If xy 0, what is the value of |xy - yz|

Question

If  xy < 0  and  yz > 0 , what is the value of  |xy - yz| - \sqrt{(yz-xz)^2} + |xy| + \sqrt{(xz)^2}

A.  -x 
 
B.  -2y 
 
C.  x + y 
 
D.  -2xy 
 
E.  y

gmatophobia · Accepted Answer

We can infer the following from the premise of the question

xy < 0   
  yz > 0

y and z share the same sign 
  x and y share opposite signs

So, if x = +ve; y and z are both -ve. On the contrary, if x = -ve, y and z are both +ve.

As no other constraints are mentioned in the question, we can assume values of x, y, and z to solve the question.

x = -1
y = 2
z = 3

|xy - yz| - \sqrt{(yz-xz)^2} + |xy| + \sqrt{(xz)^2}

|-2 - 6| - \sqrt{(6 - (-3))^2} + |-2| + \sqrt{(-3)^2}

|-8| - \sqrt{(9)^2} + |-2| + \sqrt{(-3)^2}

8 - 9 + 2 + 3

4

Answer choice elimination

A.  -x  ⇒ 1 :  Eliminate 
 
B.  -2y  ⇒ -2 :  Eliminate 
 
C.  x + y  ⇒ 1 :  Eliminate 
 
D.  -2xy  ⇒ 4 :  Answer 
 
E.  y

Option D

gmatophobia · Answer

Here you go Danish234 - Given xy < 0 yz > 0 Given the above constraint, xz will always be -ve as x and z will have opposite signs. Assume y < 0 xy < 0 ⇒ For this inequality to hold true, x > 0 yz > 0 ⇒ For this inequality to hold true, z < 0 Therefore, x and z hold opposite signs, and xz < 0 Assume y > 0 xy < 0 ⇒ For this inequality to hold true, x < 0 yz > 0 ⇒ For this inequality to hold true, z > 0 Therefore, x and z hold opposite signs, and xz < 0 |xy - yz| - \sqrt{(yz-xz)^2} + |xy| + \sqrt{(xz)^2} |xy - yz| - |yz-xz| + |xy| + |xz| xy - yz → -ve; hence |xy - yz| = -(xy -yz) = yz - xy yz-xz → +ve; hence |yz-xz| = yz-xz xy → -ve; hence |xy| = -xy xz → -ve; hence |xz| = -xz Let's substitute the above values in the equation |xy - yz| - |yz-xz| + |xy| + |xz| (yz - xy) - (yz - xz) + (-xy) + (-xz) yz - xy - yz + xz - xy - xz -2xy Option D Hope this helps !

Danish234 · Answer

Can you solve it in an algebraic format using the formula sqrt(x^2)= |x|.
Instead of assuming and plugging in values
   Posted from my mobile device

Danish234 · Answer

Thank you for the solution but how do you say that |xy-yz| is -ve And explain for the rest as well

Posted from my mobile device

gmatophobia · Answer

The premise of the question states that

xy < 0  &  yz > 0 ; hence  xy  is negative and  yz  is positive

negative - positive = negative

Hence,

xy - yz = negative

|negative number| = - (that number)

|xy - yz| = -(xy - yz)

Similar reasoning is applicable for (yz - xz) = positive.

KarishmaB · Answer

Note that: |x| = x if x >= 0 |x| = -x if x < 0 Discussed here: https://anaprep.com/algebra-the-why-beh ... questions/ So how we remove the absolute value sign from around x depends on the sign of x. xy < 0 (x and y have opposite signs) and yz > 0 (y and z have same signs) This means x and z have opposite signs too so xz <0 |xy - yz| - \sqrt{(yz-xz)^2} + |xy| + \sqrt{(xz)^2} |xy - yz| - |yz-xz| + |xy| + |xz| Since xy is negative and yz is positive, xy - yz will be negative. So |xy - yz| = - (xy - yz) Since yz is positive and xz is negative, yz - xz will be positive. So |yz - xz| = yz - xz Since xy is negative, |xy| = - xy Since xz is negative, |xz| = - xz We get - (xy - yz) - (yz - xz) - xy - xz = - 2xy Answer (D) Or simply plug in x = -1, y = 1 and z = 2 and check the options

Addu.23 · Answer

gmatophobia  I undertand what you've done and it makes sense. Where I'm stuck at is whenever the mod is removed the value is always positive, right? how come we're not doing that here? please refer to the highlighted the section 
 Bunuel

Bunuel · Answer

When a < 0, |a| = -a.

finisher009 · Answer

Take smart numbers for speed:
• Let y = 1 (positive for quick work)
• Then x < 0, z > 0

Plug y = 1, x < 0, z > 0 into the expression, Substitute for Speed:

|x1 – 1z| – |z – xz| + |x1| + |xz|
= |x – z| – |z(1 – x)| + |x| + |x*z|

With x < 0, z > 0:
• |x| = –x
• |xz| = –xz
• |x – z| = z – x (since z > x)
• |z(1 – x)| = z – z*x

Expression simplifies to: (–x) + (–xz) + (z – x) – (z – zx)
Combine terms:
= –2x

Final Answer after putting y=1
D. –2xy

bumpbot · Answer

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If xy < 0 and yz > 0, what is the value of |xy - yz|

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