Question

How do you solve `x/6=8`?

Answer 1

`x=48`

Explanation:

we need to remove the 6 from the LHS. So multiply both sides by `6`

`x/6=8`

`x/6xx6=8xx6`

`x/cancel(6)xxcancel(6)=8xx6`

`x=48`

Answer 2

`x=48`

Explanation:

`color(blue)("Shortcut method that jumps steps from first principle method")`

`color(brown)("Shortcut methods are much faster than first principles.")` `color(brown)("Which is why people use them. However the shortcut outcome is")` `color(brown)("based on the outcome of the first principle method.")`

`x/6=8`

The 6 has the process of divide applied to it in that
`x/6 = x xx1/6 = x-:6`

As the 6 has divide applied to it on the left of = it becomes the opposite of multiply when we move it to the other side of the =. So we have:

`x-:6=8" "->" "x=8xx6`

`x=48`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("3 very important facts used in first principle method")`

1. An equation by definition is stating that what is on one side of = has the same value as on the other side. It is just that they may look different. So if you change one side in some way you have to change the other side in exactly the same way

2. For multiply and divide: to move something to the other side of the = change it to 1 as `color(brown)(1xx"something = something")`.

3. For add and subtract: to move something to the other side of = change it to 0 as `color(brown)(0+" something = something")`.
   ...................................................................................................................
   `color(blue)("Answering your question using first principles")`

`x/6=8`

`x/6` is the same as `1/6xx x`

So to get `x` on its own we need to change `1/6" into "1`

Multiply both sides by `color(red)(6)`

`color(green)(x/6=8" "->" "color(red)(6xx)x/6" "=" "color(red)(6xx)8 )`

`" "color(green)((color(red)(6))/6 xx x" "=" "48`

`" "color(green)(1xx x" "=" "48)`

But `1xx x" gives just "x`

`" "color(green)(x" "=" "48`