If x + |x| + y = 7 and x + |y| - y = 6 what is x + y = ?

for x>0 ; y>0 => x=6, y=-5 ------ [1]
for x>0 ; y<0 => x=4, y=-1 => x+y = 3 -------- [2]
for x<0 ; y>0 => x =6, y=7 ----------- [3]
for x<0, y<0 => y=7, x =20 ----------- [4]

Hence, x+y = 3

Hi Bunuel
Adding both eqn we get 2x +IxI+IyI = 13
Now considering X <0 and y>0
2x-x +y = 13
we get x+y =13
Hence answer should be E

Correct me if i am wrong!!!

Archit

OA is C, not E, so yes your solution is not correct.

If x<0 and y>0, then we'll have x-x+y=7 and x+y-y=6. From the second equation x=6, so we can discard this case as y here is not more than 0.

You can also solve this question using some brute force.

Notice x+|y|-y=6
The first thing that comes to mind is that if y is positive or 0, x = 6. But when we put x = 6 in x+|x|+y=7, we get y negative. So y cannot be positive or 0. y must be negative. So, x+|y|-y=6 becomes x - 2y = 6

Since most options are positive values, it is very likely that x is positive so x+|x|+y=7 becomes 2x+y = 7. You can obtain 7 by subtracting 1 from 8.
If x = 4 and y = -1, both equations are satisfied.
So x+y = 4 - 1 = 3

Answer (C)

I don't understand the reasoning behind this statement.

If y is negative, then x+|y|-y=6 becomes x +|-y|-(-y)=6 therefore x + y + y = 6 finally becomes x +2y=6

The absolute value of a negative integer is a positive value, and the subtraction of a negative integer results in an addition.

What am I missing?

Reevak, the algebraic definition of the modulus is :

- If x > 0 then |x| = x > 0 ; (1)
- If x < 0 then |x| = -x > 0 ; (2)

The reason why your statement is incorrect is because you altered the equation.

You can't put a "-" sign to a variable to say that it's negative. In fact, in our case, since y is negative, putting a "-" sign to it will make it positive, thus altering your equation and guiding you to a wrong answer.

What you did is the same as doing the following :

y = - 3 is negative. Therefore - y (which is still negative according to your syntax) will be equal to -y = -(-3) = 3 which is NOT negative.

So be careful with your computations.

Now let's get back to your issue :

Since y is negative ( y < 0 ), then according to (2) we'll have |y| = - y.

Injecting it into our equation yields : x+|y|-y = 6 => x - y - y = 6 => x - 2y =6. Hence the result.

You should be fine if you remember the definition of the modulus. I know it can be tricky at first, but once you get used to it, it'll become second-nature to you

Hope that helped.

The first thing you should keep in mind when facing modulus (absolute value) problems is the definition of the modulus :

- If x > 0 then |x| = x > 0 (1) ;
- If x < 0 then |x| = - x > 0 (2) ;
-|x| IS ALWAYS POSITIVE NO MATTER x ! ;

Now let's solve the problem : (It's a pretty lengthy solution but I'd rather not miss anything )

We need to find the value of x+y knowing that : x+|x|+y=7 and x+|y|-y=6

Now considering the definition of the modulus, we have 4 cases :

- Case 1 : x > 0 and y > 0

According to (1) we'll have :

x+|x|+y = 7 => x+x+y = 7 => 2x + y = 7 => 12 + y = 7 => y = -5 (which is contradictory with the fact that y is positive)
x+|y|-y = 6 => x+y-y = 6 => x = 6

- Case 2 : x > 0 and y < 0

According to (1) and (2) we'll have :

x+|x|+y = 7 => x+x+y = 7 => 2x + y = 7 (E1)
x+|y|-y = 6 => x-y-y = 6 => x - 2y = 6 (E2)

Since we have two equations, we can manipulate them to find the value of x and y separately. In this case, multiplying both (E1) and (E2) by 2 then substracting each other will yield : x = 4 and y = - 1 (which is consistent with the fact that x is positive and y is negative). So x+y = 3 which is answer choice C.

But we're not stopping here

We have to make sure that the other cases don't contradict what we've found.

- Case 3 : x < 0 and y > 0

According to (1) and (2) we'll have :

x+|x|+y = 7 => x-x+y = 7 => y = 7
x+|y|-y = 6 => x+y-y = 6 => x = 6 (which is contradictory with the fact that x is negative)

- Case 4 : x < 0 and y < 0

According to (2) we'll have :

x+|x|+y = 7 => x-x+y = 7 => y = 7 (which is contradictory with the fact that y is negative)
x+|y|-y = 6 => x-y-y = 6 => x - 2y = 6

So all in all, only case 2 was valid in this case and the correct answer is C.

Note : The method may seem lengthy and daunting but when you know the modulus definition and are comfortable manipulating variables, it should be a breeze.

Hope that helped :D

Say y = -3, what is the value of |y|?
Substitute -3 in place of y

|y| = |-3| = 3
This is what you mean by mod of a negative number is positive. The number becomes positive when you remove the mod sign.
Notice that when y is negative, |y| is not equal to y. It is equal to -y (which, by the way, is a positive number since y is negative)

We define mod as:
|y| = y if y is positive
|y| = -y if y is negative

On the other hand, if I ask you: what is the value of y? It is -3 only
What is the value of -y? It is -(-3) = 3

x+|y|-y=6
If we know that y is negative, |y| = -y (which becomes positive)
x - y - y = 6 (notice that the -y in the original equation stays the same since it has no mod around it)

Thanks a lot guys! Now there is one thing I still not understand, and another that I do understand.

Now I see that if y = - 3 then |-3| = 3 ≠ y
Instead |-3| = 3 = - y (since the minus sign will turn y positive turning the equality true)
This I understand.

However, there is still one thing I don't get.
If y < 0 then shouldn't this equation x + |y| - y = 6 be equal to x - y + y = 6?

Instead, you mentioned it should be x - y - y = 6 but I don't understand why
Why - y and not + y?
Is -(-3) not 3?

You don't have to change the "-" sign to a "+" sign, since that sign is proper to the equation and not the variable.

Let's assume that y = - 3 (y is negative) and x = 0

If what you said was true, meaning that if y is negative, x + |y| - y = 6 will be equal to x - y + y = 6, we'll have for this example : 0 - (-3) + (-3) = 0 + 3 - 3 = 0 ≠ 6, which is obviously false.

My point is : the modulus is the only fonction that can change the sign of a variable or an expression based on its related sign (whether the variable or expression is positive/negative). Don't change the signs that are proper to the equation itself !!!

Hope that was clear enough.

If x+|x|+y=7 and x+|y|-y=6 what is x+y=?

A. 1
B. -1
C. 3
D. 5
E. 13

There are four possible cases here I. (x<0, y<0) II. (x<0, y>0) III. (x>0, y<0) IV. (x>0, y>0)

I.) (x<0, y<0)
x+|x|+y=7 x+(-x)+y=7 y=7
x+|y|-y=6 x+(-y)-y=6 x-2y=6
x-2(7)=6 x-14=6 x=20

II.) (x<0, y>0)
x+|x|+y=7 x+(-x)+y=7 y=7
x+|y|-y=6 x+y-y=6 x=6

III.) (x>0, y<0)
x+|x|+y=7 x+(x)+y=7 2x+y=7
x+|y|-y=6 x+(-y)-y=6 x-2y=6
x=6+2y
2(6+2y)+y=7
12+4y+y=7
12+5y=7
5y=-5
y=-1

2x+y=7
2x+(-1)=7
2x=8
x=4

IV.) (x>0, y>0)
x+|x|+y=7 x+x+y=7 2x+y=7
x+|y|-y=6 x+y-y=6 x=6
2(6)+y=7 y=-5

If you notice, III.) is the only one in which the x and y values fall within the tested ranges. (x>0, x=4 y<0, y=-1)

(C)

After struggling to understand all the explanations, I solved it my way.

x+|x|+y=7 ------ (1)
x+|y|-y=6 ------ (2)

Take (1),
Case |x| = x,
(1) => x + x + y = 7
or, 2x + y = 7 ----- (A)

Case |x| = -x,
(1) => x - x + y = 7
or, y = 7 ----- (B)

Take (2),
Case |y| = y,
(2) => x + y - y = 6
or, x = 6 ----- (C)

Case |y| = -y,
(2) => x - y - y = 6
or, x - 2y = 6 ----- (D)

Now lets look at (A), (B), (C) and (D).
We can simply ignore (B) and (C) because they can't be solutions as they were not obtained by solving the system of given equations. They are independent of the system. Even if we tried to use them they wouldn't fit the constraints given by equations.
Simple check, try using y = 7 in (2)'s case when y is positive. There is no y in the equation to substitute.

Solving the linear system of equations (A) and (D),
x = 4, y = -1,

Thus, x + y = 3

x + |x| + y = 7
x + |y| - y = 6

We can easily guess the signs of x and y.
Consider the first equation: x + |x| + y = 7
If x is negative, |x| = -x and the equation reduces to y = 7. But if y = 7, second equation gives us x = 6 (not possible since we assumed x to be negative). So x must be positive.
Consider second equation: x + |y| - y = 6
If y is positive, |y| = y and this equation reduces to x = 6. But if x = 6, first equation gives us y = -5 (not possible since we assumed y to be positive). So y must be negative.

So equations become: x + x + y = 7 and x - y - y = 6
Solving them simultaneously, we get x = 4 and y = -1 so x+y = 3
Answer (C)

Hi All,

There are some subtle Number Property rules in this question (which you'll have to do a little bit of work to define), but you can ultimately use a bit of 'brute force' math to get to the correct answer.

We're told that:
X + |X| + Y = 7
X + |Y| - Y = 6

Notice the role that the absolute value plays in each equation:
-In the first equation, IF X is negative or 0, then Y = 7.... but plugging Y=7 into the second equation gives us X = 6 (and THAT does not match X being negative or 0). Thus, X CANNOT be negative or 0.

-In the second equation, IF Y is positive or 0, then X = 6.... but plugging X = 6 into the first equations gives us Y = -5 (and THAT does not match Y being positive or 0). Thus Y CANNOT be positive or 0.

Now we know that X MUST be positive and Y MUST be negative. Since the answer choices are all simple integers, let's see what happens when we check simple integer values for X and Y (starting with the first equation)....
X=1, Y=5 --> doesn't fit the restriction that Y is negative.
X=2, Y=3 --> doesn't fit the restriction that Y is negative.
X=3, Y=1 --> doesn't fit the restriction that Y is negative.
X=4, Y=-1 --> fits BOTH equations, thus X+Y = 3

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Asked: If x+|x|+y=7 and x+|y|-y=6 what is x+y=?


]]

x>=0; y>=0	x>=0; y<0	x<0; y>=0	x<0; y<0
2x + y = 7; x = 6; y = -5; Not feasible	2x + y = 7; x-2y = 6; 2x = 7-y = 4y + 12; 5y=-5;y=-1; x= 4; x+ y= 3	y = 7; x = 6; Not feasible	y = 7; Not feasible

IMO C

We can PLUG IN THE ANSWERS, which represent the value of x+y.

Option E is not viable, since it would require |x| = -6 in the first equation, and an absolute value cannot be negative.

Generally, we should be skeptical of an ODDBALL answer choice -- one that looks different from all the rest.
The reason?
The eye is drawn to the oddball, making it an attractive option for a test-taker who wants to guess.
Since the goal of the question-writer is to create attractive answer choices that SEEM correct but are actually WRONG, an oddball answer is unlikely to be correct.
Here, B is the only negative answer choice and thus constitutes an oddball.
The correct answer is probably A, C, or D.

When the correct answer is plugged in, x + |y| - y = 6.

A: x+y = 1
Plugging x+y = 1 into x + |x| + y = 7, we get:
|x| = 6 --> x=6, y=-5 or x=-6, y=7, with the result that x+y=1
If x=6 and y=-5, then x + |y| - y = 16
If x=-6 and y=7, then x + |y| - y = -6
Doesn't work.

C: x+y = 3
Plugging x+y = 3 into x + |x| + y = 7, we get:
|x| = 4 --> x=4, y=-1 or x=-4, y=7, with the result that x+y=3
If x=4 and y=-1, then x + |y| - y = 6
Success!

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