Cyufjy

A little of a stretch but

$87 - 32 = 55$

by

moving the upright + matchstick to the left of the 3 in 37

Technically, you could change this to

$37 - 32 neq 55$

by

moving the cross stick in the plus sign to go diagonally across the equals sign.

Super easy:

37 - 32 = 5 | 5

Where:

the | is bitwise OR operator

By:

moving the vertical match from the plus sign to the middle of the 55

Just to prove that this is impossible (without some creativity), I wrote a python script to solve these:

subs = '1':,'2':,'3':,'4':,'5':,'6':['5'],'7':['1'],'8':['6','9','0'],'9':['3','5'],'0':,'+':['-'],'-':,'=':['-']

adds = '1':['7'],'2':,'3':['9'],'4':,'5':['6','9'],'6':['8'],'7':,'8':,'9':['8'],'0':['8'],'+':,'-':['+'],'=':

noop = '1':,'2':['3'],'3':['2','5'],'4':,'5':['3'],'6':['9','0'],'7':,'8':,'9':['6','0'],'0':['9','6'],'+':['='],'-':,'=':['+']

ts = list("37+32=55")
alt_strings = 

for i, c in enumerate(ts):
 for new_char in noop[c]:
 alt_strings.append(ts[:i]+[new_char]+ts[i+1:])

 for sub_c in subs[c]:
 for i2, c2 in enumerate(ts):
 for add_c in adds[c2]:
 s = [x for x in ts]
 s[i] = sub_c
 s[i2] = add_c
 alt_strings.append(s)

alt_strings = [''.join(x) for x in alt_strings]
print alt_strings

for alt_string in alt_strings:
 split_strings = alt_string.split('=')
 if len(split_strings) != 2:
 continue

 left = eval(split_strings[0])
 right = eval(split_strings[1])
 if left == right:
 print alt_string

The possible combinations I got were:

['27+32=55', '57+32=55', '91+32=55', '31+92=55', '31+32=65', '31+32=95', '31+32=56', '31+32=59', '37=32=55', '97-32=55', '37-92=55', '37-32=65', '37-32=95', '37-32=56', '37-32=59', '37+22=55', '37+52=55', '37+33=55', '37+32+55', '97+32-55', '37+92-55', '37+32-65', '37+32-95', '37+32-56', '37+32-59', '37+32=35', '37+32=53']

And there were no matches (hehe).

Outside the box solution!

Remove the match on the + to make it a -, then eat it. Then crop the picture so that it cuts out the last 5. 37 - 32 = 5, only one match (and the frame of the picture) has moved.

37+32>55

There you go,

just change the equals sign to a more than sign!

-edit

You take the

vertical match from the plus sign

and

use it to burn the first five in 55

to make

37 - 32 = 5

If only the first 5 would be a 6. That would help a lot.
Please ask your sister for the solution and double check if it's actually possible. Because I have a hunch that this puzzle is impossible.

I saw this in "hot network questions" and tried to solve it on paper before I clicked to avoid getting spoiled. So I scribbled down the equation. When I couldn't find a way to solve it, I finally opened the question and saw, that my 7 had one less matchstick (the very left one). So I wondered about the display of numbers we don't get to see. Would a 9 without an underscore be legal for example? It clearly would be a distinct nine, wheter the bottom stick is there or not. You wouldn't return your 80's alarm clock because of such a nine, anyway.

Along this line of thinking I came up with this solution:

37 + 32 = 69
by taking the bottom matchstick from the second five to make the first five a six. Granted, the resulting nine is somewhat weird, but you clearly would not assume another number instead of it. Maybe for arguments sake just now, but not if some hot girl wrote the nines of her phone number in that way. That would probably just be a-okay for you. So just give me the correct flag now. Thanks.

It gives a negative value but I would like to share it anyways:

37-92 = 55

How?

Just remove the vertical match from the "+" sign and add onto 32, now it is 92.

Along the lines of El-Guest and Alto:

37 + 3P = 55 where P = 6.

I know that's not the answer, however; from my testing I have found that the only (to my knowledge) possible (true) combinations are:

37 + 22 = 59, 27 + 32 = 59, 57 - 22 = 35, 97 - 32 = 65, 34 + 22 = 56

But all of those require more than one match. I'll keep at it, if I solve it I'll update. All in all, I have found 210 total numeric combinations. None of which are achieved with a single move.

My answer, similar to @alto's, is a little bit more elegant.

Take the leftmost downstroke from the 7 and place it at an angle above and and touching the top bar of the = sign to form 'a greater-than-or-equal' (or less-than-or-equal") sign.

BTW, this brute force & ignorance approach (in Python 3.x.) shows that it is not possible without modifying the + or = chararacter:

''' Dictionary of all possible "matchstick' substitutions. Note that
 some "matchstick" characters can be transformed into variants of 
 themselves by adding or removing one stroke/matchstick, e.g. 7 (by removal of 
 leftmost downstroke); 6 and 9 (by adding a horizontal top/bottom stroke). 

SubstitutionList = {
 '+':['+','-'',='], '-':['+','-'',='], '=':['+','-'',='],
 '1':['1'], '2':['2','3','6'], '3':['2','3',5','9'], 
 '4':['4','9'], '5':['3','5','6','9'], '6':['6','9','0'],
 '7':['7','9'], '8':['8'], '9':['9','6','0'], 
 '0':['0','9','6'], 
''' 
''' Only these values are needed (after allowing for character variations) '''

SubstitutionList = '2':['2','3','6'], '3':['2','3','5','9'],
 '5':['3','5','6','9'], '7':['7','9'], 
 '+':['+','-','='], '-':['+','-','='],
 '=':['+','-','='] 

TestEquation = '37+32=55' 
PossibleSolutions = # None yet

''' For clarity the code that extracts values from "TestEquation" has been omitted '''

for I in SubstitutionList.get( '3' ) : 
 for J in SubstitutionList.get( '7' ) :
 for K in SubstitutionList.get( "+" ) : 
 for L in SubstitutionList.get( "3" ) :
 for M in SubstitutionList.get( "2" ) :
 for N in SubstitutionList.get( "=" ) :
 for O in SubstitutionList.get( "5" ) :
 for P in SubstitutionList.get( "5" ) : 
 Equation = I+J+K+L+M+N+O+P # concatinate the letters into a string
 FirstNumber = int( Equation[ : 2 ] )
 SecondNumber = int( Equation[ 3 : 5 ] )
 ThirdNumber = int( Equation[ 6 : ] )
 if ( (Equation[2] == '=' ) and (Equation[5] == '+' ) ) : 
 if ( FirstNumber == (SecondNumber + ThirdNumber) ) :
 PossibleSolutions += [ Equation ] 
 elif ( (Equation[2] == '=' ) and (Equation[5] == '-' ) ) : 
 if ( FirstNumber == (SecondNumber - ThirdNumber) ) :
 PossibleSolutions += [ Equation ]
 elif ( (Equation[2] == '+' ) and (Equation[5] == '=' ) ) : 
 if ( (FirstNumber + SecondNumber) == ThirdNumber ) :
 PossibleSolutions += [ Equation ]
 elif ( (Equation[2] == '-' ) and (Equation[5] == '=' ) ) : 
 if ( (FirstNumber - SecondNumber) == ThirdNumber ) :
 PossibleSolutions += [ Equation ]

OneCharacterMoves = # match changed position inside one character, e.g. change "3' to '2', "6" to "9" etc. 
TwoCharacterMoves = # match moved from one character to another
for ValidEquation in PossibleSolutions : # valid solutions change one or two characters
 DifferenceCount = 0 
 for i, _ in enumerate( ValidEquation ) :
 if ValidEquation[ i ] != TestEquation[ i ] :
 DifferenceCount += 1 
 if ( DifferenceCount == 1 ) : 
 OneCharacterMoves += [ ValidEquation ]
 elif ( DifferenceCount == 2 ) :
 TwoCharacterMoves += [ ValidEquation ]

print( 'Original : ', TestEquation )
print( 'Involving One Character: ', OneCharacterMoves ) 
print( 'Involving Two Characters: ', TwoCharacterMoves )

The program produces :

Original : 37+32=55
Involving One Character: 
Involving Two Characters: ['27+32=59', '37+22=59', '37+32=69']

Inspecting the values shows that none of them can be produced by moving one matchstick.

What about:

You can either interpret that as a zero or a five that's been slashed out.

Move the lower match of the plus and the top match moves too! Make $37times32=1184=550times2+84$, and the picture has been (unfairly in my opinion) cropped so that you can't see the last bit!

Take a match from the equal sign and put it anywhere else where it creates a number. For example: 37 + 92 - 55. It's neither true nor false and it's left to the reader to calculate!

Here's a boolean logic solution:

Take one of the $=$ matches, break it in half, and put it on the right side of the remaining = match to make a right facing arrow $rightarrow$. Then, the LHS is $69$, interpreted as a boolean is True, and the RHS is $55$, interpreted as a boolean is True. Then, the statment True $rightarrow$ True is True. Thus, the statement becomes True.

Here is a solution exploiting that the algebraic operations are not specified in the question.

Take the vertical part of the plus sign and place it horizontally above the equality sign to obtain 37 "minus" 32 "is defined as" 55.