Solve systems of two linear equations in two variables algebraically by substitution and elimination (NC.M1.A-REI.6).

What this topic is asking

NC.M1.A-REI.6 asks you to solve a system of two linear equations in two variables algebraically and to estimate solutions graphically. The two algebraic methods are substitution and elimination. A solution is an ordered pair $(x, y)$ that satisfies both equations at once, the point where the two lines meet.

What a solution means

A solution must satisfy both equations simultaneously.

This is why checking requires substituting into both equations: a pair that fits only one is not a solution.

The substitution method

Substitution shines when one equation is already solved for a variable (like $y = \ldots$).

The elimination method

Elimination shines when a variable has matching or opposite coefficients (or can be made so).

When no coefficients match, multiply one or both equations by a constant first so a variable cancels.

No solution and infinitely many solutions

As with single equations, watch for the variables canceling:

• All variables cancel with a false statement ($0 = 5$): no solution, the lines are parallel.
• All variables cancel with a true statement ($0 = 0$): infinitely many solutions, the equations are the same line.

How the NC Math 1 EOC examines this topic

• Gridded response. Enter the $x$- or $y$-value of the solution.
• Multiple choice. Choose the ordered-pair solution, or identify a system with no solution.
• Calculator-active. Many system items appear in the calculator-active section, often as context problems.

Algebraic solving complements solving by graphing, where the same solution appears as an intersection point, and underlies modeling with systems, where two real conditions become two equations.

Why two equations are needed for two unknowns

A single equation in two variables, like $x + y = 10$, has infinitely many solutions (a whole line of points). To pin down one pair, you need a second independent condition, a second equation. The intersection of the two lines is the single point that satisfies both, which is why a well-posed system has exactly one solution unless the lines happen to be parallel (none) or identical (infinitely many). This is the same counting principle throughout algebra: one equation per unknown. Recognizing it tells you in advance whether a problem is solvable and warns you when a system is degenerate.

Choosing a method

• Use substitution when a variable is already isolated or easy to isolate (coefficient $1$).
• Use elimination when coefficients line up to cancel, or can be scaled to.

Both give the same answer; the choice is about which is less arithmetic.

Try this

Q1. Solve $y = 3x$ and $x + y = 8$. [2 points]

• Cue. Substitute: $x + 3x = 8 \Rightarrow x = 2$, $y = 6$. Solution $(2, 6)$.

Q2. Solve $x + y = 7$ and $x - y = 1$. [2 points]

• Cue. Add: $2x = 8 \Rightarrow x = 4$; then $y = 3$. Solution $(4, 3)$.