Question 559095
So far, so good. You got to
{{{-5=(x)sqrt(2)-3x}}} and all your x's are on one side of the equal sign.
If you had {{{2x-3x}}} on the right side, you would simplify it to {{{-x}}},
because {{{2x-3x}}} = {{{2*x-3*x}}} = {{{(2-3)*x}}} = {{{(-1)*x}}} = {{{-x}}}
You are really applying the distributive property there, but we do not need to explain that much.
It looks simple and easy with {{{2x}}}.
With {{{sqrt(2)*x}}} it looks more intimidating, but it works just the same way:
{{{sqrt(2)*x-3*x}}} = {{{(sqrt(2)-3)*x}}}.
So we simplify
{{{x*sqrt(2)-3x=-5}}} --> {{{(sqrt(2)-3)*x=-5}}}
To find {{{x}}}, we have to divide by {{{(sqrt(2)-3)}}} both sides of the equation, to get
{{{x=-5/(sqrt(2)-3)}}}
There's not much more that can be done, but there are equivalent ways to express it, and you may see another expression as the solution.
It is customary to convert something like that into an eqivalent expression with no square roots in the denominator. It is called "rationalizing". I would do it like this:
{{{x=(-5/(sqrt(2)-3))*((sqrt(2)+3)/(sqrt(2)+3))}}} --> {{{x=-5(sqrt(2)+3)/((sqrt(2)-3)(sqrt(2)+3))}}} --> {{{x=-5(sqrt(2)+3)/((sqrt(2))^2-3^2))}}} --> {{{x=-5(sqrt(2)+3)/(2-9))}}} --> {{{x=-5(sqrt(2)+3)/(-7))}}} --> {{{x=5(sqrt(2)+3)/7}}}