Solve systems of two linear equations by graphing, substitution, and elimination, and interpret the solution as the point where the lines meet (A.PAR, Patterning and Algebraic Reasoning).

What this topic is asking

This Patterning and Algebraic Reasoning (A.PAR) standard asks you to solve a system of two linear equations by three methods, graphing, substitution, and elimination, and to interpret the solution as the point where the lines meet. The Georgia Milestones EOC tests all three methods and, crucially, the ability to choose the efficient one for a given system. It also tests the special cases: parallel lines (no solution) and identical lines (infinitely many). Systems are a frequent source of two-point items because they reward a clean multi-step process.

Method 1: graphing

Graph both lines; the intersection point is the solution. This is the method behind the meaning of a system, and the EOC uses graphing tools for it, but it is exact only when the intersection lands on a lattice point. For non-integer solutions, an algebraic method is more reliable.

Method 2: substitution

Substitution works best when one equation is already solved for a variable (or easy to solve).

Method 3: elimination

Elimination works best when a variable has matching or easily matched coefficients. Add or subtract the equations to cancel that variable.

For $2x + 3y = 12$ and $2x - y = 4$, both have $2x$, so subtracting eliminates $x$: $4y = 8$, $y = 2$; then $x = 3$. When coefficients do not match, multiply one or both equations by a constant first: to solve $3x + 2y = 7$ and $2x + 5y = 12$, scale to make the $x$ (or $y$) coefficients opposite, then add.

The special cases

The number of solutions matches the geometry of two lines.

• One solution. The lines have different slopes and cross once.
• No solution. The lines are parallel (same slope, different y-intercept) and never meet. Algebraically, the variables cancel and a false statement remains.
• Infinitely many solutions. The lines are identical (same slope and intercept). Algebraically, the variables cancel and a true statement remains.

This mirrors the one-variable cases: a vanished variable leaving true means infinitely many, leaving false means none.

How the Milestones examines this topic

• Numeric entry. Solve a system and type the ordered pair.
• Multiple choice. Identify the solution, or determine whether a system has one, none, or infinitely many solutions.
• Constructed response. Solve by a named method and justify the choice, or set up and solve a system from a context.

Why all three methods agree

The three methods look different but solve the same problem: finding the single point that lies on both lines. Graphing locates that point by sight; substitution finds it by reducing two equations to one equation in one variable; elimination finds it by combining the equations so one variable disappears. Because each method is just a different bookkeeping route to "the point on both lines," they must give the same answer, and that is why checking the solution in both original equations is the right verification. If your point satisfies only one equation, you have the intersection of that line with something else, a sign of an arithmetic slip. Seeing the methods as three paths to one geometric idea also tells you when each is efficient: pick the one whose algebra is already half-done by the form the system is given in.

Systems from context

Word problems often hide a system. "Two adult tickets and three child tickets cost $34; one adult and four child tickets cost$28" becomes $2a + 3c = 34$ and $a + 4c = 28$. Solving (elimination after scaling, or substitution from the second equation) gives the per-ticket prices. The modeling skill is assigning a variable to each unknown and writing one equation per condition, then solving by whichever method the equations invite. Defining the variables explicitly ("let $a$ be the adult price") and labeling the answer with units is part of the constructed-response credit.

Try this

Q1. Solve $y = x - 2$ and $2x + y = 10$. [1 point]

• Cue. Substitute: $2x + (x - 2) = 10$, so $3x = 12$, $x = 4$, $y = 2$; $(4, 2)$.

Q2. How many solutions does $\begin{cases} y = 3x + 1 \\ y = 3x - 4 \end{cases}$ have? [1 point]

• Cue. Same slope 3, different intercepts, so parallel, no solution.