How do you solve #4sqrt(7 + 3z) = 2#?

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Answer 1 · 2015-08-19T12:47:00

#color(blue)(z=-9/4#

#4sqrt(7+3z)=2#

Squaring the L.H.S and R.H.S:

#(4sqrt(7+3z))^2=2^2#

#16(7+3z)=4#

#7+3z=4/16#

#7+3z=1/4#

#7*4+3z*4=1/4*4#

#28+12z=1#

#12z=1-28#

#z=-27/12#

#color(blue)(z=-9/4#