Question

How do you solve for x: `6(x+1)=12(x-3)`?

Answer 1

See the entire solution process below:

Explanation:

First, expand all the terms in exponents:

`(6 xx x) + (6 xx 1) = (12 xx x) - (12 xx 3)`

`6x + 6 = 12x - 36`

Next, subtract `color(red)(6x)` and add `color(blue)(36)` to each side of the equation to isolate the `x` term while keeping the equation balanced:

`6x + 6 - color(red)(6x) + color(blue)(36) = 12x - 36 - color(red)(6x) + color(blue)(36)`

`6x - color(red)(6x) + 6 + color(blue)(36) = 12x - color(red)(6x) - 36 + color(blue)(36)`

`0 + 42 = 6x - 0`

`42 = 6x`

Now, divide each side of the equation by `color(red)(6)` to solve for `x` while keeping the equation balanced:

`42/color(red)(6) = (6x)/color(red)(6)`

`7 = (color(red)(cancel(color(black)(6)))x)/cancel(color(red)(6))`

`7 = x`

`x = 7`

Answer 2

`x=7`

Explanation:

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

`:.6x+6=12x-36`

`:.6x-12x=-36-6`

`:.-6x=-42`

multiply both sides by`-1`

`:.6x=42`

`:.x=cancel42^7/cancel6^1`

`:.x=7`

substitute `x=7`

`:.6(7+1)=12(7-3)`

`:.6 xx 8=12 xx 4`

`:.48=48`