5.4 Systems of Two-Variable Linear Equations & Inequalities

5.4 Systems of Two-Variable Linear Equations & Inequalities

Skill 6 of Competency 2 asks for systems solved algebraically, graphically, and in real-world contexts, and it includes inequalities as well as equations.

What a solution is

A solution to a system is an ordered pair (x, y) that satisfies every equation simultaneously. Graphically, it is the point where the lines intersect. Checking a candidate solution means substituting into both equations — satisfying only one is not enough, and items exploit that.

Classification by solution count

+---------------------------------------------------------------------------+
|  Lines intersect once   ->  ONE solution        (consistent, independent) |
|     different slopes                                                      |
|  Lines are parallel     ->  NO solution         (inconsistent)            |
|     same slope, different y-intercept                                     |
|  Lines coincide         ->  INFINITELY many     (consistent, dependent)   |
|     same slope, same y-intercept                                          |
+---------------------------------------------------------------------------+

You can classify without solving by comparing slopes and intercepts. For 2x + 6y = 12 and x + 3y = 9: dividing the first by 2 gives x + 3y = 6, which has the same left side as the second but a different constant, so the lines are parallel and the system has no solution.

Solving by graphing

Skill 6 names graphing explicitly, so expect both directions: reading a solution off a graph, and matching a supplied graph to the system it represents.

Put each equation in slope-intercept form y = mx + b, plot the y-intercept, step off the slope, and read the intersection. For y = 2x − 1 and y = −x + 5 the lines cross at (2, 3), and substituting confirms both: 2(2) − 1 = 3 and −(2) + 5 = 3. Equations already in standard form graph fastest from their intercepts: 3x + 4y = 12 passes through (4, 0) and (0, 3).

Graphing has one real limitation, and items are written to test it. A graph gives an exact answer only when the intersection lands on lattice points you can actually read. If the true solution is (7/3, −1/6), the picture shows only roughly where the lines meet, and algebra is required for the exact pair. Because test 025 supplies an on-screen scientific calculator rather than a graphing one, graph-based items typically hand you the picture and ask what it means: which system it represents, what the intersection represents in context, or which shaded region satisfies the constraints.

Choosing a method

Every correct method reaches the same ordered pair, so a poor choice costs time rather than credit — and on a 50-item, 150-minute test that averages three minutes per question, time is the scarce resource.

Substitution

Best when a variable is already isolated or has coefficient 1.

y = 2x − 3 and 4x + 3y = 21 Substitute: 4x + 3(2x − 3) = 21 → 4x + 6x − 9 = 21 → 10x = 30 → x = 3 Back-substitute: y = 2(3) − 3 = 3 Solution: (3, 3). Check in the second equation: 4(3) + 3(3) = 21 ✓

Elimination

Best when coefficients already match or can be matched by scaling one or both equations. The goal is to make one variable's coefficients opposites, so that adding the equations eliminates it.

When the coefficients are already opposites, add immediately:

4x + 3y = 27 and 2x − 3y = −9 The y-coefficients are already opposites, so add directly: 6x = 18 → x = 3 Substitute: 4(3) + 3y = 27 → 3y = 15 → y = 5 Solution (3, 5). Check: 2(3) − 3(5) = 6 − 15 = −9 ✓

When neither variable is ready, scale both equations to a common multiple:

2x + 5y = 16 and 3x − 2y = 5 Multiply the first by 2 and the second by 5: 4x + 10y = 32 and 15x − 10y = 25 Add: 19x = 57 → x = 3. Then 2(3) + 5y = 16 → 5y = 10 → y = 2.

Clearing fractions and decimals first

Fractional or decimal coefficients are a scaling problem, not a separate method. Multiply each equation through by the least common denominator, or by a power of ten, before choosing substitution or elimination.

(1/2)x + (1/3)y = 4 becomes 3x + 2y = 24 after multiplying every term by 6. 0.25x + 0.1y = 1.5 becomes 25x + 10y = 150 after multiplying by 100, and then 5x + 2y = 30 after dividing by 5.

Multiplying every term, the constant included, is the step students skip most often, and the resulting system is wrong from that point forward.

Recognizing special cases algebraically

If both variables vanish during solving, read the remaining statement:

• 0 = 12 (false) → no solution, parallel lines
• 0 = 0 (true) → infinitely many solutions, the same line twice

This mirrors the single-variable rule from section 5.1.

Modeling with systems

The standard setup pairs a count equation with a value equation.

A school sells 320 tickets, adult tickets at $8 and student tickets at $5, collecting $2,020. How many of each? Count: a + s = 320. Value: 8a + 5s = 2,020. Substitute s = 320 − a: 8a + 5(320 − a) = 2,020 → 8a + 1,600 − 5a = 2,020 → 3a = 420 → a = 140, so s = 180. Check: 140 + 180 = 320 ✓ and 8(140) + 5(180) = 1,120 + 900 = 2,020 ✓

Mixture and rate problems follow the same two-equation shape: one equation for total quantity, one for total value, concentration, or distance.

Break-even and plan-comparison items

The most common real-world system on this test compares two pricing plans.

A club charges a $30 membership fee plus $2 per visit; a drop-in gym charges $10 plus $6 per visit. When do the two cost the same? Set 30 + 2v = 10 + 6v → 20 = 4v → v = 5 visits, at a cost of 30 + 2(5) = $40.

Interpreting that answer is the graded part. At five visits the plans cost the same $40; below five visits the drop-in gym is cheaper, and above five the membership is cheaper. Note that "when is the membership cheaper?" is answered by an inequality, v > 5, and not by the intersection point itself. That gap between solving a system and interpreting it is exactly what skill 6 asks for.

Systems of linear inequalities

Graph each inequality as a half-plane, then take the overlap.

y ≤ −x + 6 and y > 2x − 3 The first has a solid boundary shaded below; the second has a dashed boundary shaded above. The solution is the wedge where both shadings overlap.

Testing a point is the reliable method: (0, 0) satisfies 0 ≤ 6 and 0 > −3, so the origin lies in the solution region, which confirms the shading.

In applications the overlap is called the feasible region. Constraints such as x ≥ 0 and y ≥ 0 restrict it to the first quadrant, and the corner points of the region are where a linear quantity like cost or profit reaches its extreme values. Items ask you to identify which corner satisfies a constraint set, or whether a specific ordered pair lies inside the feasible region — answered by substituting the pair into every inequality.

A system of inequalities can also have no solution: y > x + 4 and y < x − 1 have parallel boundary lines whose half-planes never overlap, so no ordered pair satisfies both.

Errors these items are built to catch

• Dropping a negative while distributing during substitution, as in 5(320 − a).
• Adding two equations before any coefficients are opposites, which eliminates nothing.
• Solving correctly but reporting the wrong variable; distractors are usually the other value.
• Failing to reverse the inequality symbol when an inequality is multiplied or divided by a negative number during scaling.
• Drawing a dashed boundary for ≤ or ≥, or a solid boundary for < or >; strict inequalities exclude their boundary line.
• Treating a non-lattice intersection read off a graph as an exact answer.