Question 1208260: If 4x + 4xy + 4y = 260,
find all the possible values of (x + y).
Found 3 solutions by ikleyn, Edwin McCravy, greenestamps:
Answer by ikleyn(52748)   (Show Source):
You can put this solution on YOUR website! .
If 4x + 4xy + 4y = 260,
find all the possible values of (x + y).
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The problem in the post is worded/printed/presented incorrectly.
In real numbers, there is a continuum set of possible solutions,
and the question can not be answered.
In order for the problem would make sense, it should be re-edited.
It should say that in the given equation x and y are non-negative integer numbers.
In my solution below I will assume it.
Starting equation is
4x + 4xy + 4y = 260.
Divide both sides by 4
x + y + xy = 65.
Add 1 to both sides
1 + x + y + xy = 66.
Factor left side
(1+x)*(1+y) = 66.
In integer numbers, it means that "EITHER OR"
(a) 1+x = 1, 1+y = 66 ---> x = 0, y = 65, x+y = 65;
(b) 1+x = 2, 1+y = 33 ---> x = 1, y = 32, x+y = 33;
(c) 1+x = 3, 1+y = 22 ---> x = 2, y = 21, x+y = 23;
(d) 1+x = 6, 1+y = 11 ---> x = 5, y = 10, x+y = 15
plus 4 other symmetric cases that do not add new values of x+y.
So, the answer to the problem question is that possible values of x+y are 15, 23, 33, and 65.
Solved.
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Today, I see significant flow of crippled "Math problems" at this forum.
Answer by Edwin McCravy(20054)  (Show Source):
You can put this solution on YOUR website!
The clever trick of adding ±1 to trinomials with terms
±x, ±y, and ±xy
to get something factorable as (x @ 1)(y # 1), where
@ and # represent either + or -, has been used several
times on here. So it may be wise to add this to your
bag of tricks.
Since integers are not necessarily positive or 0, you could have these
negative values also, using Ikleyn's format above.
(a) x+1 = -1, y+1 = -66 ---> x = -2, y = -67, x+y = -69;
(b) x+1 = -2, y+1 = -33 ---> x = -3, y = -34, x+y = -37;
(c) x+1 = -3, y+1 = -22 ---> x = -4, y = -23, x+y = -27;
(d) x+1 = -6, y+1 = -11 ---> x = -7, y = -12, x+y = -19
Edwin
Answer by greenestamps(13195)  (Show Source):
You can put this solution on YOUR website!
 --> 
Treat this as a Diophantine equation -- one equation with two variables, in which both variables have integer values.
Note that the equation has an infinite number of solutions if we don't make the restriction that the variables have integer values.
Note also that, while many Diophantine equations have the restriction that the variable values be non-negative, in general the solution can be negative integers as well as positive integers or zero.
Following is a solution to the problem using standard procedures for solving linear Diophantine equations.
Solve the equation for one variable in terms of the other.
x must be an integer, and -1 is an integer; that means 66/(y+1) must be an integer, and that means (y+1) must be a (positive or negative) factor of 66.
Make a table showing the possible values for y+1 and the resulting values for y and x.
Note, in making the table, that the given equation is symmetrical in x and y, which will reduce the amount of work required to find all the solutions.
y+1 y x x+y
-----------------------------
66 65 0 65
33 32 1 33
22 21 2 23
11 10 5 15
-66 -67 -2 -69
-33 -34 -3 -37
-22 -23 -4 -27
-11 -12 -7 -19
ANSWERS (if x and y are non-negative): 65, 33, 23, 15
Additional ANSWERS (if x and y can be negative): -69, -37, -27, -19
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