Question

How do you solve `3=-2|1/4s-5|+3`?

Answer 1

`s=color(red)(20)`

Explanation:

`3=-2abs(1/4s-5)+3`

`rarr0=-2abs(1/4s-5)`

`rarr abs(1/4s-5)=0`

`rarr 1/4s=5`

`rarr s=20`

Answer 2

See a solution process below:

Explanation:

First, subtract `color(red)(3)` from each side of the equation to isolate the absolute value term while keeping the equation balanced:

`3 - color(red)(3) = -2abs(1/4s - 5) + 3 - color(red)(3)`

`0 = -2abs(1/4s - 5) + 0`

`0 = -2abs(1/4s - 5)`

Next, divide each side of the equation by `color(red)(-2)` to isolate the absolute value function while keeping the equation balanced:

`0/color(red)(-2) = (-2abs(1/4s - 5))/color(red)(-2)`

`0 = (color(red)(cancel(color(black)(-2)))abs(1/4s - 5))/cancel(color(red)(-2))`

`0 = abs(1/4s - 5)`

The absolute value function takes any term and transforms it to its non-negative form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent. However, because `-0` equals `0` we can just solve the term within the absolute value function once for `0`:

`0 = 1/4s - 5`

Add `color(red)(5)` to each side of the equation to isolate the `s` term while keeping the equation balanced:

`0 + color(red)(5) = 1/4s - 5 + color(red)(5)`

`5 = 1/4s - 0`

`5 = 1/4s`

Now, multiply each side of the equation by `color(red)(4)` to solve for `s` while keeping the equation balanced:

`color(red)(4) xx 5 = color(red)(4) xx 1/4s`

`20 = cancel(color(red)(4)) xx 1/color(red)(cancel(color(black)(4)))s`

`20 = s`

`s = 20`