Question

Question #5b8b4

Answer 1

The `y` intercept of both lines is `-1`.

Explanation:

First, you'll want to convert the equations to standard form, which is

`y = mx + b`

Then, `b` will be the `y` intercept.

`16x - 10y = 10 => 10y = 16x - 10 => y = 16/10x - 1`

`-1` is the `y` intercept here, and the point is `(0,-1)`.

Here is a graph of `16x - 10y = 10`:

graph{16x-10y=10 [-10, 10, -5, 5]}

`-8x -6y = 6 => 6y = -8x -6 => y = -8/6x -1`

`-1` is the `y` intercept again, and the point is `(0,-1)`.

Here is a graph of `-8x-6y = 6`:

graph{-8x-6y=6 [-10, 10, -5, 5]}

Answer 2

`"y-intercept "=-1" for both equations"`

Explanation:

The equation of a line in `color(blue)"slope-intercept form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(y=mx+b)color(white)(2/2)|)))`
where m represents the slope and b, the y-intercept.

`"for " 16x-10y=10`

Rearrange into this form to obtain y-intercept.

subtract 16x from both sides of the equation.

`cancel(16x)cancel(-16x)-10y=10-16x=-16x+10`

`rArr-10y=-16x+10`

divide ALL terms on both sides by - 10

`(cancel(-10) y)/cancel(-10)=(-16)/(-10)x+10/(-10`

`rArry=8/5x-1larrcolor(red)" in form y = mx + b"`

`rArrb=-1=" y-intercept"`

Repeat the process for `-8x-6y=6`

to obtain `y=-4/3x-1rArr" y-intercept "=-1`