Question

8th grade midterm study guide for math 2025-26

1. graph the following two linear equations on the same graph.

a. $y = \frac{1}{3}x - 3$
b. $y = -\frac{1}{3}x + 3$
graph

1. graph the following two linear equations on the same graph. (hint: get in $y = mx + b$ form)

c. $y - 4 = 3(x - 2)$
d. $2x - 4y = 8$
graph

Explanation:

Step1: Analyze Equation a ($y = \frac{1}{3}x - 3$)

This is in slope - intercept form ($y=mx + b$), where the slope $m=\frac{1}{3}$ and the y - intercept $b=-3$. To graph this line, start by plotting the y - intercept at the point $(0, - 3)$. Then, use the slope to find another point. The slope $\frac{1}{3}$ means for every 3 units we move to the right (increase in $x$ by 3), we move up 1 unit (increase in $y$ by 1). So from $(0,-3)$, moving 3 units right to $x = 3$ and 1 unit up gives the point $(3,-2)$.

Step2: Analyze Equation b ($y=-\frac{1}{3}x + 3$)

This is also in slope - intercept form with slope $m =-\frac{1}{3}$ and y - intercept $b = 3$. Plot the y - intercept at $(0,3)$. Using the slope, for every 3 units we move to the right (increase in $x$ by 3), we move down 1 unit (decrease in $y$ by 1). So from $(0,3)$, moving 3 units right to $x = 3$ and 1 unit down gives the point $(3,2)$.

Step3: Analyze Equation c ($y - 4=3(x - 2)$)

First, convert it to slope - intercept form. Expand the right - hand side: $y-4 = 3x-6$. Then add 4 to both sides: $y=3x - 6 + 4=3x-2$. The slope $m = 3$ and the y - intercept $b=-2$. Plot the y - intercept at $(0,-2)$. Using the slope (rise over run, $m = 3=\frac{3}{1}$), for every 1 unit we move to the right (increase in $x$ by 1), we move up 3 units (increase in $y$ by 3). So from $(0,-2)$, moving 1 unit right to $x = 1$ and 3 units up gives the point $(1,1)$.

Step4: Analyze Equation d ($2x-4y = 8$)

Convert to slope - intercept form. Subtract $2x$ from both sides: $-4y=-2x + 8$. Divide every term by $-4$: $y=\frac{-2x}{-4}+\frac{8}{-4}=\frac{1}{2}x-2$. The slope $m=\frac{1}{2}$ and the y - intercept $b = - 2$. Plot the y - intercept at $(0,-2)$. Using the slope, for every 2 units we move to the right (increase in $x$ by 2), we move up 1 unit (increase in $y$ by 1). So from $(0,-2)$, moving 2 units right to $x = 2$ and 1 unit up gives the point $(2,-1)$.

(Note: Since the problem just asks to graph the equations, the final answer is the graphical representation of the four lines as described above. However, if we were to describe the graphs:

• For 18a and 18b: Two lines with slopes $\frac{1}{3}$ and $-\frac{1}{3}$ (negative reciprocals? No, $\frac{1}{3}\times(-\frac{1}{3})=-\frac{1}{9}

eq - 1$, they are not perpendicular, but they have slopes that are negative of each other with the same magnitude) intersecting at some point (solving $\frac{1}{3}x-3=-\frac{1}{3}x + 3$ gives $\frac{2}{3}x=6\Rightarrow x = 9$, $y=0$).

• For 19c and 19d: Equation c has a steeper slope (slope 3) than equation d (slope $\frac{1}{2}$), and both have a y - intercept at $(0,-2)$? Wait, no, equation c has $y = 3x-2$ (y - intercept $(0,-2)$) and equation d has $y=\frac{1}{2}x - 2$ (y - intercept $(0,-2)$). So they intersect at the y - intercept? Let's check: set $3x-2=\frac{1}{2}x-2$. Subtract $\frac{1}{2}x$ and add 2: $\frac{5}{2}x=0\Rightarrow x = 0$, $y=-2$. So they intersect at $(0,-2)$.)

Answer:

The graphs are drawn as follows:

• For $y=\frac{1}{3}x - 3$: A line with slope $\frac{1}{3}$ passing through $(0,-3)$ and $(3,-2)$.
• For $y =-\frac{1}{3}x + 3$: A line with slope $-\frac{1}{3}$ passing through $(0,3)$ and $(3,2)$.
• For $y = 3x-2$ (from $y - 4=3(x - 2)$): A line with slope 3 passing through $(0,-2)$ and $(1,1)$.
• For $y=\frac{1}{2}x-2$ (from $2x - 4y=8$): A line with slope $\frac{1}{2}$ passing through $(0,-2)$ and $(2,-1)$.

(To actually draw the graphs on the provided coordinate planes, plot the points as calculated and draw a straight line through them for each equation.)