Question

How do you solve for y in `k(y+3z)=4(y-5)`?

Answer 1

See a solution process below:

Explanation:

First, expand the terms in parenthesis on each side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:

`color(red)(k)(y + 3z) = color(red)(4)(y - 5)`

`(color(red)(k) xx y) + (color(red)(k) xx 3z) = (color(red)(4) xx y) - ((color(red)(4) xx 5)`

`ky + 3kz = 4y - 20`

Next, subtract `color(red)(3kz)` and `color(blue)(4y)` from each side of the equation to isolate the `y` terms while keeping the equation balanced:

`ky - color(blue)(4y) + 3kz - color(red)(3kz) = 4y - color(blue)(4y) - 20 - color(red)(3kz)`

`ky - 4y + 0 = 0 - 20 - 3kz`

`ky - 4y = 20 - 3kz`

The factor out the common term on the left side of the equation:

`kcolor(red)(y) - 4color(red)(y) = 20 - 3kz`

`(k - 4)y = 20 - 3kz`

Now, divide both sides of the equation by `color(red)((k - 4))` to solve for `y` while keeping the equation balanced:

`((k - 4)y)/color(red)((k - 4)) = (20 - 3kz)/color(red)((k - 4))`

`(color(red)(cancel(color(black)((k - 4))))y)/cancel(color(red)((k - 4))) = (20 - 3kz)/(k - 4)`

`y = (20 - 3kz)/(k - 4)`