through \((4,0)\) and \((0,-2)\).
    Slope \(m = \frac{-2 - 0}{0 - 4} = \frac{1}{2}\).
    Equation: \(y = \frac{1}{2}x - 2\). The region is above this solid line, so \(y \ge \frac{1}{2}x - 2\).
  • Line \(L_3\): Passes through \((-1,0)\) and \((0,2)\).
    Slope \(m = \frac{2 - 0}{0 - (-1)} = 2\).
    Equation: \(y = 2x + 2\). The region is below this dashed line, so \(y < 2x + 2\).

System of Inequalities:

\[ \begin{cases} y < -x + 3 \\ y \ge \frac{1}{2}x - 2 \\ y < 2x + 2 \end{cases} \]

Example 4: Quadrilateral Region

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Find the system of inequalities whose solution set is represented by the region below.

Quadrilateral ABCD bounded by four lines with vertices at A(2,2), B(8,-2), C(9,6), and D(5,6)

Solution

We find the equation for each of the four boundary lines bounding the quadrilateral \(ABCD\):

1. Equation of solid line \(AB\):
Using points \(A(2,2)\) and \(B(8,-2)\), the slope is:
\[ m = \frac{-2 - 2}{8 - 2} = -\frac{4}{6} = -\frac{2}{3} \] Equation: \(y - 2 = -\frac{2}{3}(x - 2) \implies y = -\frac{2}{3}x + \frac{10}{3}\).
The region is above this solid line: \(y \ge -\frac{2}{3}x + \frac{10}{3}\).

2. Equation of dashed line \(BC\):
Using points \(B(8,-2)\) and \(C(9,6)\), the slope is:
\[ m = \frac{6 - (-2)}{9 - 8} = 8 \] Equation: \(y - 6 = 8(x - 9) \implies y = 8x - 66\).
The region is above this dashed line: \(y > 8x - 66\).

3. Equation of dashed line \(CD\):
This is a horizontal line passing through \(y = 6\).
The region is below this dashed line: \(y < 6\).

4. Equation of solid line \(DA\):
Using points \(D(5,6)\) and \(A(2,2)\), the slope is:
\[ m = \frac{6 - 2}{5 - 2} = \frac{4}{3} \] Equation: \(y - 2 = \frac{4}{3}(x - 2) \implies y = \frac{4}{3}x - \frac{2}{3}\).
The region is below this solid line: \(y \le \frac{4}{3}x - \frac{2}{3}\).

System of Inequalities:

\[ \begin{cases} y \ge -\frac{2}{3}x + \frac{10}{3} \\ y > 8x - 66 \\ y < 6 \\ y \le \frac{4}{3}x - \frac{2}{3} \end{cases} \]

Example 5: Region Bounded by a Line and a Parabola

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Find the system of inequalities whose solution set is shown below.

Region bounded above by a solid line through (1,0) and (0,-2) and below by a dashed parabola y = x² - 4

Solution

The boundary consists of an upward-opening parabola and a straight line:

  • Parabola: The dashed curve represents the standard parabola shifted down by 4 units. Its equation is \(y = x^2 - 4\). Because the region is inside (above) the dashed parabola, we write \(y > x^2 - 4\).
  • Line: The solid line passes through the points \((1,0)\) and \((0,-2)\). The slope is \(m = \frac{-2 - 0}{0 - 1} = 2\). Its equation is \(y = 2x - 2\). The shaded region is below this solid line, so we write \(y \le 2x - 2\).

System of Inequalities:

\[ \begin{cases} y > x^2 - 4 \\ y \le 2x - 2 \end{cases} \]

Example 6: Region Bounded by a Semicircle and a Parabola

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What system of inequalities produces the region shown below?

Region bounded above by the lower semicircle y = -sqrt(4 - x²) and below by the downward parabola y = -x² - 4

Solution

The boundary consists of two non-linear curves:

  • Lower Curve (Parabola): This is a downward-opening parabola with a vertex at \((0, -4)\). Its equation is \(y = -x^2 - 4\). The shaded region is above the solid boundary, yielding \(y \ge -x^2 - 4\).
  • Upper Curve (Semicircle): The upper boundary corresponds to the lower half of a circle centered at the origin with a radius of \(2\). The equation of the full circle is \(x^2 + y^2 = 4\). Solving for the lower half gives \(y = -\sqrt{4 - x^2}\). The shaded region is below this solid boundary, yielding \(y \le -\sqrt{4 - x^2}\).

System of Inequalities:

\[ \begin{cases} y \le -\sqrt{4 - x^2} \\ y \ge -x^2 - 4 \end{cases} \]

Domain consideration: The semicircle \(y = -\sqrt{4 - x^2}\) is defined only for \(-2 \le x \le 2\), so the solution set is naturally restricted to that interval.

Frequently Asked Questions

How do you find the system of inequalities from a graph?

To find the system of inequalities from a graph: (1) Identify each boundary line or curve. (2) Find the equation of each boundary. (3) Determine which side of the boundary is shaded by testing a point. (4) Use a solid line for \( \le \) or \( \ge \) (boundary included) and a dashed line for \( < \) or \( > \) (boundary excluded).

How do you know if the inequality sign is \( < \) or \( > \) versus \( \le \) or \( \ge \)?

If the boundary line is solid, the inequality includes equality (\( \le \) or \( \ge \)). If the boundary line is dashed, the inequality is strict (\( < \) or \( > \)), meaning points on the boundary are not included in the solution set.

How do you determine which side of a line to shade?

Choose any test point not on the boundary line (the origin \((0,0)\) is usually easiest). Substitute the coordinates into the equation of the boundary. If the result satisfies the inequality, shade the side containing the test point. Otherwise, shade the opposite side.

What is the difference between a linear and a non-linear inequality?

A linear inequality involves a straight-line boundary (e.g., \( y \le 2x + 1 \)). A non-linear inequality involves a curved boundary such as a parabola (\( y > x^2 - 4 \)), circle (\( x^2 + y^2 \le 4 \)), or semicircle (\( y \le -\sqrt{4 - x^2} \)).

Can a system of inequalities have no solution?

Yes. If the shaded regions defined by the individual inequalities do not overlap, the system has no solution. Graphically, this means the intersection of the half-planes or curved regions is empty.

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