SOLUTION: Can you please help me understand this one, solve for x. My work is typed below the problem. 3x-4=1+(x)sqrt(2) -1 -1 3x-5=(x)sqrt(2) -3x -3x -5=(x)sqrt(2)-3X

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Can you please help me understand this one, solve for x. My work is typed below the problem. 3x-4=1+(x)sqrt(2) -1 -1 3x-5=(x)sqrt(2) -3x -3x -5=(x)sqrt(2)-3X      Log On


   

Question 559095: Can you please help me understand this one, solve for x. My work is typed below the problem.
3x-4=1+(x)sqrt(2)
-1 -1
3x-5=(x)sqrt(2)
-3x -3x
-5=(x)sqrt(2)-3X
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
So far, so good. You got to
-5=%28x%29sqrt%282%29-3x 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%2Ax-3%2Ax = %282-3%29%2Ax = %28-1%29%2Ax = -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%282%29%2Ax it looks more intimidating, but it works just the same way:
sqrt%282%29%2Ax-3%2Ax = %28sqrt%282%29-3%29%2Ax.
So we simplify
x%2Asqrt%282%29-3x=-5 --> %28sqrt%282%29-3%29%2Ax=-5
To find x, we have to divide by %28sqrt%282%29-3%29 both sides of the equation, to get
x=-5%2F%28sqrt%282%29-3%29
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=%28-5%2F%28sqrt%282%29-3%29%29%2A%28%28sqrt%282%29%2B3%29%2F%28sqrt%282%29%2B3%29%29 --> x=-5%28sqrt%282%29%2B3%29%2F%28%28sqrt%282%29-3%29%28sqrt%282%29%2B3%29%29 --> x=-5%28sqrt%282%29%2B3%29%2F%28%28sqrt%282%29%29%5E2-3%5E2%29%29 --> x=-5%28sqrt%282%29%2B3%29%2F%282-9%29%29 --> x=-5%28sqrt%282%29%2B3%29%2F%28-7%29%29 --> x=5%28sqrt%282%29%2B3%29%2F7