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

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.


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

\(|xy - yz| - |yz-xz| + |xy| + |xz|\)

\(xy - yz\) → -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 !

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

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.

Note that:

|x| = x if x >= 0
|x| = -x if x < 0

Discussed here: https://anglesandarguments.com/blog-details/133

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

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