Open File Using Variable in Excel Macro

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Can someone tell me how to open a file from within a macro where the file name needs to be a variable? Currently using

Workbooks.Open (ThisWorkbook.path & "\Ridgemark 12-1-17.xls")

but want to use a variable in place of the "Ridgemark 12-1-17" file name.

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Sub fileOpen()

'Name of the file within quotation marks
yourVariable = "test.xlsx"

'Open the file
Workbooks.Open (ThisWorkbook.Path & "\" & yourVariable)

End Sub

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The answer resolved the issue however another related issue arose later in the macro. Again it is how to use the variable, "iNewFile", to activate the open file.
The code line that fails is [Code]Windows(iNewFile).Activate"[\Code]. The error is Run-time error '9': Subscript out of range. More assistance would be greatly appreciated. Thanks
NormZ (rep: 2) Dec 18, '17 at 5:49 pm
Basically, you are not actually telling Excel what to activate due to a number of potential reasons. Post a new question with more/all of the code and it will be easier to help. Also, just click the CODE tags button to add the code when you make a post and it will then work.
don (rep: 1382) Dec 18, '17 at 5:59 pm
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