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.Com
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)   (Show Source): You can put this solution on YOUR website!
So far, so good. You got to
and all your x's are on one side of the equal sign.
If you had on the right side, you would simplify it to ,
because = = = =
You are really applying the distributive property there, but we do not need to explain that much.
It looks simple and easy with .
With it looks more intimidating, but it works just the same way:
= .
So we simplify
-->
To find , we have to divide by both sides of the equation, to get

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:
--> --> --> --> -->

RELATED QUESTIONS

Can you please help me solve this problem x/4-3x/2=-1/2 (answered by jim_thompson5910)
I kind of understand how functions work but this is just confusing and I dont understand... (answered by Jk22)
can you please help me solve this algebra 1 problem?... (answered by Fombitz)
can you help me solve this problem 20 5 ____ + ______ = 1 3x-2... (answered by Poohbear272)
How do you solve this problem. This is hard can someone please help me 3/4 x 5 +... (answered by Fombitz)
Can you please help me solve this problem or see if I'm doing the work correctly? (answered by josgarithmetic)
I need help with the following 2 questions... 1. Solve by extracing square roots:... (answered by Cintchr)
I need help with the following 2 questions... 1. Solve by extracing square roots:... (answered by Cintchr)
Can you please help me with this problem... (answered by ankor@dixie-net.com)