Respuesta :

[tex]\bf wxy+xyz=wx+yz\implies wxy+xyz-yz=wx\leftarrow \begin{array}{llll} taking\\ common\\ factor \end{array} \\\\\\ y(wx+xz-z)=wx\implies y=\cfrac{wx}{wx+xz-z}[/tex]

I suppose you're not expected to get a constant for "y", since the fact that the other variables are >1 won't give us enough mileage for a constant value for "y".
MrRoyal

To solve for a variable, we start by isolating the variable terms and then solve for the variable. The equivalent of y is: [tex]y = \frac{wx}{wx + xz - z}[/tex]

Given that:

[tex]wxy + xyz = wx + yz[/tex]

Collect like terms of y

[tex]wxy + xyz - yz= wx[/tex]

Factor out y

[tex]y \times (wx + xz - z)= wx[/tex]

Make y the subject

[tex]y = \frac{wx}{(wx + xz - z)}[/tex]

Remove the brackets

[tex]y = \frac{wx}{wx + xz - z}[/tex]

Hence, the equivalent of y is: [tex]y = \frac{wx}{wx + xz - z}[/tex]

Read more about equations at:

brainly.com/question/953809

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