1. **Problem:** Find the x- and y-intercepts of the linear equation $3x + 4y = 12$. 2. **Formula and rules:** - The x-intercept is found by setting $y=0$ and solving for $x$. - The y-intercept is found by setting $x=0$ and solving for $y$. 3. **Find x-intercept:** Set $y=0$ in $3x + 4y = 12$: $$3x + 4(0) = 12$$ $$3x = 12$$ Divide both sides by 3: $$\cancel{3}x = \frac{12}{\cancel{3}}$$ $$x = 4$$ 4. **Find y-intercept:** Set $x=0$ in $3x + 4y = 12$: $$3(0) + 4y = 12$$ $$4y = 12$$ Divide both sides by 4: $$\cancel{4}y = \frac{12}{\cancel{4}}$$ $$y = 3$$ 5. **Answer:** - The x-intercept is at $(4, 0)$. - The y-intercept is at $(0, 3)$. This means the graph crosses the x-axis at 4 and the y-axis at 3.