Answer:
[tex]b=\frac{z(2y+1)}{a(3-4y)}[/tex] for [tex]a(3-4y)\neq 0[/tex].
Step-by-step explanation:
The given equation is
[tex]3ab-z=y(4ab+2z)[/tex]
We need to solve the above equation for b.
Using distributive property we get
[tex]3ab-z=y(4ab)+y(2z)[/tex]
[tex]3ab-z=4aby+2yz[/tex]
Subtract 4aby from both sides.
[tex]3ab-z-4aby=2yz[/tex]
Add z on both sides.
[tex]3ab-4aby=2yz+z[/tex]
Taking out the common factor.
[tex]ab(3-4y)=z(2y+1)[/tex]
Divide both sides by a(3-4y).
[tex]b=\frac{z(2y+1)}{a(3-4y)}[/tex]
Therefore the value of b is [tex]b=\frac{z(2y+1)}{a(3-4y)}[/tex] for [tex]a(3-4y)\neq 0[/tex].
