Solve systems of two linear equations in two variables algebraically using substitution and elimination (LA A1: A-REI.C.6).

What this topic is asking

Standard A1: A-REI.C.6 asks you to solve a system of two linear equations in two variables algebraically, with substitution and elimination. On LEAP 2025 these are Type I Major Content items worth multiple points, often equation-response. The solution is an ordered pair $(x, y)$, the point where the two lines meet.

Substitution

Use substitution when one equation is already solved for a variable (or is easy to solve).

Elimination

Use elimination when a variable has matching or opposite coefficients, or can be made so by multiplying an equation.

When coefficients do not match, multiply one or both equations first. To eliminate $x$ from $2x + y = 7$ and $3x - y = 8$, the $y$ terms are already opposites, so add; but to eliminate $x$ you would scale to $6x$ and $6x$ and subtract.

No solution and infinitely many

As with one-variable equations, if both variables cancel:

• A true statement (like $0 = 0$) means the two equations are the same line: infinitely many solutions.
• A false statement (like $0 = 5$) means the lines are parallel: no solution.

How LEAP examines this topic

• Equation response. Solve a system and enter the ordered pair.
• Multiple choice. Pick the solution, or identify a system with no solution or infinitely many.
• Type III modeling. Set up a system from a context and solve it (see the modeling topic).

A clarifying idea: substitution and elimination give the same answer, because both find the single point that satisfies both equations. Choose whichever the system makes easier, substitution when one variable is isolated, elimination when coefficients line up.

Why both methods find the intersection

Substitution and elimination are two routes to the same geometric fact: the solution is where the two lines cross, the one point lying on both. Substitution enforces this directly, it says "at the solution, $y$ has the same value in both equations," so it replaces $y$ in one equation with its expression from the other, leaving a single equation in $x$ whose answer is the crossing point's $x$-coordinate. Elimination enforces it differently, by combining the equations so one variable disappears; adding or subtracting true equations produces another true equation, and choosing the combination that cancels a variable isolates the other coordinate of the intersection. Because both methods are just valid algebra applied to "both equations hold at once," they cannot disagree. The special cases fall out naturally: parallel lines never cross, so the algebra collapses to a false statement (no point works), and identical lines cross everywhere, so it collapses to a true statement (every point on the line works). Seeing the solution as an intersection point is what ties this algebraic topic to solving systems by graphing.

Try this

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

• Cue. $x + 3x = 8 \Rightarrow x = 2$, $y = 6$; solution $(2, 6)$.

Q2. Solve $x + y = 10$ and $x - y = 2$ by elimination. [3 points]

• Cue. Add: $2x = 12 \Rightarrow x = 6$, then $y = 4$; solution $(6, 4)$.