How do you solve #\sqrt { 7u + 6} = \sqrt { 5u + 16}#?

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Answer 1 · 2017-03-13T11:13:52

#u = 5#

#sqrt (7 u + 6) = sqrt (5 u + 16)#

square both sides

#(sqrt (7 u + 6))^2 = (sqrt (5 u + 16))^2#

#7 u + 6 = 5 u + 16#

move #5 u# to left hand side and #6# to right hand side

#7 u - 5 u = 16 - 6#

#2 u = 10#

divide both sides with #2#

#(2 u)/2 = 10/2#

#u = 5#

Answer 2 · 2017-03-13T11:34:28

#u=5#

#color(blue)(sqrt(7u+6)=sqrt(5u+16)#

To find the value of #u#, we need to isolate it. We should balance both sides (applying same operations)

Square both sides to remove the radical signs

#rarr(sqrt(7u+6))^color(red)(2)=(sqrt(5u+16))^color(red)(2)#

#rarr7u+6=5u+16#

Subtract #6# both sides

#rarr7u+6-color(red)(6)=5u+16-color(red)(6)#

#rarr7u=5u+10#

Subtract #5u# both sides

#rarr7u-color(red)(5u)=5u+10-color(red)(5)#

#rarr2u=10#

Divide both sides by #2#

#rarr(cancel2u)/(color(red)(cancel2))=10/(color(red)(2))#

#color(green)(rArru=5#

Hope this helps! :)