Question

How do you write the equation `y+3=-5(x+1)` in standard form?

Answer 1

`5x+y=-8`

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+3=-5(x+1)" into this form"`

`"distribute bracket"`

`y+3=-5x-5`

`"add 5x to both sides"`

`5x+y+3=cancel(-5x)cancel(+5x)-5`

`"subtract 3 from both sides"`

`5x+ycancel(+3)cancel(-3)=-5-3`

`rArr5x+y=-8larrcolor(red)" in standard form"`

Answer 2

See the 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, expand the terms on the right side of the equation:

`y + 3 = color(red)(-5)(x + 1)`

`y + 3 = (color(red)(-5) * x) + (color(red)(-5) * 1)`

`y + 3 = -5x + -5`

`y + 3 = -5x - 5`

Now, add `color(red)(5x)` and subtract `color(blue)(3)` from each side of the equation to isolate the `x` and `y` terms on the left side of the equation and the constant on the right side of the equation while keeping the equation balanced:

`color(red)(5x) + y + 3 - color(blue)(3) = color(red)(5x) - 5x - 5 - color(blue)(3)`

`5x + y + 0 = 0 - 8`

`color(red)(5)x + color(blue)(1)y = color(green)(-8)`