How do you solve #2+sqrt(5x-1)>5#?

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Answer 1 · 2018-05-07T11:11:38

#color(blue)((2,oo)#

#2+sqrt(5x-1) > 5#

Subtract 2 from both sides of the inequality sign:

#2-color(red)(2)+sqrt(5x-1) > 5-color(red)(2)#

#sqrt(5x-1) > 3#

Square both sides:

#(sqrt(5x-1))^2 > 3^2#

#5x-1 > 9#

Add 1 to both sides:

#5x-1+color(red)(1) > 9+color(red)(1)#

#5x >10#

Divide both sides by 5:

#(5x)/5 > 10/5#

Cancelling:

#(cancel(5)x)/(cancel(5)) > (cancel(10)2)/(cancel(5))#

#x > 2#

We can write the range of solutions in interval notation as:

#(2,oo)#