How do you name the slope and y-intercept of $4 x - 8 y = 16$ $4x-8y=16$ ?

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Answer 1 · 2016-12-13T14:43:18

The slope for the equation in this problem is $\frac{1}{2}$ $1/2$ and the y-intercept is $- 2$ $-2$

To solve this problem first put the equation into standard form by solving for $y$ $y$ :

$4 x - 8 y + 8 y - 16 = 16 - 16 + 8 y$ $4x - 8y + 8y - 16 = 16 - 16 + 8y$

$4 x - 0 - 16 = 0 + 8 y$ $4x - 0 - 16 = 0 + 8y$

$4 x - 16 = 8 y$ $4x - 16 = 8y$

$\frac{4 x - 16}{8} = \frac{8 y}{8}$ $(4x - 16)/8 = (8y)/8$

$(4x)/8 - 16/8 = (cancel(8)y)/cancel(8)$

$1/2x - 2 = y$ or $y = 1/2x - 2$

The standard form is also:

$color(red)(y = mx - y_1)$ where $m$ is the slope and $y_1$ is the y-intercept.

Therefore the slope for the equation in this problem is $1/2$ and the y-intercept is $-2$

How do you name the slope and y-intercept of 4x-8y=164x−8y=164x-8y=16?