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Re: Interesting Absolute value Problem [#permalink]
13 Jun 2010, 06:01

8

This post received KUDOS

Expert's post

Hussain15 wrote:

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

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

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. _________________

Re: Interesting Absolute value Problem [#permalink]
18 Jun 2010, 06:23

Bunuel wrote:

Hussain15 wrote:

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

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

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.

Re: Interesting Absolute value Problem [#permalink]
18 Jun 2010, 06:28

Expert's post

GMATBLACKBELT720 wrote:

Bunuel wrote:

Hussain15 wrote:

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

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

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.

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

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

Re: Interesting Absolute value Problem [#permalink]
18 Jun 2010, 07:17

^ 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...

Re: If x + |x| + y = 7 and x + |y| - y =6 , then x + y = [#permalink]
02 Jul 2013, 05:38

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

Re: Interesting Absolute value Problem [#permalink]
05 Jul 2013, 10:19

Bunuel wrote:

GMATBLACKBELT720 wrote:

Bunuel wrote:

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.

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

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

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?

Re: If x + |x| + y = 7 and x + |y| - y =6 , then x + y = [#permalink]
03 Sep 2014, 10:18

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