Question

How do you write the equation `y+4=-1/3(x-12)` in standard form?

Answer 1

x+3y=0#

Explanation:

The equation of a line in `color(blue)"standard form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(Ax+By=C)color(white)(2/2)|)))`
where A is a positive integer and B, C are integers.

`"Rearrange " y+4=-1/3(x-12)" into this form"`

`"multiply ALL terms by 3 to eliminate the fraction"`

`3y+12=(cancel(3)^1xx-1/cancel(3)^1(x-12))`

`rArr3y+12=-x+12`

`"add x to both sides"`

`x+3y+12=cancel(-x)cancel(+x)+12`

`rArrx+3y+12=12`

`"subtract 12 from both sides"`

`x+3ycancel(+12)cancel(-12)=12-12`

`rArrx+3y=0larrcolor(red)" in standard form"`

Answer 2

See the entire solution process below:

Explanation:

The standard form of a linear equation is: `color(red)(A)x + color(blue)(B)y = color(green)(C)`

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

First, multiply each side of the equation by `color(red)(3)` to eliminate the fractions while keeping the equation balanced:

`color(red)(3)(y + 4) = color(red)(3) xx -1/3(x - 12)`

`(color(red)(3) xx y) + (color(red)(3) xx 4) = cancel(color(red)(3)) xx -1/color(red)(cancel(color(black)(3)))(x - 12)`

`3y + 12 = -1(x - 12)`

`3y + 12 = -1x + 12`

Now, subtract `color(red)(12)` and add `color(red)(1x)` to each side of the equation to put in standard form while keeping the equation balanced:

`color(red)(1x) + 3y + 12 - color(red)(12) = color(red)(1x) - 1x + 12 - color(red)(12)`

`1x + 3y + 0 = 0 + 0`

`color(red)(1)x + color(blue)(3)y = color(green)(0)`