Solve systems of two linear equations in two variables exactly and approximately by graphing, substitution, and elimination, and justify the elimination method (TN A1.A.REI.C.5, A1.A.REI.C.6).

What this topic is asking

A system of equations is two (or more) equations considered together; its solution is the ordered pair $(x, y)$ that satisfies both. Standard A1.A.REI.C.6 asks you to solve by graphing, substitution, and elimination, and A1.A.REI.C.5 asks you to understand why elimination works (replacing one equation with a sum of multiples gives an equivalent system).

Three methods

• Graphing gives an approximate (or exact, if integer) solution: graph both lines and read the intersection. Best when a graph is provided or the numbers are friendly.
• Substitution is best when one equation is already solved for a variable (or easily can be). Solve for that variable, substitute into the other equation, solve, then back-substitute.
• Elimination is best when adding or subtracting the equations cancels a variable. Scale one or both equations so a variable's coefficients are opposites, then add.

One, none, or infinitely many

The number of solutions matches how the lines sit:

• One solution: different slopes, lines cross once.
• No solution: same slope, different intercept, parallel lines.
• Infinitely many: same slope and same intercept, the equations describe one line.

Algebraically, if the variables cancel and you get a false statement, there is no solution; a true statement means infinitely many, exactly the same logic as single linear equations.

How TNReady examines this topic

• Numeric response. Solve a system and enter one coordinate or the ordered pair.
• Multiple choice. Choose the solution or the number of solutions, with parallel-line and arithmetic distractors.
• Graphing. Plot both lines and mark the intersection.

A clarifying idea is that A1.A.REI.C.5 justifies elimination: adding a multiple of one equation to another does not change the solution set, because any pair that satisfies both original equations also satisfies their combination. That is why you are allowed to scale and add.

Choosing the most efficient method

All three methods give the same answer, so the skill is picking the fastest for the given system. If one equation already reads $y = \dots$ or $x = \dots$, substitution is quickest. If a variable has matching or opposite coefficients (or can get them with one multiplication), elimination is cleanest. If the item supplies a graph or asks for an approximate solution, graphing is intended. A common time sink is forcing substitution when it creates ugly fractions, for instance solving $4x + 3y = 12$ for $y$ gives $y = \frac{12 - 4x}{3}$, which then fractions up the second equation; elimination would avoid that. Reading the system's structure before committing to a method saves both time and arithmetic errors on the calculator subparts.

Try this

Q1. Solve $\begin{cases} y = x + 2 \\ 2x + y = 11 \end{cases}$. [2 points]

• Cue. Substitute: $2x + x + 2 = 11 \Rightarrow 3x = 9 \Rightarrow x = 3$, $y = 5$. Solution $(3, 5)$.

Q2. How many solutions does $\begin{cases} x - y = 4 \\ 2x - 2y = 8 \end{cases}$ have? [1 point]

• Cue. The second is twice the first: same line, infinitely many solutions.