Solve for x

frctl said:
Yes I have, are you able to show me step by step?
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No, that is not what we are about here. No one ever learned any mathematics by watching it be done.
On the contrary. if you post what you think should be done then we can see what help you need.
So post away.
 
frctl said:
x + 5x - √2 = 1
6x = √2x + 1
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After combining like terms, you could arrange as:

[MATH]6x=\sqrt{2}+1[/MATH]
Do you see why the number \(\sqrt{2}\) on the RHS would not be multiplied by \(x\)?
 
Could you show me what the next line would be after this operation?
 
There is no such thing as a "next line" or "next step". The "next" anything is what 1) Works correctly and 2) Makes sense to you.

One guideline might be to ask yourself, "How are things attached to or related to that variable I'm trying to find?"

If the answer is addition, x + 1, then use subtraction to better isolate the variable.
If the answer is multiplication, 5x, then use division to better isolate the variable.
etc.

You have to think about how to proceed. There should not ever be a choice between knowing the steps and doing nothing. Just do what comes to mind and makes sense and try not to break it. JOKES!! You can't break it.
 
frctl said:
Could you show me what the next line would be after this operation?
Click to expand...

If you are trying to solve for \(x\) here, then I would undo the multiplcation by 6 with a division by 6:

[MATH]\frac{6x}{6}=\frac{\sqrt{2}+1}{6}[/MATH]
Divide out common factors:

[MATH]\frac{\cancel{6}x}{\cancel{6}}=\frac{\sqrt{2}+1}{6}[/MATH]
Thus:

[MATH]x=\frac{\sqrt{2}+1}{6}[/MATH]
 
frctl said:
I wish to use subtraction, can I do 6x - 1?
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In another post, you talked about isolating the variable. Do you know what that means?

If x is the variable to be solved for, "isolating the variable" means getting all the terms that involve x onto one side of the equation and all terms that do not involve x on the other side of the equation. You then figure out what simplifications and operations are needed to get the side with the terms involving x reduced to x without introducing any terms involving x into the other side of the equation.

As Mark says, from

[MATH]6x = \sqrt{2} + 1[/MATH],

the variable has already been isolated, and the obvious and direct way to reduce the LHS to x is simply to divide by 6.

[MATH]6x = \sqrt{2} + 1 \implies \dfrac{\cancel {6} x}{\cancel {6}} \implies x = \dfrac{\sqrt{2} + 1}{6}.[/MATH]
 
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