Lesson 7.3 Linear Inequalities in Two Variables Answer Key
Linear inequalities in two variables extend the concept of solving equations to situations where the relationship between x and y is not exact but rather falls within a range of possible values. Mastering this topic is essential for interpreting real‑world constraints such as budget limits, production capacities, or speed limits, and it forms a foundation for later studies in linear programming and optimization. The following sections break down the theory, provide a step‑by‑step method for graphing and solving these inequalities, highlight common pitfalls, and include a complete answer key for the practice problems typically found in Lesson 7.3.
Understanding Linear Inequalities in Two Variables
A linear inequality in two variables looks like one of the following forms:
- Ax + By < C
- Ax + By ≤ C
- Ax + By > C
- Ax + By ≥ C
where A, B, and C are real numbers, and A and B are not both zero. Unlike an equation, which yields a single line of points, an inequality defines a half‑plane—all the points on one side of the boundary line that satisfy the condition.
Key ideas to remember:
- The boundary line is the line obtained by replacing the inequality symbol with an equals sign (Ax + By = C).
- If the inequality is strict (< or >), the boundary line is dashed, indicating that points on the line are not part of the solution set.
- If the inequality is inclusive (≤ or ≥), the boundary line is solid, showing that points on the line are included.
- To decide which side of the line to shade, pick a test point (commonly the origin (0,0) unless it lies on the boundary) and substitute its coordinates into the original inequality.
Steps to Solve and Graph Linear Inequalities in Two Variables
Follow this systematic procedure for any inequality of the form Ax + By < C (or ≤, >, ≥):
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Rewrite in slope‑intercept form (if helpful)
Solve for y to obtain y < mx + b (or the appropriate symbol). This makes it easy to identify the slope (m) and y‑intercept (b) of the boundary line. -
Graph the boundary line
- Plot the y‑intercept (0, b).
- Use the slope to find a second point (rise over run).
- Draw a dashed line for < or >, and a solid line for ≤ or ≥.
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Select a test point
The origin (0,0) works unless it lies exactly on the boundary line. If the origin is on the line, choose another point such as (1,0) or (0,1) Not complicated — just consistent.. -
Test the inequality
Substitute the test point’s coordinates into the original inequality.- If the statement is true, shade the half‑plane that contains the test point.
- If the statement is false, shade the opposite half‑plane.
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Label the solution region
Optionally, shade lightly and add arrows or the inequality symbol to indicate the direction of the solution set Small thing, real impact. No workaround needed..
Example Walk‑through
Problem: Graph the solution set of 2x − 3y ≥ 6 That's the part that actually makes a difference..
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Solve for y
[ -3y \ge -2x + 6 \quad\Rightarrow\quad y \le \frac{2}{3}x - 2 ]
(Notice the inequality flips when dividing by a negative.) -
Boundary line: y = (2/3)x − 2
- y‑intercept: (0, −2)
- Slope: 2/3 → from (0,−2) go up 2, right 3 to point (3, 0).
- Because the inequality is ≥ (original) which became ≤ after solving for y, we use a solid line.
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Test point: Choose (0,0).
Substitute into original: 2(0) − 3(0) ≥ 6 → 0 ≥ 6 → false. -
Shade: Since the test point fails, shade the side opposite the origin, i.e., the region below the line (consistent with y ≤ (2/3)x − 2).
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Final graph: Solid line through (0,−2) and (3,0); shading below the line.
Common Mistakes and How to Avoid Them
| Mistake | Why It Happens | Correct Approach |
|---|---|---|
| Forgetting to reverse the inequality sign when multiplying/dividing by a negative number | Treating inequality like an equation | Always check the sign of the coefficient you are dividing/multiplying by; flip the direction if it is negative. So |
| Picking a test point that lies on the boundary line | Substituting yields equality leads to 0 = C, giving no information | If the origin (or any convenient point) is on the line, select another point like (1,0) or (0,1). |
| Using a dashed line for ≤ or ≥ | Confusing strict vs. | |
| Shading the wrong side after a successful test | Misreading the direction of shading | Double‑check: if test point makes inequality true, shade the side containing the test point; otherwise, shade the opposite side. Here's the thing — inclusive symbols |
| Not labeling axes or scaling incorrectly | Leads to inaccurate graphs | Use a consistent scale, label both axes, and plot at least two points to define the line precisely. |
Practice Problems with Answer Key
Below are six problems similar to those found in Lesson 7.In real terms, 3. Attempt each one, then compare your work with the provided answer key.
Problem Set
- Graph the solution of x + 2y < 4.
- Graph the solution of −3x + y ≥ −6.
- Graph the solution of 5x − y > 10.
- Graph the solution of **−x − 3
−x − 3y ≤ 9.
5. Graph the solution of 4x + 5y > 20.
6. Graph the solution of y < −½x + 3.
Answer Key
1. x + 2y < 4
- Boundary: Dashed line (strict inequality).
- Intercepts: (0, 2) and (4, 0).
- Slope‑intercept: y < −½x + 2.
- Test (0,0): 0 < 4 → True. Shade toward the origin (below the line).
2. −3x + y ≥ −6
- Boundary: Solid line (inclusive).
- Slope‑intercept: y ≥ 3x − 6.
- Intercepts: (0, −6) and (2, 0).
- Test (0,0): 0 ≥ −6 → True. Shade toward the origin (above the line).
3. 5x − y > 10
- Boundary: Dashed line.
- Slope‑intercept: −y > −5x + 10 → y < 5x − 10 (flip sign).
- Intercepts: (0, −10) and (2, 0).
- Test (0,0): 0 > 10 → False. Shade away from the origin (below the steep line).
4. −x − 3y ≤ 9
- Boundary: Solid line.
- Slope‑intercept: −3y ≤ x + 9 → y ≥ −⅓x − 3 (flip sign).
- Intercepts: (0, −3) and (−9, 0).
- Test (0,0): 0 ≤ 9 → True. Shade toward the origin (above the line).
5. 4x + 5y > 20
- Boundary: Dashed line.
- Slope‑intercept: 5y > −4x + 20 → y > −⅘x + 4.
- Intercepts: (0, 4) and (5, 0).
- Test (0,0): 0 > 20 → False. Shade away from the origin (above the line).
6. y < −½x + 3
- Boundary: Dashed line (already solved for y).
- y‑intercept: (0, 3); slope −½ → next point (2, 2).
- Test (0,0): 0 < 3 → True. Shade toward the origin (below the line).
Conclusion
Graphing linear inequalities is a foundational skill that bridges algebraic manipulation and geometric visualization. By consistently following the four‑step process—rewrite, draw the boundary, test a point, and shade—you transform abstract symbols into a clear picture of all possible solutions.
Key habits to carry forward:
- Flip the inequality whenever you multiply or divide by a negative.
Which means - Match the line style to the symbol (solid for ≤/≥, dashed for </>). - Verify with a test point every time; it takes seconds and prevents shading errors.
As you progress to systems of inequalities, these same principles apply—each inequality contributes a half‑plane, and the solution set is simply the overlap of those regions. Keep practicing with varied slopes and intercepts, and soon the mechanics will become second nature, allowing you to focus on modeling real‑world constraints with confidence Surprisingly effective..
