How do you solve #2/5(z+1)=y# for #z#?

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Answer 1 · 2018-06-13T07:47:39

By arranging the equation

Get rid of the paranthesis first:

#(2z)/5 + 2/5 = y#

#(2z)/5 = y - 2/5#

#2z = 5y - (2times5)/5#

#2z = 5y - 2#

#z = (5y)/2 - 1#

Your answer is:

#z= (5y)/2 - 1#