To graph the inequality $x + 3y > 3$, follow these steps:

Step 1: Rewrite the inequality in slope-intercept form

Start with: $x + 3y > 3$

Subtract $x$ from both sides: $3y > -x + 3$

Divide through by $3$ to isolate $y$: $y > -\frac{1}{3}x + 1$

This is now in slope-intercept form: $y > -\frac{1}{3}x + 1$.

Step 2: Graph the boundary line

• The boundary line is $y = -\frac{1}{3}x + 1$. This is a straight line with:
  • Slope: $-\frac{1}{3}$
  • y-intercept: $1$

To draw the line:

1. Start at $(0, 1)$, the y-intercept.
2. Use the slope $-\frac{1}{3}$: Go down $1$ unit and right $3$ units to get another point $(3, 0)$.
3. Connect the points with a dashed line (since the inequality is strict, $>$, not $\geq$).

Step 3: Shade the appropriate region

• Since $y > -\frac{1}{3}x + 1$, shade the region above the line (where $y$ values are greater).

Would you like me to assist further with a visual graph or clarify any steps?

Related Questions

1. How do you determine which side of the line to shade for inequalities?
2. What changes in the graph if the inequality is $x + 3y \geq 3$?
3. How do you test a point to confirm the shaded region?
4. What is the role of the slope in determining the direction of the line?
5. How would the graph look if the inequality were $x + 3y < 3$?

Tip

Always test a point (e.g., $(0, 0)$) to confirm whether it lies in the solution region.