Comprehensive Guide to Identifying Linear Equations and Graphs

Selecting Test Points for Differentiation:
- When multiple graphs are plotted on the same coordinate plane, it is essential to choose an $x$ -value that yields unique $y$ -values for each function.
- Warning regarding the origin: In the first set of equations ( $y = x$ , $y = 2x$ , and $y = -3x$ ), one must avoid choosing $x = 0$ as a test point. Because all three equations have a $y$ -intercept of $0$ , all three graphs intersect at the point $(0, 0)$ . Choosing $x = 0$ would result in $y = 0$ for all calculations, making it impossible to distinguish between the lines.
The Substitution Method ( $x = 1$ ):
- A common and effective strategy is to choose $x = 1$ as the test point to determine the corresponding $y$ -coordinate for each graph.
The Slope-Intercept Method ( $y = mx + b$ ):
- Equations can be rewritten or interpreted in the form $y = mx + b$ , where $m$ represents the slope ( $\text{rise}/\text{run}$ ) and $b$ represents the $y$ -intercept.
- Analyzing the vertical shift ( $b$ ) and the steepness/direction of the line ( $m$ ) allows for visual identification without plotting multiple points.

Equation 1: $y = x$
- Point Plotting Analysis: Using the test point $x = 1$ , we find $y = 1$ . This yields the coordinate pair $(1, 1)$ . Based on this point, the equation corresponds to Graph C.
- Slope-Intercept Analysis: The equation $y = x$ is equivalent to $y = \frac{1}{1}x + 0$ . This indicates a $y$ -intercept of $0$ and a slope where the rise is $1$ and the run is $1$ . This confirms the identification of Graph C.
Equation 2: $y = 2x$
- Point Plotting Analysis: Using the test point $x = 1$ , the calculation is $y = 2(1) = 2$ . This yields the coordinate pair $(1, 2)$ . Based on this point, the equation corresponds to Graph A.
- Slope-Intercept Analysis: The equation $y = 2x$ is equivalent to $y = \frac{2}{1}x + 0$ . This indicates a $y$ -intercept of $0$ and a steeper slope where the rise is $2$ and the run is $1$ . This confirms the identification of Graph A.
Equation 3: $y = -3x$
- Point Plotting Analysis: Using the test point $x = 1$ , the calculation is $y = -3(1) = -3$ . This yields the coordinate pair $(1, -3)$ . Based on this point, the equation corresponds to Graph B.
- Slope-Intercept Analysis: The equation $y = -3x$ is equivalent to $y = \frac{-3}{1}x + 0$ . This indicates a $y$ -intercept of $0$ and a negative slope where the rise is $-3$ (a drop of $3$ ) and the run is $1$ . This confirms the identification of Graph B.

Initial Observation of Intercepts:
- Before calculating specific points, note the $y$ -intercepts: two equations share a $y$ -intercept of $3$ , while one equation has a $y$ -intercept of $-3$ .
Equation 4: $y = 3x + 3$
- Point Plotting Analysis: Choosing $x = 1$ , the calculation is $y = 3(1) + 3 = 3 + 3 = 6$ . This yields the coordinate pair $(1, 6)$ . This corresponds to Graph C.
- Slope-Intercept Analysis: This equation has a $y$ -intercept of $3$ . The slope ( $m = 3$ ) indicates a rise of $3$ and a run of $1$ . This confirms Graph C.
Equation 5: $x + y = 3$
- Point Plotting Analysis: Using $x = 1$ , the equation becomes $1 + y = 3$ . Solving for $y$ gives $y = 2$ . This yields the coordinate pair $(1, 2)$ . This corresponds to Graph A.
- Slope-Intercept Analysis: Converting the equation to slope-intercept form results in $y = -x + 3$ (or $y = \frac{-1}{1}x + 3$ ). The $y$ -intercept is $3$ , but the slope is negative, meaning a "drop" of $1$ and a run of $1$ . This confirms Graph A.
Equation 6: $y = 3x - 3$
- Point Plotting Analysis: Using $x = 1$ , the calculation is $y = 3(1) - 3 = 3 - 3 = 0$ . This yields the coordinate pair $(1, 0)$ . This corresponds to Graph B.
- Slope-Intercept Analysis: The equation has a $y$ -intercept of $-3$ . The slope ( $m = 3$ ) indicates a rise of $3$ and a run of $1$ . This confirms Graph B.