If x + |x| + y = 7 and x + |y| - y =6 , then x + y =

Question

A. 3
B. 4
C. 5
D. 6
E. 9

Bunuel · Accepted Answer

If x\leq{0} , then x + |x| + y = 7 becomes: x-x+y=7 --> y=7>0 , but then x + |y| - y =6 becomes: x+y-y=6 --> x=6>0 , which contradicts the initial assumption x\leq{0} . So x can not be \leq{0} --> hence x>0 . Similarly if y\geq{0} , then x + |y| - y =6 becomes: x+y-y=6 --> x=6>0 , but then x + |x| + y = 7 becomes: x+x+y=12+y=7 --> y=-5<0 , which contradicts the initial assumption y\geq{0} . So y can not be \geq{0} --> hence y<0 . So x>0 and y<0 : x+|x|+y=7 becomes: x+x+y=7 --> 2x+y=7 ; x+|y|-y=6 becomes: x-y-y=6 --> x-2y=6 . Solving: x=4 and y=-1 --> x+y=3 . Answer: A. I feel there is an easier way, but world cup makes it harder to concentrate.

AbhayPrasanna · Answer

I did it by brute force, considering all possibilities. But I'm sure someone can come up with a way to quickly identify the signs of x and y.

x + |x| + y = 7 ....1

x + |y| - y = 6 .....2

Consider x < 0
=> |x| = -x
Substitute in 1

x - x + y = 7
y = 7

Substitute in 2

x + 7 - 7 = 6
x = 6

But this violates x < 0 So our assumption was incorrect.

Consider x > 0 and y > 0
=> |x| = x and |y| = y

Substitute in 2

x + y - y = 6
x = 6

Substitute in 1

6 + 6 + y = 7
y = -5

This violates y > 0 so our assumption was incorrect.

Consider x > 0 and y < 0
=> |x| = x and |y| = -y

Substituting in 1 and 2:

2x + y = 7 .....3
x - 2y = 6 .....4

Solving we get x = 4 and y = -1
x + y = 3

Pick A.

GMATBLACKBELT720 · Answer

Why when y<0 do we get -2y?

Bunuel · Answer

When  y<0 , then  |y|=-y  and  x+|y|-y=6  becomes:  x-y-y=6  -->  x-2y=6 .

GMATBLACKBELT720 · Answer

^
Wow... confusing as heck. Essentially saying that (hypothetical number here) |-3| = -(-3), thats fine. But I kept thinking you would apply this to -y and essentially make it +y since and make them both +y...

amit2k9 · Answer

for x>0 and x < 0 2x+ y = 7 and y = 7 for y>0 and y<0 x=6 and x-2y=6 for x,y<0 solution exists. solving x=4 and y = -1 3

s8388 · Answer

If x + |x| + y = 7 and x + |y| - y =6 , then x + y =

can be done in 2.3 mins :

there are 4 cases to be tested :
1) x is -ve and y is -ve 
 substituting in the equation , we get x-x+y=7 and x-y-y=6 solve for x and y we get x=20 and y=7 , so x+y=27 REJECT

2)x is +ve and y is +ve

substitute in the equation, we ger x+x+y=7 and x+y-y=6 solve for x and y we get x=6 and y=-5 ,therefore x+y=1 not on list so REJECT

3) x is -ve and y is +ve

substitute , we get x-x=y=7 and x+y-y=6 solve fo x and y we get x=6 and y=7, x+y=13 not on list so REJECT

4) x is +ve and y is -ve

substitute , we get x+x=y=7 and x-y-y=6 solve for x and y , we get x=4 and y= -1 ,x+y=3 , ANSWER CHOICE

swati007 · Answer

I solved in 1.36 mins

jjack0310 · Answer

Sorry, still confusing me. 
I understand the first y i.e. 
|y| = -y
But y < 0, 
so wouldn't x + |y| - y = x - y - (-y), which would make it just x?

Bunuel · Answer

When y<0, then |y|=-y: correct. But y must stay as it is.

Consider this, suppose we have only x-y=6, and I tell you that y is negative would you rewrite the equation as x-(-y)=6?

Hope it's clear.

WholeLottaLove · Answer

If x + |x| + y = 7 and x + |y| - y =6 , then x + y =

x>0, y>0

x + (x) + y = 7
2x+y=7

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

2x+y=7
2(6)+y=7
y=-5

INVALID as 6>0 but -5 is not > 0

x>0, y<0

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 - 1 = 7
2x=8
x=4

VALID as 4>0 and -1<0

therefore;

x+y = (4)+(-1) = 3

(A)

KarishmaB · Answer

In such a question, you can use some brute force and get to the answer too. How long it takes depends on how quickly you observe the little things.

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

Both equations yield about the same result though in one y is positive and in the other it is negative. |x| and |y| are positive and assuming x is positive, a negative y would pull down the first equation and pump up the second one to give almost equal values. The difference is very small also signifies that the negative variable might have a very small value. 
Since the options give the value of x + y as 3/4/5... etc, it is likely that we are dealing with small number pairs such as (4, 2), (3, 2), (4, 1) etc. 
Since the first equation has 7 as the result, both variables will not be even.
A couple of quick iterations brought me to (4, -1).
So x + y = 3

sagar2911 · Answer

Hi Karishma,

I tried another approach, but got stuck - can you hep me solve by this method?
x + |x| + y = 7 => x + y = 7 - |x| => Whatever be the value of x, |x| will be always non-negative => x + y has to be less than 7 => This eliminates option E
x + |y| - y = 6 => Now, if y is positive, then this equation becomes x = 6 => on substituting in above eqn, we get y = -5 => x+y = 1 => Not present in any of the options => y is negative => x - y - y = 6 => x - 2y = 6 => x = 6 + 2y
Now x + y = 7 - |6 + 2y|
If we take y = -1, we get x + y = 3 = Option A
If we take y = -2, we get x + y = 5 = Option C
I am getting both options here - where am I going wrong here? Or this approach incorrect?

ENGRTOMBA2018 · Answer

Let me try to answer. You are making a mistake with using |6+2y| After you get x=6+2y and substitute in equation (1), you get 6+2y+|6+2y|+y=7 ---->6+3y+|6+2y| = 7 ---> 3y+|6+2y| = 1. Now the point to note is that the 'nature' of |6+2y| changes at y=-3 and thus you need to evaluate |6+2y| for values smaller than -3 and for values greater than -3. You have already assumed that y<0 in order to get x=6+2y, so your ranges to consider become -3 \le q y < 0 and y < -3 Case 1: -3 \le y < 0, giving you |6+2y| \geq 0 ----> 3y+|6+2y| = 1 ---> 3y+6+2y = 1 ---> 5y=-5 ---> y=-1. Acceptable value giving you x=6+2y = 4 --> x+y = 3 Case 2: y<-3 ---> |6+2y| = -(6+2y) ---> 3y+|6+2y| = 1 ---> 3y-6-2y = 1 ---> y=7 this contradicts the assumption that y<-3 , making this out of scope. Thus the only value of x+y = 3.

sagar2911 · Answer

Excellent. Thank you dude!

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If x + |x| + y = 7 and x + |y| - y =6 , then x + y =

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