6 6 Practice Systems Of Linear Inequalities

Mastering Linear Inequalities: 6 Practice Systems to Boost Your Problem‑Solving Skills

Linear inequalities are the building blocks of algebraic modeling, and mastering them opens doors to higher‑level math, economics, engineering, and data science. By practicing with a variety of systems—pairs, triples, and even more complex arrangements—you’ll learn to visualize solution sets, translate verbal statements into algebra, and apply critical thinking to real‑world scenarios. Below are six carefully designed practice systems that cover the spectrum of difficulty and application, complete with step‑by‑step solutions and key take‑aways.

Quick note before moving on.

Introduction to Linear Inequalities

A linear inequality has the form
[ ax + by \leq c,\quad ax + by \geq c,\quad ax + by < c,\quad \text{or}\quad ax + by > c, ] where (a, b,) and (c) are constants. Also, unlike equations, inequalities describe a region of the coordinate plane, not just a single line. Solving a system of inequalities means finding all points ((x, y)) that satisfy every inequality simultaneously.

Why practice?

• Visualization: Graphing teaches you to interpret “above,” “below,” “left,” and “right.”
• Critical thinking: You’ll learn to filter extraneous solutions and identify feasible regions.
• Real‑world relevance: From budgeting constraints to engineering tolerances, inequalities model limitations everywhere.

1. Simple Two‑Variable System

Problem
Solve the system:
[ \begin{cases} 2x + y \geq 4 \ x - 3y \leq 1 \end{cases} ]

Solution Steps

1. Rearrange each inequality for y
   [ y \geq -2x + 4 \quad\text{(1)} \ y \geq \frac{x-1}{3} \quad\text{(2)} ] (Note: multiply the second inequality by (-1) and flip the sign.)

2. Graph the boundary lines

   • Line 1: slope (-2), y‑intercept (4).
   • Line 2: slope (\frac{1}{3}), y‑intercept (-\frac{1}{3}).

3. Shade the feasible regions

   • For (1), shade above the line.
   • For (2), shade above the line as well.

4. Find intersection
   Solve the two equalities simultaneously:
   [ \begin{cases} 2x + y = 4 \ x - 3y = 1 \end{cases} ] From the first, (y = 4 - 2x). Substitute into the second:
   [ x - 3(4 - 2x) = 1 \Rightarrow x - 12 + 6x = 1 \Rightarrow 7x = 13 \Rightarrow x = \frac{13}{7} ] Then (y = 4 - 2\left(\frac{13}{7}\right) = \frac{12}{7}) Small thing, real impact..

5. Describe the solution set
   All points ((x, y)) with (x \geq \frac{13}{7}) and (y \geq \frac{12}{7}) that also satisfy both inequalities. The feasible region is the intersection of the two half‑planes, bounded by the two lines and extending infinitely outward.

Key Take‑away

When both inequalities are “greater than or equal to,” the feasible region lies in the upper part of the plane. Visualizing the intersection early prevents algebraic missteps Small thing, real impact..

2. System with Mixed Inequalities

Problem
Solve:
[ \begin{cases} x + 2y \leq 8 \ -3x + y > 2 \ x \geq 0 \end{cases} ]

Solution Steps

1. Rewrite for y
   [ y \leq \frac{8 - x}{2} \quad\text{(1)} \ y > 3x + 2 \quad\text{(2)} \ x \geq 0 \quad\text{(3)} ]

2. Graph the boundaries

   • Line (1): slope (-\frac{1}{2}), intercept (4).
   • Line (2): slope (3), intercept (2).
   • Line (3): vertical line at (x = 0).

3. Shade the regions

   • (1): below the line.
   • (2): above the line (strict inequality, use dashed line).
   • (3): right of the y‑axis.

4. Find intersection points

   • Intersection of (1) and (2): solve
     [ \frac{8 - x}{2} = 3x + 2 \Rightarrow 8 - x = 6x + 4 \Rightarrow 5x = 4 \Rightarrow x = \frac{4}{5} ] Then (y = 3\left(\frac{4}{5}\right) + 2 = \frac{12}{5} + 2 = \frac{22}{5}).
   • Intersection of (1) with (x = 0): (y = 4).
   • Intersection of (2) with (x = 0): (y = 2).

5. Describe solution set
   The feasible region is a bounded triangle with vertices at ((0,2)), ((0,4)), and (\left(\frac{4}{5}, \frac{22}{5}\right)). Since (2) is strict, the boundary line is excluded; the solution set is the interior of the triangle (excluding the edge along (2)).

Key Take‑away

Mixed inequalities often produce a bounded feasible region. Pay close attention to strict versus non‑strict inequalities; they determine whether boundary lines belong to the solution set Small thing, real impact. Which is the point..

3. Three‑Variable System (Projected onto the Plane)

Problem
Solve the system for (x, y, z):
[ \begin{cases} x + y + z \leq 6 \ 2x - y + 3z \geq 4 \ x - 2y + z \leq 2 \end{cases} ] Find the projection of the feasible region onto the (xy)-plane.

Solution Steps

1. Eliminate (z)
   Treat (z) as a parameter and express inequalities in terms of (x) and (y).
   From the first: (z \leq 6 - x - y).
   Substitute into the second:
   [ 2x - y + 3(6 - x - y) \geq 4 \Rightarrow 2x - y + 18 - 3x - 3y \geq 4 \Rightarrow -x - 4y \geq -14 \Rightarrow x + 4y \leq 14 ] From the third:
   [ x - 2y + (6 - x - y) \leq 2 \Rightarrow x - 2y + 6 - x - y \leq 2 \Rightarrow -3y \leq -4 \Rightarrow y \geq \frac{4}{3} ]

2. Collect projected inequalities
   [ \begin{cases} x + y \leq 6 - z \quad (\text{but } z \text{ is eliminated}) \ x + 4y \leq 14 \ y \geq \frac{4}{3} \end{cases} ] Since (z) can vary within its bounds, the tightest constraints on (x, y) are: [ \begin{cases} x + 4y \leq 14 \ y \geq \frac{4}{3} \end{cases} ] Additionally, from the first inequality with (z \geq 0): (x + y \leq 6).

3. Graph the 2‑D projection
   Plot the lines (x + 4y = 14), (x + y = 6), and the horizontal line (y = \frac{4}{3}).
   Shade the region satisfying all three: below both slanted lines and above the horizontal line Most people skip this — try not to..

4. Describe feasible set
   The projection is a convex polygon bounded by the intersection points:

   • Intersection of (x + 4y = 14) and (y = \frac{4}{3}): (x = 14 - 4 \cdot \frac{4}{3} = 14 - \frac{16}{3} = \frac{26}{3}).
   • Intersection of (x + y = 6) and (y = \frac{4}{3}): (x = 6 - \frac{4}{3} = \frac{14}{3}).
   • Intersection of (x + y = 6) and (x + 4y = 14): solve simultaneously to find (y = \frac{8}{3}), (x = \frac{10}{3}).

   The resulting triangle has vertices (\left(\frac{10}{3}, \frac{8}{3}\right)), (\left(\frac{14}{3}, \frac{4}{3}\right)), and (\left(\frac{26}{3}, \frac{4}{3}\right)) Simple, but easy to overlook. Simple as that..

Key Take‑away

Projecting a higher‑dimensional system onto a plane can simplify visualization, but always check that the eliminated variable’s bounds do not restrict the projection further.

4. Non‑Standard Coefficients

Problem
Solve:
[ \begin{cases} -5x + 2y \geq -10 \ 3x - 4y < 12 \ x + y \leq 5 \end{cases} ]

Solution Steps

1. Rearrange
   [ y \geq \frac{5x - 10}{2} \quad\text{(1)} \ y > \frac{3x - 12}{4} \quad\text{(2)} \ y \leq 5 - x \quad\text{(3)} ]

2. Graph boundaries

   • (1): slope (2.5), intercept (-5).
   • (2): slope (0.75), intercept (-3).
   • (3): slope (-1), intercept (5).

3. Shade

   • (1): above the line.
   • (2): above the dashed line (strict).
   • (3): below the line.

4. Find intersection points

   • Intersection of (1) and (3):
     [ \frac{5x - 10}{2} = 5 - x \Rightarrow 5x - 10 = 10 - 2x \Rightarrow 7x = 20 \Rightarrow x = \frac{20}{7},\ y = 5 - \frac{20}{7} = \frac{15}{7} ]
   • Intersection of (2) and (3):
     [ \frac{3x - 12}{4} = 5 - x \Rightarrow 3x - 12 = 20 - 4x \Rightarrow 7x = 32 \Rightarrow x = \frac{32}{7},\ y = 5 - \frac{32}{7} = \frac{3}{7} ]
   • Intersection of (1) and (2):
     [ \frac{5x - 10}{2} = \frac{3x - 12}{4} \Rightarrow 2(5x - 10) = 3x - 12 \Rightarrow 10x - 20 = 3x - 12 \Rightarrow 7x = 8 \Rightarrow x = \frac{8}{7},\ y = \frac{5(8/7) - 10}{2} = \frac{40/7 - 70/7}{2} = \frac{-30/7}{2} = -\frac{15}{7} ] But (y = -\frac{15}{7}) violates (3) ((y \leq 5 - x) is satisfied, but check (2) strictness? It meets equality, but (2) requires (y >) line, so this point is excluded).

5. Feasible region
   The solution set is the convex quadrilateral bounded by the points (\left(\frac{8}{7}, -\frac{15}{7}\right)) (excluded), (\left(\frac{20}{7}, \frac{15}{7}\right)), (\left(\frac{32}{7}, \frac{3}{7}\right)), and the intersection of (1) and (3) again. Since the first point is excluded, the feasible region is a triangle with vertices (\left(\frac{20}{7}, \frac{15}{7}\right)), (\left(\frac{32}{7}, \frac{3}{7}\right)), and the intersection of (1) and (3) (which is the same as the first point). Thus the feasible region collapses to a line segment? Actually, because the strict inequality in (2) removes the lower corner, the remaining region is the segment between (\left(\frac{20}{7}, \frac{15}{7}\right)) and (\left(\frac{32}{7}, \frac{3}{7}\right)).

Key Take‑away

Non‑standard coefficients can flip the orientation of inequalities. Carefully solve for (y) and check each intersection’s validity relative to every inequality, especially strict ones.

5. Real‑World Budget Constraint

Problem
A student has $120 to spend on textbooks and supplies. Textbooks cost $40 each; supplies cost $15 each. Additionally, the student wants to buy at least 2 textbooks and no more than 5 supplies. Represent and solve this as a system of inequalities Small thing, real impact..

Solution Steps

1. Define variables
   Let (t) = number of textbooks, (s) = number of supplies Easy to understand, harder to ignore..

2. Translate constraints

   • Budget: (40t + 15s \leq 120).
   • Minimum textbooks: (t \geq 2).
   • Maximum supplies: (s \leq 5).
   • Non‑negativity: (t \geq 0, s \geq 0).

3. Graph

   • Plot (40t + 15s = 120) (line through ((3,0)) and ((0,8))).
   • Shade below this line (budget).
   • Shade right of (t=2).
   • Shade below (s=5).

4. Find feasible integer points
   Since (t) and (s) must be integers:

   • (t = 2): budget allows (s \leq \frac{120 - 80}{15} = \frac{40}{15} \approx 2.67) → (s \leq 2).
   • (t = 3): budget allows (s \leq \frac{120 - 120}{15} = 0) → (s = 0).
     No higher (t) possible because budget would be exceeded.
     So feasible integer solutions: ((t,s) = (2,0), (2,1), (2,2), (3,0)).

5. Interpretation
   The student can purchase either 2 textbooks and up to 2 supplies, or 3 textbooks and no supplies, staying within the $120 budget It's one of those things that adds up..

Key Take‑away

Systems of inequalities naturally model resource allocation problems. Always remember to include integrality constraints when variables represent discrete items.

6. Optimization with Inequalities (Linear Programming)

Problem
Maximize profit (P = 5x + 3y) subject to:
[ \begin{cases} 2x + y \leq 10 \ x + 3y \leq 12 \ x, y \geq 0 \end{cases} ]

Solution Steps

1. Graph the feasible region

   • Line 1: (y = 10 - 2x).
   • Line 2: (y = \frac{12 - x}{3}).
   • Shade below both lines and in the first quadrant.

2. Find corner points

   • Intersection of lines: solve
     [ 10 - 2x = \frac{12 - x}{3} \Rightarrow 30 - 6x = 12 - x \Rightarrow -5x = -18 \Rightarrow x = \frac{18}{5} = 3.6 ] Then (y = 10 - 2(3.6) = 10 - 7.2 = 2.8).
   • Intersections with axes:
     • (x)-axis: ((5, 0)) from line 1.
     • (y)-axis: ((0, 4)) from line 1.
     • From line 2: ((12, 0)) and ((0, 4)).

   Feasible corner points: ((0,0), (5,0), (0,4), (3.6, 2.8)).

3. Evaluate objective function at corners

   • ((0,0)): (P = 0).
   • ((5,0)): (P = 25).
   • ((0,4)): (P = 12).
   • ((3.6, 2.8)): (P = 5(3.6) + 3(2.8) = 18 + 8.4 = 26.4).

4. Optimal solution
   The maximum profit (P_{\max} = 26.4) occurs at ((x, y) = (3.6, 2.8)) Practical, not theoretical..

Key Take‑away

Linear programming reduces to evaluating the objective function at the vertices of the feasible region. In practice, integer constraints may require rounding or more sophisticated methods.

FAQ

Conclusion

Mastering linear inequalities involves more than just algebraic manipulation; it requires a blend of geometric intuition, logical rigor, and contextual awareness. By working through these six practice systems—from simple two‑variable cases to multi‑dimensional projections and real‑world optimization—you’ll develop a versatile toolkit that applies across mathematics, science, economics, and engineering. Keep practicing, keep visualizing, and soon the solution space will feel as natural as a familiar playground.

Not the most exciting part, but easily the most useful.