2.4 Systems of Two Linear Equations

2.4 Systems of Two Linear Equations

The last Table 11 skill in the linear-equations slice is solving a system of two linear equations in two variables. Each equation is a line in the plane. Two lines either cross once, run parallel and distinct, or lie on top of each other. Those three geometric facts are exactly the three algebraic outcomes: one ordered pair, no solution, or infinitely many solutions.

/practice/accuplacer-advanced-algebraPractice questions with detailed explanations

The three solution cases

Rewrite each equation as y = m₁x + b₁ and y = m₂x + b₂ (when neither line is vertical) and compare.

• Different slopes (m₁ ≠ m₂): the lines intersect at exactly one point. Unique solution.
• Same slope, different intercepts (m₁ = m₂ but b₁ ≠ b₂): parallel distinct lines. No solution.
• Same slope and same intercept (m₁ = m₂ and b₁ = b₂): coincident lines. Infinitely many solutions — every point on the line.

If one equation is vertical, x = k, it is not a function, but it still makes a system: substitute x = k into the other equation and read y. Two distinct vertical lines (x = 2 and x = 5) are parallel and have no solution. The same vertical line written twice is coincident.

Worked example: elimination with integers

Solve

x + y = 10

2x − y = 5.

The y terms are already opposites (+y and −y). Add the equations term by term:

(x + 2x) + (y − y) = 10 + 5 → 3x = 15 → x = 5.

Back-substitute into x + y = 10: 5 + y = 10 → y = 5.

The ordered pair is (5, 5). Check the second equation: 2(5) − 5 = 5, true. Graphically, y = 10 − x and y = 2x − 5 meet at (5, 5) because the slopes −1 and 2 differ.

Worked example: substitution

Solve

y = 3x − 1

2x + y = 9.

The first equation already isolates y. Substitute 3x − 1 for y in the second:

2x + (3x − 1) = 9 → 5x − 1 = 9 → 5x = 10 → x = 2.

Then y = 3(2) − 1 = 5. Solution: (2, 5).

Notice that y = 3x − 1 is the same line you wrote from the table (2, 5), (4, 11) in Section 2.1. The second equation 2x + y = 9 is another line through (2, 5). Unique intersection, as the slopes 3 and −2 differ.

Substitution is fastest when one variable is already isolated or has coefficient ±1. Elimination is fastest when coefficients of one variable are equal or opposite, or when a small integer multiplier makes them so.

Worked example: multiply the entire equation

Solve

3x + 2y = 16

x − y = 2.

To eliminate y, multiply the second equation by 2:

2(x − y) = 2(2) → 2x − 2y = 4.

The trap is writing 2x − y = 4 or 2x − 2y = 2 — scaling the variables but not the constant, or scaling the constant but not both variable terms. Every term, including the right-hand side, must be multiplied.

Now add to the first equation? Better: align y opposites by using 3x + 2y = 16 and 2x − 2y = 4:

Add: 5x = 20 → x = 4.

From x − y = 2: 4 − y = 2 → y = 2. Solution (4, 2).

Check: 3(4) + 2(2) = 12 + 4 = 16. Both equations hold.

If the item had asked you to eliminate x instead, you would multiply the second equation by 3 to get 3x − 3y = 6, then subtract from the first: (3x + 2y) − (3x − 3y) = 16 − 6 → 5y = 10 → y = 2, same pair. Choose the variable whose coefficients are easier to match.

No solution and infinitely many

No solution (parallel). Compare

2x + 4y = 10

x + 2y = 8.

Divide the first by 2: x + 2y = 5. The second says x + 2y = 8. The left sides are identical and the right sides are not: 5 = 8 is false. After elimination the variables vanish into a contradiction. In slope-intercept form both lines have slope −1/2, but intercepts 5/2 and 4. Parallel and distinct: no solution.

Infinitely many (coincident). Compare

2x + 4y = 10

x + 2y = 5.

The second is exactly half the first. Elimination produces 0 = 0, an identity. Every point on x + 2y = 5 — for example (5, 0), (3, 1), (1, 2) — satisfies both equations. Report infinitely many solutions. If an item asks for the solution set, you may write the line itself: all (x, y) with x + 2y = 5.

AAF options often list a specific pair such as (1, 2) next to “infinitely many” and “no solution.” A pair that happens to lie on a coincident system is true but incomplete if the item asks for the full solution set. Read whether the stem wants a solution or the solution set.

Graphical interpretation in one paragraph

Reading a graph versus predicting from slopes

Each equation is a line. Solving the system is finding intersection points. One crossing: one ordered pair. Two distinct parallels: empty intersection. One line drawn twice: the whole line. You do not need a perfect sketch on AAF, but you should be able to read a graph of two lines and name the case, and you should be able to predict the case from slopes without graphing. That prediction is faster than substitution when the question is qualitative (“how many solutions?”).

Skills Insight band 237–249 is the integer-coefficient systems band: expect systems like the four worked above, not Greek-letter parameters. Higher bands add nonlinear partners — the 263–275 band relates the solutions of a linear-plus-nonlinear system (a line and a parabola) to the graphs, which sit in later chapters. For this chapter, master multiply every term, classify 0 = 0 versus 0 = 5, and name the ordered pair when it exists.

A compact playbook

1. If one variable is isolated, substitute.
2. Otherwise pick a variable and multiply entire equations so that one pair of coefficients is equal or opposite, then add or subtract.
3. If both variables vanish, classify: true → infinitely many; false → none.
4. If you get a number for x, find y from the simpler original equation and check both originals.
5. Optional speed check: compare slopes. Different → exactly one; same and different intercepts → none; identical lines → infinitely many.

Elimination errors are almost always incomplete multiplication. Substitution errors are almost always dropping a sign when you replace y with an expression such as 3x − 1. Both are the same arithmetic traps as Section 2.2, now running on two lines at once. Slow down for one extra second on the multiplier, then the CAT item is a 237–249 gift rather than a low-band stall.