1. **State the problem:** We need to sketch the graph of the linear equation $x + 3y = 14$. 2. **Rewrite the equation in slope-intercept form:** The slope-intercept form is $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept. Starting with: $$x + 3y = 14$$ Subtract $x$ from both sides: $$3y = 14 - x$$ Divide both sides by 3: $$y = \frac{14 - x}{3}$$ Show canceling for clarity: $$y = \frac{\cancel{14} - x}{\cancel{3}}$$ Simplify: $$y = \frac{14}{3} - \frac{1}{3}x$$ 3. **Identify slope and intercept:** - Slope $m = -\frac{1}{3}$ - Y-intercept $b = \frac{14}{3}$ 4. **Find x-intercept:** Set $y=0$ and solve for $x$: $$0 = \frac{14 - x}{3}$$ Multiply both sides by 3: $$0 = 14 - x$$ Solve for $x$: $$x = 14$$ 5. **Plot points:** - Y-intercept at $(0, \frac{14}{3})$ - X-intercept at $(14, 0)$ 6. **Draw the line:** Connect these two points with a straight line. The slope $-\frac{1}{3}$ means the line goes down 1 unit for every 3 units it moves to the right. **Final answer:** The graph is a straight line crossing the y-axis at $\frac{14}{3}$ and the x-axis at 14 with slope $-\frac{1}{3}$.