16.1 Systems of Linear Equations

16.1 Systems of Linear Equations

Quick Answer: A system of linear equations is a pair of equations in the same two unknowns (usually x and y). The solution is the ordered pair (x, y) that makes both equations true at once—the intersection of two lines. Solve by substitution (solve one equation for a variable, plug into the other) or elimination (add multiples so one variable cancels). Graphically: one intersection → one solution; parallel distinct lines → none; same line → infinitely many. McCann’s PERT Mathematics list explicitly includes simultaneous linear equations in two variables, so this skill is high-yield free points when the algebra is clean.

PERT Math builds from single linear equations into two equations working together. You already know how to isolate a variable and clear coefficients. Systems add one new idea: every candidate answer must satisfy both equations, not just one. On the adaptive Math subtest you will see clean integer or simple-fraction systems—no graphing calculator required, just careful paper work and a final check.

What a System Looks Like

A standard two-variable system is written:

2x + y = 11
 x − y = 1

Or in slope-intercept form:

y = −2x + 11
y = x − 1

Either presentation is the same math. Multiple-choice options may be ordered pairs like (4, 3), “no solution,” or “infinitely many solutions.”

You do not need to graph on PERT. Classification comes from the algebra after substitution or elimination.

Method 1: Substitution

Goal: Express one variable from Equation A, then substitute that expression into Equation B so you temporarily have a one-variable equation.

Workflow:

1. Choose the equation where a variable already has coefficient ±1 (easiest isolation).
2. Solve that equation for that variable.
3. Substitute the expression into the other equation.
4. Solve for the remaining variable.
5. Back-substitute to find the other variable.
6. Check both original equations.

Worked Example 1 — Substitution (clean isolation)

Solve:

x + y = 10     (1)
2x − y = 5     (2)

Step 1: From (1), y = 10 − x.

Step 2: Substitute into (2): 2x − (10 − x) = 5 2x − 10 + x = 5 3x − 10 = 5 3x = 15 x = 5

Step 3: y = 10 − 5 = 5.

Step 4 — Check: (1) 5 + 5 = 10 ✓ (2) 2(5) − 5 = 10 − 5 = 5 ✓

Answer: (5, 5).

Trap: Forgetting the minus sign when substituting: 2x − (10 − x) becomes 2x − 10 + x, not 2x − 10 − x. Distributing the minus is the most common arithmetic error on substitution items.

Worked Example 2 — Substitution when y is given in slope-intercept form

Solve:

y = 3x − 4
y = −x + 8

Both already solved for y. Set them equal:

3x − 4 = −x + 8 3x + x = 8 + 4 4x = 12 x = 3

y = 3(3) − 4 = 9 − 4 = 5

Check second equation: −3 + 8 = 5 ✓

Answer: (3, 5). This is the “set expressions equal” shortcut when both lines are y = mx + b form.

Worked Example 3 — Substitution with fractions

Solve:

(1/2)x + y = 4     (1)
 x − 2y = 2        (2)

Step 1: From (1), y = 4 − (1/2)x.

Step 2: Plug into (2): x − 2[4 − (1/2)x] = 2 x − 8 + x = 2 2x − 8 = 2 2x = 10 x = 5

y = 4 − (1/2)(5) = 4 − 2.5 = 1.5 = 3/2

Check (2): 5 − 2(3/2) = 5 − 3 = 2 ✓

Answer: (5, 3/2). Prefer simplified fractions on paper; match the form used in the options.

Method 2: Elimination (Addition Method)

Goal: Multiply one or both equations by constants so that adding (or subtracting) the equations cancels one variable.

Workflow:

1. Align like terms (x under x, y under y, constants under constants).
2. Choose a variable to eliminate.
3. Multiply equations so that variable’s coefficients become opposites (for example +4 and −4).
4. Add the equations; solve for the remaining variable.
5. Back-substitute; check both originals.

Worked Example 4 — Elimination with opposite coefficients ready

Solve:

3x + 2y = 16     (1)
 x − 2y = 0      (2)

The y-coefficients are already +2 and −2. Add (1) + (2):

(3x + x) + (2y − 2y) = 16 + 0 4x = 16 x = 4

From (2): 4 − 2y = 0 → −2y = −4 → y = 2

Check (1): 3(4) + 2(2) = 12 + 4 = 16 ✓

Answer: (4, 2).

Worked Example 5 — Elimination after multiplying one equation

Solve:

2x + 3y = 7      (1)
4x − y = 11      (2)

Eliminate y: multiply (2) by 3 so the y-coefficients are +3 and −3:

2x + 3y = 7
12x − 3y = 33     (2)×3

Add: 14x = 40 x = 40/14 = 20/7

From (2): 4(20/7) − y = 11 80/7 − y = 11 −y = 11 − 80/7 = (77 − 80)/7 = −3/7 y = 3/7

Check (1): 2(20/7) + 3(3/7) = 40/7 + 9/7 = 49/7 = 7 ✓

Answer: (20/7, 3/7).

Trap: Multiplying only one side of an equation. Every term—including the constant—must be scaled.

Worked Example 6 — Elimination after multiplying both equations

Solve:

3x + 4y = 10     (1)
5x + 6y = 16     (2)

Eliminate x: LCM of 3 and 5 is 15. Multiply (1) by 5 and (2) by 3:

15x + 20y = 50
15x + 18y = 48

Subtract the second from the first: (20y − 18y) = 50 − 48 2y = 2 y = 1

From (1): 3x + 4(1) = 10 → 3x = 6 → x = 2

Check (2): 5(2) + 6(1) = 10 + 6 = 16 ✓

Answer: (2, 1).

Special Cases: None or Infinitely Many

Worked Example 7 — No solution (parallel lines)

Solve:

2x + y = 5       (1)
4x + 2y = 8      (2)

Multiply (1) by 2: 4x + 2y = 10.

Subtract from (2): (4x + 2y) − (4x + 2y) = 8 − 10 → 0 = −2, which is false.

Conclusion: no solution. The second equation is not a multiple of the first with matching constants (left sides scale by 2, but 5×2 = 10 ≠ 8).

Worked Example 8 — Infinitely many solutions (same line)

Solve:

x − 3y = 4       (1)
2x − 6y = 8      (2)

Multiply (1) by 2: 2x − 6y = 8, which is identical to (2).

Elimination: 0 = 0 (true for all x, y on that line).

Conclusion: infinitely many solutions. Every point on the line x − 3y = 4 works (for example (4, 0), (7, 1), (1, −1)).

Choosing Substitution vs Elimination on PERT

Both methods always work when applied carefully. Pick the path with the least arithmetic risk under time pressure—even though PERT Math is untimed, cleaner algebra still protects accuracy and stamina.

Check Protocol (Do Not Skip)

Before submitting:

1. Plug (x, y) into equation 1.
2. Plug the same pair into equation 2.
3. If either fails, redo the algebra—do not “adjust” only one equation.
4. If the item is a special case, confirm the contradiction or identity explicitly rather than guessing from slopes alone.

Worked Example 9 — Full PERT-style mixed system

Which ordered pair solves the system?

5x − 2y = 4
3x + y = 11

Elimination on y: Multiply second equation by 2:

5x − 2y = 4
6x + 2y = 22

Add: 11x = 26 → x = 26/11

From 3x + y = 11: y = 11 − 3(26/11) = (121 − 78)/11 = 43/11

Check first equation: 5(26/11) − 2(43/11) = (130 − 86)/11 = 44/11 = 4 ✓

Answer: (26/11, 43/11). If options are decimals, 26/11 ≈ 2.36 and 43/11 ≈ 3.91—prefer exact fractions when they match a choice.

Common Traps

• Solving only one equation and stopping (one equation has infinitely many points; the system needs both).
• Sign errors when distributing a negative in substitution.
• Multiplying an equation but forgetting the constant term.
• Confusing “no solution” with “x = 0” or “y = 0.”
• Reporting only x when the question asks for the ordered pair.

Putting It Together on Test Day

1. Scan both equations — is a variable free or nearly free? Use substitution.
2. Otherwise eliminate — make opposite coefficients, add, solve, back-substitute.
3. Watch for 0 = k or 0 = 0 — classify no solution or infinitely many.
4. Check both originals — two quick plugs prevent most wrong answers.

Mastering systems closes the McCann simultaneous-equations competency and sets up two-variable word problems in the next section, where the hard part is writing the system and the solving part reuses exactly these methods.