If xy + x = z, is |x + y| z?

Question

If xy + x = z, is |x + y| > z?

(1) x < 0
(2) y > 0

Sindbad · Answer

Answer: C
Re-writing xy+x=z as x(y+1)=z --(a)
Now looking at the question we can definitely say
|x+y|>z if a) z is -ve as mod of any number is always +ve
OR b) it can be proved that z is positive and > x+y
Taking i) x< 0 our modified equation (a) suggests that sign of z depends on sign of (y+1) so it can be both +ve or -ve. There is no data on the actual numeric value of x ,y or z. so. Not sufficient
Taking ii) y>0 our modified equation (a) suggests that sign of z depends on sign of x so it can be both +ve or -ve. There is no data on the actual numeric value of x ,y or z. so. Not sufficient

Combining i and ii and looking at eq -(a)
x< 0 and y>0 => (y+1)>0. Hence x(y+1) <0 => z <0. From our earlier inference if z<0 then |x+y|>z. Sufficient.

sedreddin · Answer

neither statement can give us much
but if we combine both here is the interesting thing we can find: whether |x+y|> xy+x

if x is negative and y is positive, then xy will be negative and plus negative will always be negative, and if we sum negative and negative the result will be negative so, xy+x is negative if we combine statements 1) and 2)

|x+y| though will be positive no matter what, so the answer will be a definite yes, so, the answer is C (both 1 &2 are sufficient together)

neilphilip10 · Answer

By manipulating the condition we get x+y>z? or x+y<-z
Therefore x+y>xy+x => is y(x-1)<0?
x+y<-(xy+x)=> is y(x+1)+2x<0?
We clearly need the signs of x and y
1) No sign of y.Insufficient
2)No sign of X. Insufficient

1)+2)=>Sufficient to prove above statements

niks18 · Answer

Given x(y+1)=z , we need the values of x , y & z to determine whether |x+y|>z , where LHS is always positive Statement 1 : implies x is negative so x(y+1) i.e. z can be negative or it can be positive depending upon the value of (y+1) . But nothing is mentioned about y . So insufficient Statement 2 : implies y is positive so x(y+1) i.e. z can be negative or it can be positive depending upon the value of x . But nothing is mentioned about x . So insufficient Combining 1 & 2 we get x<0 & (y+1)>0 so x(y+1)<0 i.e. z<0 but |x+y| is always positive hence we get a Yes for our question stem. Sufficient Option C

Krishaa12 · Answer

According me E is correct since y is less than zero so y=-1 x= 1 therefore z can be neget ive or +'ve while combining plz correct me if I am wrong

niks18 · Answer

Hi  Krishaa12

y>0, hence it is positive, rather x<0 i.e negative

Posted from my mobile device

ishan261288 · Answer

An alternate explanation

x(y+1) - z = 0
First take 1 and 2 both, if x is negative and y is positive, then term x(y+1) is negative, so for x(y+1) - z =0 to be true, z has to be negative so that -z is positive and x(y+1) - z =0. So from this we can deduce that z is also negative--So question stem---> |x+y| is always positive and if z is negative, so |x+y| >z is true.
Now C is correct unless we can make one out of 1 and 2 individually correct or both individually correct. Let's analyze

Take 1) x < 0, if x is negative (and y is positive or negative), nothing can be said about sign of z so we can never know |x+y| >z.
Take 2) y > 0, if y is positive (x can be positive or negative), nothing can be said about sign of z so we can never know |x+y| >z.

So answer is C

anon7164 · Answer

I didn't find any of the solutions very intuitive or doable in 2 minutes, so I came up with a method but I'm not entirely sure it is correct. Here it is:

Given: x(y + 1) = z
Question: Is |x+y| > z?

Restate the question by opening the absolute value bars and taking both cases.

Is |x+y| > z? is restated as
Is x + y > z --> Is x + y - z > 0
AND
Is -x - y > z --> Is -x - y - z > 0?

If the conditions given satisfy both cases above, we should be able to say that |x+y| > z.

So combine both inequalities by adding them: x + y - z + (- x - y - z) > 0? ---> -2z > 0? ---> z < 0?

We have finally restated the question as, "is z < 0?"

1) x < 0
Tells us nothing about z, insufficient.

2) y > 0
Tells us nothing about z, insufficient.

1) & 2) x < 0 and y > 0 means that x(y+1) = z is <0. Sufficient.

Not sure if I've assumed something wrong here or if there is any other error. Bunuel chetan2u would be grateful if you guys could help me with this - is this approach correct? If not, could you suggest a better method? Thanks in advance

If xy + x = z, is |x + y| > z?

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If xy + x = z, is |x + y| > z?

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