How do you solve for b in #3ab–z = y(4ab +2z)#?

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Answer 1 · 2016-02-14T20:58:45

You must first distribute the parentheses.

#3ab - z = 4aby + 2yz#

Put the terms containing b on one side of the equation.

#3ab - 4aby = 2yz + z#

Factor out the b.

#b(3a - 4ay) = 2yz + z#

#b = (2yz + z)/(3a - 4ay)# or # b = (z(2y + 1))/(a(3 - 4y))#

Practice exercises:

a) #(2x + 3y)/(4x - y) = (x + 3y)/(2x - 5y)#

b). #(2x + 3y + 2z) / x + (4x - 5y + 3z)/y = (x - 3y - 2z)/z#

Good luck!