Solve systems of two linear equations in two variables by graphing, substitution, and elimination, and determine whether a system has one solution, no solution, or infinitely many (TEKS A.5C, A.3E).

What this topic is asking

A system of two linear equations asks for the point that satisfies both equations at once. TEKS A.5C and A.3E ask you to solve such systems by graphing, substitution, and elimination, and to determine the number of solutions. Systems are a reliable part of Reporting Category 3 and feed directly into the modeling questions.

Method 1: graphing

Graph both lines and read the intersection point. This is quick when the lines are easy to graph and the intersection is a lattice point, and it is the method the calculator supports directly. Its weakness is precision: a non-integer intersection is hard to read exactly, so confirm by substitution.

Method 2: substitution

Use substitution when one equation has a variable already isolated (or easy to isolate). Solve one equation for a variable, substitute that expression into the other, solve, then back-substitute.

For $y = 3x + 1$ and $2x + y = 11$: substitute to get $2x + (3x + 1) = 11$, so $5x + 1 = 11$, $x = 2$, and $y = 3(2) + 1 = 7$, giving $(2, 7)$.

Method 3: elimination

Use elimination when adding or subtracting the equations cancels a variable. Scale one or both equations first if needed so a pair of coefficients becomes opposite.

Number of solutions

The relationship between the two lines determines the count.

• One solution. Different slopes; the lines cross once.
• No solution. Same slope, different $y$-intercept; the lines are parallel. Algebraically, the variables cancel and a false statement remains.
• Infinitely many. Same slope and same intercept; the equations are the same line. Algebraically, the variables cancel and a true statement remains.

So $y = 2x + 1$ and $y = 2x - 4$ have no solution (parallel), while $y = 2x + 1$ and $2y = 4x + 2$ have infinitely many (identical).

How STAAR examines this topic

• Multiple choice. Solve a system and pick the ordered pair, with the coordinates-reversed distractor.
• Equation editor and number entry. Enter the solution pair, or a single coordinate.
• Inline choice. Classify the number of solutions from the slopes and intercepts.

A clarifying idea is that all three methods find the same intersection point, so choosing the method that fits the system, substitution when a variable is isolated, elimination when coefficients align, saves time, while graphing offers a visual check.

Checking a proposed solution

A fast way to handle a multiple-choice system, especially on the no-calculator portions of practice, is to test the answer choices. A candidate ordered pair is the solution only if it satisfies both equations, so substitute it into each. The pair $(3, 5)$ solves $y = 2x - 1$ and $x + y = 8$ because $5 = 2(3) - 1$ and $3 + 5 = 8$; a pair that fits only one equation is not a solution. This back-check is also the safeguard against the coordinate-reversal error, since $(5, 3)$ would fail the first equation immediately.

Reading the number of solutions from the equations

You can often classify a system without fully solving it by comparing slopes and intercepts. Put both equations in slope-intercept form: different slopes guarantee exactly one solution; equal slopes with different intercepts give no solution; equal slopes and equal intercepts give infinitely many. This is quicker than solving when an inline-choice item only asks for the count, and it reinforces the geometric picture of two lines crossing, running parallel, or coinciding.

Try this

Q1. Solve $y = x + 2$ and $3x + y = 10$. [1 point]

• Cue. $3x + (x + 2) = 10 \Rightarrow 4x = 8 \Rightarrow x = 2$, $y = 4$: $(2, 4)$.

Q2. How many solutions do $y = 4x - 3$ and $y = 4x + 5$ have? [1 point]

• Cue. Same slope, different intercept: parallel, no solution.