7.2 Systems of Linear Equations and Inequalities

Quick Answer: A system of two linear equations has one solution (intersecting lines, different slopes), no solution (parallel lines, same slope but different y-intercepts), or infinitely many solutions (the same line written two ways). Solve algebraically with substitution (best when one variable has coefficient 1) or elimination (best when coefficients can be matched). For systems of inequalities, graph each inequality, shade each half-plane, and the overlapping region is the solution set.

Classifying Systems by Solution Count (MA.912.AR.9.1)

Before computing, classify the system to predict what you will find.

Worked classification: y = 2x + 5 and y = 2x − 1. Slopes are equal (both 2), intercepts differ (5 vs −1), so the system is inconsistent—no solution. Graphically the lines never intersect.

Solving by Substitution

When to use: one equation already has a variable isolated or can be isolated cleanly.

System: y = 3x + 1 and 2x + y = 17.

1. Substitute 3x + 1 for y in the second equation: 2x + (3x + 1) = 17.
2. Combine like terms: 5x + 1 = 17.
3. Subtract 1: 5x = 16.
4. Divide by 5: x = 16/5 = 3.2.
5. Back-substitute: y = 3(3.2) + 1 = 10.6.

The solution is (3.2, 10.6). Verify in the unused equation: 2(3.2) + 10.6 = 6.4 + 10.6 = 17. ✓

Solving by Elimination

When to use: coefficients of one variable can be matched by multiplying one or both equations.

System: 3x + 2y = 12 and 5x − 2y = 20.

The y-coefficients are already opposites (+2 and −2), so adding the equations eliminates y: 8x = 32, so x = 4. Substitute back: 3(4) + 2y = 12 → 12 + 2y = 12 → 2y = 0 → y = 0. The solution is (4, 0).

If no pair of coefficients is already matched, multiply one or both equations. For 2x + 3y = 8 and x − y = 1, multiply the second equation by 2 to get 2x − 2y = 2, then subtract from the first: 5y = 6, so y = 6/5 and x = 11/5.

Systems of Linear Inequalities (MA.912.AR.9.4)

A system of linear inequalities is two or more inequalities considered together. The solution is the set of points satisfying every inequality simultaneously—the overlap of the individual shaded half-planes.

Graphing Procedure

1. Graph each inequality's boundary line (solid for ≤/≥, dashed for </>).
2. Shade each half-plane by testing a point (typically the origin) in each inequality.
3. The solution region is where all shaded regions overlap. Darken or label that region.

Worked example: y > x − 2 and y ≤ −x + 4.

• First boundary: y = x − 2, dashed (strict). Test (0, 0): 0 > −2 true, shade above this line.
• Second boundary: y = −x + 4, solid (non-strict). Test (0, 0): 0 ≤ 4 true, shade below this line.
• Solution region: the overlap, a wedge-shaped area above the first line and below the second.

A point is a solution of the system only if it satisfies both inequalities. A point on a dashed boundary is never a solution of that inequality, even if it satisfies the other.

Real-World System Applications (MA.912.AR.9.6)

Florida EOC word problems build systems from two constraints. Common templates:

Worked example: A museum sells adult tickets at $12 and child tickets at $7. On a certain day, 250 tickets are sold for total revenue of $2,400. How many of each type were sold?

Let a = adult tickets, c = child tickets.

• Count equation: a + c = 250.
• Revenue equation: 12a + 7c = 2400.

Solve by substitution: c = 250 − a. Substitute: 12a + 7(250 − a) = 2400 → 12a + 1750 − 7a = 2400 → 5a = 650 → a = 130. Then c = 250 − 130 = 120. Check: 12(130) + 7(120) = 1560 + 840 = 2400. ✓ 130 adult and 120 child tickets were sold.

Florida EOC System Traps

• Answer choices with the variables swapped (adult tickets reported as 120 instead of 130). Always label variables and check which quantity the question asks for.
• Negative solutions in a context that forbids them (negative tickets, negative time). If your algebra gives a = −10, the system is set up incorrectly—re-read the constraints.
• 'Infinite solutions' disguised as 'no solution.' Eliminating both variables and getting 0 = 5 means no solution (parallel lines). Getting 0 = 0 means infinitely many (the same line). Example: 2x + 4y = 10 and x + 2y = 5 are the same line, so infinitely many solutions.
• Forgetting to substitute back after solving for one variable. An answer choice listing only the x-value is wrong when the question asks for the ordered pair.

Pick the method that minimizes fractions; verify in the unused equation.