Linear Equations, Slope-Intercept Form, and Graphing Techniques

The slope-intercept form of a linear equation is algebraically expressed as: $y = mx + b$
Variable definitions:
- $y$ : Dependent variable representing the vertical coordinate.
- $x$ : Independent variable representing the horizontal coordinate.
- $m$ : The slope of the line, defining the rate of change and the ratio of vertical change to horizontal change ( $\frac{\text{rise}}{\text{run}}$ ).
- $b$ : The $y$ -intercept, defining the point $\left(0, b\right)$ where the line crosses the vertical $y$ -axis.

Line A:
- Base equation: $y = 2 + x$
- Standard slope-intercept rearrangement: $y = x + 2$
- Identified slope: $m = 1$
- Identified $y$ -intercept: $b = 2$
Line B:
- Base equation: $y = -4 + 2x$
- Standard slope-intercept rearrangement: $y = 2x - 4$
- Identified slope: $m = 2$
- Identified $y$ -intercept: $b = -4$
- Transcription variant noted: $X\,2 - 4 + 2x$
Line C:
- Base equation: $y = 3 - x$
- Standard slope-intercept rearrangement: $y = -x + 3$
- Identified slope: $m = -1$
- Identified $y$ -intercept: $b = 3$
- Transcription variant noted: $8 = 3 - X$

Problem 34:
- Given equation: $y = -3x + 10$
- Current format: Already in standard slope-intercept form ( $y = mx + b$ )
- Identified slope: $m = -3$
- Identified $y$ -intercept: $b = 10$
- Evaluation status: Confirmed correct ("good")
Problem 33:
- Given equation in standard form: $5x + 3y = 15$
- Step 1 (Isolate the variable term containing $y$ by subtracting $5x$ from both sides): $5x - 5x + 3y = -5x + 15$ $3y = -5x + 15$
- Step 2 (Divide every term on both sides by the coefficient $3$ ): $\frac{3y}{3} = \frac{-5x}{3} + \frac{15}{3}$ $y = -\frac{5}{3}x + 5$
- Identified slope: $m = -\frac{5}{3}$
- Identified $y$ -intercept: $b = 5$
Problem 37:
- Given equation: $2y = x + 12$
- Step 1 (Isolate $y$ by dividing each term by $2$ ): $\frac{2y}{2} = \frac{1}{2}x + \frac{12}{2}$ $y = \frac{1}{2}x + 6$
- Identified slope: $m = \frac{1}{2}$
- Identified $y$ -intercept: $b = 6$
Additional Linear Form ( $7x - 8 = y$ ):
- Given equation: $7x - 8 = y$
- Algebraic manipulation notes: $-7x - 7x$ , yielding $-8 = -7x + y$ or rewritten as $-8 = -7x - 4$
- Symmetric property rearrangement to standard slope-intercept form: $y = 7x - 8$
- Identified slope: $m = 7$
- Identified $y$ -intercept: $b = -8$

Problem 34 (Graphing):
- Equation: $y = -x - 5$
- Slope: $m = -1 = -\frac{1}{1}$
- $y$ -intercept: $b = -5$ , corresponding to coordinate point $\left(0, -5\right)$
- Graphing trajectory: From $\left(0, -5\right)$ , move down $1$ unit and right $1$ unit (or up $1$ unit and left $1$ unit)
Problem 35:
- Equation: $y = 2x + 8$
- Slope: $m = 2 = \frac{2}{1}$
- $y$ -intercept: $b = 8$ , corresponding to coordinate point $\left(0, 8\right)$
- Graphing trajectory: From $\left(0, 8\right)$ , move up $2$ units and right $1$ unit (or down $2$ units and left $1$ unit)
- Evaluation status: Confirmed correct ("good")
Problem 38:
- Given equation: $-2x + 2y = -8$
- Step 1 (Add $2x$ to both sides of the equation): $-2x + 2x + 2y = 2x - 8$ $2y = 2x - 8$
- Step 2 (Divide every term by $2$ ): $\frac{2y}{2} = \frac{2x}{2} - \frac{8}{2}$ $y = 1x - 4$ $y = x - 4$
- Identified slope: $m = 1 = \frac{1}{1}$
- Identified $y$ -intercept: $b = -4$ , corresponding to coordinate point $\left(0, -4\right)$
Problem 40:
- Equation: $y = -2x + 4$
- Slope: $m = -2 = -\frac{2}{1}$
- $y$ -intercept: $b = 4$ , corresponding to coordinate point $\left(0, 4\right)$
- Graphing trajectory: From $\left(0, 4\right)$ , move down $2$ units and right $1$ unit

Negative scale markers present on the Cartesian grid axes:
- $-8$ , $-5$ , $-4$ , $-3$ , $-2$ , $-1$
Neutral origin marker:
- $0$
Positive scale markers present on the Cartesian grid axes:
- $1$ , $2$ , $3$ , $4$ , $5$ , $6$ , $7$
Additional raw numerical annotations and coordinate references recorded:
- Sequence: $7, 7, 7, 7, 4, 13, 3, 9, 557$
- Value: $7054$
- Incremental coordinate runs: $1, 2, 3, 4, 5, 6, 7$ and $-8, -5, -4, -3, -2$