Linear Inequalities in Two Variables: A Complete Answer Key for Lesson 7.3
Linear inequalities in two variables are a cornerstone of algebra, bridging the gap between algebraic equations and geometric visualization. Consider this: this lesson focuses on graphing inequalities, understanding solution regions, and applying these concepts to real‑world problems. The answer key below not only provides step‑by‑step solutions to typical exercises but also offers insights into common pitfalls and strategies for mastering the material And that's really what it comes down to..
Introduction
In Lesson 7.3, students learn to solve linear inequalities of the form
(ax + by ; (\leq,; \geq,; <,; >) ; c),
where (a), (b), and (c) are constants.
Key objectives include:
- Converting inequalities into equalities for boundary lines.
- Determining whether boundary lines are solid or dashed.
- Testing points to identify the correct shading region.
- Interpreting the solution set in both algebraic and geometric terms.
The answer key below follows the typical structure of the textbook’s problem set, ensuring that each solution is transparent and logically sound.
Step‑by‑Step Solution Guide
1. Identify the Boundary Line
-
Rewrite the inequality as an equation by replacing the inequality sign with an equals sign.
Example: (3x - 2y \ge 6) → (3x - 2y = 6). -
Solve for one variable (usually (y)) to get the slope–intercept form (y = mx + b).
Example: (3x - 2y = 6) → (-2y = -3x + 6) → (y = \frac{3}{2}x - 3).
2. Draw the Boundary Line
- Plot the intercepts or use two points to draw the line.
- Solid line if the inequality is (\leq) or (\geq);
dashed line if the inequality is (<) or (>).
3. Test a Point
- Choose a simple point not on the line (commonly ((0,0)) unless it lies on the line).
- Substitute into the original inequality to see if it satisfies the condition.
- If it does, shade the side of the line containing the point; if not, shade the opposite side.
4. Write the Solution Set
- Algebraically: e.g., (y \le \frac{3}{2}x - 3).
- Graphically: the shaded region including the boundary line for “(\leq)” or “(\geq)”.
Sample Problem Set with Answers
Problem 1
Solve and graph: (2x + 5y \le 10).
Solution
- Boundary: (2x + 5y = 10).
Solve for (y): (5y = -2x + 10) → (y = -\frac{2}{5}x + 2). - Plot points:
- (x = 0) → (y = 2).
- (x = 5) → (y = 0).
- Solid line (≤).
- Test ((0,0)): (2(0)+5(0)=0 \le 10) ✔️ → shade below the line.
- Answer: All points ((x,y)) such that (y \le -\frac{2}{5}x + 2).
Problem 2
Solve and graph: (-3x + 4y > 12).
Solution
- Boundary: (-3x + 4y = 12).
Solve for (y): (4y = 3x + 12) → (y = \frac{3}{4}x + 3). - Plot points:
- (x = 0) → (y = 3).
- (x = 4) → (y = 6).
- Dashed line (>).
- Test ((0,0)): (-3(0)+4(0)=0 > 12) ❌ → shade opposite side (above the line).
- Answer: ({(x,y) | y > \frac{3}{4}x + 3}).
Problem 3
Determine the solution set for the system: [ \begin{cases} x + y \ge 4 \ 2x - y < 0 \end{cases} ]
Solution
- Graph both inequalities as described above.
- Intersection of the two shaded regions is the solution set.
- The resulting region is a convex polygon (often a triangle or unbounded area).
- Answer: All points lying in the intersection of the half‑planes defined by (x + y \ge 4) (solid line, shading above/right) and (2x - y < 0) (dashed line, shading below/right).
Problem 4
Interpretation Question
A company can produce two products, A and B. Production constraints are:
[
\begin{aligned}
3A + 2B &\le 30 \quad (\text{material constraint})\
A + 4B &\ge 10 \quad (\text{labor constraint})\
A, B &\ge 0
\end{aligned}
]
Identify the feasible production region.
Solution
- Plot each inequality.
- Include the non‑negativity constraints (axes).
- The feasible region is the common shaded area bounded by the two lines and the first quadrant.
- Answer: The set of ((A,B)) pairs that satisfy all three inequalities, typically a quadrilateral in the first quadrant.
Common Mistakes & How to Avoid Them
| Mistake | Why It Happens | Fix |
|---|---|---|
| Using a solid line for “<” or “>” | Forgetting the strictness of the inequality. | Draw a dashed line whenever the inequality is strict. |
| Shading the wrong side | Misinterpreting the test point or using |
Continuing the discussion ongraphing linear inequalities and systems, we now turn our attention to solving and interpreting systems of inequalities, which is crucial for modeling real-world constraints and finding feasible regions.
Systems of Linear Inequalities
A system of linear inequalities consists of two or more inequalities involving the same variables. The solution set is the set of all points that satisfy all inequalities simultaneously. Graphically, this is the region where the shaded areas of each individual inequality overlap.
Key Steps for Solving Systems:
- Graph Each Inequality Separately:
- Convert each inequality to slope-intercept form ((y = mx + b)) for easier graphing.
- Draw the boundary line (solid for (\leq) or (\geq), dashed for (<) or (>)).
- Test a point (e.g., ((0,0))) to determine the correct shading side.
- Identify the Intersection:
- The solution region is the overlapping area of all shaded regions.
- This region may be bounded (e.g., a polygon) or unbounded (extending infinitely).
- Verify the Solution:
- Select a test point within the intersection and substitute it into all original inequalities to confirm it satisfies them.
Example:
Solve the system:
[
\begin{cases}
x + y \ge 4 \
2x - y < 0
\end{cases}
]
- Graph (x + y = 4) (solid line, shade above).
- Graph (2x - y = 0) (dashed line, shade below).
- The solution region is the area where both shaded regions overlap, forming a wedge in the first quadrant.
Real-World Applications
Systems of inequalities model constraints in fields like economics, engineering, and logistics. For instance:
- Production Planning: Constraints on materials, labor, and demand (as in Problem 4) define feasible production levels.
- Resource Allocation: Budget, time, or capacity limits are expressed as inequalities.
Problem 4 Recap:
The company’s constraints ((3A + 2B \le 30), (A + 4B \ge 10), (A, B \ge 0)) define a feasible region in the (A)-(B) plane. This region represents all producible combinations of products A and B that meet resource and demand requirements Not complicated — just consistent..
Common Pitfalls and Solutions
- Shading the Wrong Side: Always test a point (e.g., ((0,0))) to confirm shading direction.
- Ignoring Boundary Lines: Remember that solid lines include boundary points; dashed lines exclude them.
- Misinterpreting Systems: Graph each inequality carefully; the solution is the intersection, not the union, of shaded regions.
Conclusion
Graphing linear inequalities and systems provides a visual and analytical framework for solving problems with multiple constraints. By mastering boundary lines, shading techniques, and solution verification, you can efficiently model real-world scenarios—from production planning to resource allocation. The key is systematic graphing, careful testing, and recognizing that the solution lies in the shared region of all constraints. This approach transforms abstract inequalities into actionable insights, bridging mathematical theory and practical application.
