To reduce the given equation $2x + 3y - 6 = 0$ into slope-intercept form ($y = mx + b$) and standard intercept form ($x/a + y/b = 1$), follow these steps:

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1. Slope-Intercept Form ($y = mx + b$):

The general slope-intercept form of a line is: $y = mx + b$

Start with the given equation: $2x + 3y - 6 = 0$

Rearrange to isolate $y$: $3y = -2x + 6$

Divide through by 3: $y = -\frac{2}{3}x + 2$

Here:

• Slope ($m$) = $-\frac{2}{3}$
• Y-intercept ($b$) = 2

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2. Intercept Form ($\frac{x}{a} + \frac{y}{b} = 1$):

The general intercept form of a line is: $\frac{x}{a} + \frac{y}{b} = 1$

Start with the given equation: $2x + 3y - 6 = 0$

Solve for the intercept form: $3y = -2x + 6$

Divide through by 6 to normalize: $\frac{-2x}{6} + \frac{3y}{6} = 1$

Simplify the fractions: $\frac{x}{-3} + \frac{y}{2} = 1$

Here:

• X-intercept ($a$) = $-3$
• Y-intercept ($b$) = $2$

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Final Results:

1. Slope-Intercept Form: $y = -\frac{2}{3}x + 2$

2. Intercept Form: $\frac{x}{-3} + \frac{y}{2} = 1$

Would you like a detailed plot or further clarifications?

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Expanding Questions:

1. How can the slope be interpreted geometrically in the graph of a line?
2. What happens to the intercepts if the equation coefficients change?
3. How would you convert intercept form back to slope-intercept form?
4. What is the geometric significance of having one intercept at zero?
5. How do parallel and perpendicular lines relate to the slope of this line?

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Tip: Always ensure the line equation is simplified before extracting slope or intercepts to avoid calculation errors.