How do you solve `2root3(10-3x)=root3(2-x)`?

First, cube each side of the equation to eliminate the radicals while keeping the equation balanced:

`(2root(3)(10 - 3x))^3 = (root(3)(2 - x))^3`

`2^3(10 - 3x) = 2 - x`

`8(10 - 3x) = 2 - x`

Next, eliminate the parenthesis on the left side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:

`color(red)(8)(10 - 3x) = 2 - x`

`(color(red)(8) xx 10) - (color(red)(8) xx 3x) = 2 - x`

`80 - 24x = 2 - x`

Then, add `color(red)(24x)` and subtract `color(blue)(2)` from each side of the equation to isolate the `x` term while keeping the equation balanced:

`-color(blue)(2) + 80 - 24x + color(red)(24x) = -color(blue)(2) + 2 - x + color(red)(24x)`

`78 - 0 = 0 - 1x + color(red)(24x)`

`78 = (-1 + color(red)(24))x`

`78 = 23x`

Now, divide each side of the equation by `color(red)(23)` to solve for `x` while keeping the equation balanced:

`78/color(red)(23) = (23x)/color(red)(23)`

`78/23 = (color(red)(cancel(color(black)(23)))x)/cancel(color(red)(23))`

`78/23 = x`

`x = 78/23`

`2root3(10-3x)=root3(2-x)`

Cube both sides

`:.(2root3(10-3x))^3=(root3(2-x))^3`

`:.root3(a-b)*2root3(a-b)*root3(a-b)=a-b`

`:.2^3(10-3x)=2-x`

`:.8(10-3x)=2-x`

`:.80-24x=2-x`

`:.-24x+x=2-80`

`-23x=-78`

Multiply both sides by `-1`

`:.23x=78`

`:.color(magenta)(x=78/23` or`color(magenta)(3.391304348`

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`color(magenta)(check:`

`:.2root3(10-3(color(magenta)3.391304348))=root3(2-(color(magenta)3.391304348))`

`:.2root3(10-10.17391304))=root3(2-(3.391304348))`

`:.2root3(-0.17391304))=root3(-1.391304348)`

`:.2 xx -0.558183998=-1.116368004`

`:.color(magenta)(-1.116367997=-1.116368004`