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On 27 Jan 2014 17:02, "saurabh singh" <saurab...@gmail.com> wrote:

> ^ No its not invalid. It just represents an equation with infinitely many
> correct solutions depending on the domain of x.
>
> Saurabh Singh
> B.Tech (Computer Science)
> MNNIT
> blog:geekinessthecoolway.blogspot.com
>
>
> On Mon, Jan 27, 2014 at 4:21 PM, Amol Sharma <amolsharm...@gmail.com>wrote:
>
>> i din't get ur question.
>>
>> isn't the equation "*(x - 7) + 7 = (x + 1) - 5*"  invalid ?
>>
>> --
>> Thanks and Regards,
>> Amol Sharma
>>
>>
>>
>> On Wed, Jan 15, 2014 at 3:34 AM, Arpit Sood <soodfi...@gmail.com> wrote:
>>
>>> Equivalent to solving an infix expression using stack with a pair (first
>>> variable, second constant) as the element
>>>
>>>
>>> On Sat, Jan 11, 2014 at 6:50 AM, atul anand <atul.87fri...@gmail.com>wrote:
>>>
>>>> Hello,
>>>>
>>>> How to solve an equation with one unknown variable ?
>>>> operator allowed are : + , -
>>>>
>>>> for eg .  input could be :-
>>>>             x + ( 5 + 4 ) = 6
>>>>             (x - 7) + 7 = (x + 1) - 5
>>>>
>>>> If operator also allows " * " (multiply) , then what change in
>>>> algorithm is required.
>>>>
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