Solve for y in $y + \frac{1}{2} = - \frac{1}{3} (x + \frac{1}{2})$ $y + 1/2 = -1/3(x+1/2)$ ?

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Solve for y in $y + \frac{1}{2} = - \frac{1}{3} (x + \frac{1}{2})$ $y + 1/2 = -1/3(x+1/2)$ ?

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Answer 1 · 2015-04-07T05:44:33

The solution is $y = - \frac{1}{3} x - \frac{2}{3}$ $y=-1/3x-2/3$ .

$y + \frac{1}{2} = - \frac{1}{3} (x + \frac{1}{2})$ $y+1/2=-1/3(x+1/2)$

Perform the distributive property first.

$y + \frac{1}{2} = - \frac{1}{3} x - \frac{1}{6}$ $y+1/2=-1/3x-1/6$

Subtract $\frac{1}{2}$ $1/2$ from both sides. The common denominator for $\frac{1}{2}$ $1/2$

and $\frac{1}{6}$ $1/6$ is $6$ $6$ . So $\frac{1}{2} \cdot \frac{3}{3} = \frac{3}{6}$ $1/2*3/3=3/6$ .

$y = - \frac{1}{3} x - \frac{1}{6} - \frac{3}{6}$ $y=-1/3x-1/6-3/6$ =

$y = \frac{1}{3} x - \frac{4}{6}$ $y=1/3x-4/6$ =

Reduce $\frac{4}{6}$ $4/6$ to $\frac{2}{3}$ $2/3$ .

$y = - \frac{1}{3} x - \frac{2}{3}$ $y=-1/3x-2/3$

If this is supposed to be the slope-intercept equation for a line, the slope is $- \frac{1}{3}$ $-1/3$ and the y-intercept is $- \frac{2}{3}$ $-2/3$ .

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Solve for y in $y + \frac{1}{2} = - \frac{1}{3} (x + \frac{1}{2})$ $y + 1/2 = -1/3(x+1/2)$ ?

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