How do you find the slope of this equation: x/3 + y/4=1?

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How do you find the slope of this equation: x/3 + y/4=1?

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Answer 1 · 2017-04-26T17:55:46

See the solution process below:

Explanation:

First, multiply each side of the equation by $12$ $color(red)(12)$ to put the equation in Standard Form for a linear equation. The standard form of a linear equation is: $A x + B y = C$ $color(red)(A)x + color(blue)(B)y = color(green)(C)$

Where, if at all possible, $A$ $color(red)(A)$ , $B$ $color(blue)(B)$ , and $C$ $color(green)(C)$ are integers, and A is non-negative, and, A, B, and C have no common factors other than 1

$12 (\frac{x}{3} + \frac{y}{4}) = 12 \cdot 1$ $color(red)(12)(x/3 + y/4) = color(red)(12) * 1$

$(12 \cdot \frac{x}{3}) + (12 \cdot \frac{y}{4}) = 12$ $(color(red)(12) * x/3) + (color(red)(12) * y/4) = 12$

$4 x + 3 y = 12$ $color(red)(4)x + color(blue)(3)y = color(green)(12)$

The slope of an equation in standard form is: $m = - \frac{A}{B}$ $m = -color(red)(A)/color(blue)(B)$

Substituting for $A$ $color(red)(A)$ and $B$ $color(blue)(B)$ gives:

$m = - \frac{4}{3}$ $m = -color(red)(4)/color(blue)(3)$

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How do you find the slope of this equation: x/3 + y/4=1?

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