How do you solve #24/(5z+4)=4/(z-1)#?

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Answer 1 · 2016-08-30T06:13:25

#z = 10#

Given:

#24/(5z+4) = 4/(z-1)#

Multiply both sides by #(5z+4)# to get:

#24 = (4(5z+4))/(z-1)#

Multiply both sides by #(z-1)# to get:

#24(z-1) = 4(5z+4)#

Expand both sides to get:

#24z-24 = 20z+16#

Subtract #20z# from both sides to get:

#4z-24 = 16#

Add #24# to both sides to get:

#4z = 40#

Divide both sides by #4# to get:

#z = 10#

Answer 2 · 2016-08-30T08:29:03

#z=10#

The equation has one fraction on each side of the equal side.

One way to get rid of the denominators is to cross-multiply:

#24/(5z+4) = 4/(z-1)#

#24(z-1) = 4(5z+4) color(white)(xxx)larr# use distributive law

#24z-24 = 20z+16 color(white)(xxxx)larr# re-arrange terms

#24z-20z = 16+24 color(white)(xxxx)larr# simplify

#4z = 40color(white)(xxxxxxxxxxxxx)larr div 4#

#z = 10#