The exercise in the image involves graphing linear equations and finding the x- and y-intercepts of the linear equations in questions 7 and 8. Here's a breakdown of how to approach these problems:

Exercises 3-6: Graphing Linear Equations

• 3. $x = 4$: This is a vertical line at $x = 4$, passing through all points where the x-coordinate is 4. The graph will be a straight vertical line crossing the x-axis at $x = 4$.

• 4. $y = 2$: This is a horizontal line at $y = 2$, passing through all points where the y-coordinate is 2. The graph will be a straight horizontal line crossing the y-axis at $y = 2$.

• 5. $y = -3$: This is a horizontal line at $y = -3$, passing through all points where the y-coordinate is -3. The graph will be a straight horizontal line crossing the y-axis at $y = -3$.

• 6. $x = -1$: This is a vertical line at $x = -1$, passing through all points where the x-coordinate is -1. The graph will be a straight vertical line crossing the x-axis at $x = -1$.

Exercises 7-8: Finding x- and y-Intercepts

To find the intercepts, we set $y = 0$ to find the x-intercept and $x = 0$ to find the y-intercept.

• 7. $2x + 3y = 12$

  • x-intercept: Set $y = 0$. $2x + 3(0) = 12 \quad \Rightarrow \quad x = 6$ The x-intercept is $(6, 0)$.

  • y-intercept: Set $x = 0$. $2(0) + 3y = 12 \quad \Rightarrow \quad y = 4$ The y-intercept is $(0, 4)$.

• 8. $3x + 6y = 24$

  • x-intercept: Set $y = 0$. $3x + 6(0) = 24 \quad \Rightarrow \quad x = 8$ The x-intercept is $(8, 0)$.

  • y-intercept: Set $x = 0$. $3(0) + 6y = 24 \quad \Rightarrow \quad y = 4$ The y-intercept is $(0, 4)$.

Would you like a more detailed explanation or have any questions?

Here are 5 related questions:

1. How do you find the slope of a line given two points?
2. Can you explain how to graph a linear equation in slope-intercept form?
3. How do vertical and horizontal lines differ in terms of slope?
4. What is the general form of a linear equation?
5. How do you solve a system of linear equations graphically?

Tip: For vertical lines like $x = 4$, the slope is undefined, while for horizontal lines like $y = -3$, the slope is zero.