How do you solve #sqrt25 - 3x -6 = 1#?

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Answer 1 · 2015-07-24T18:29:30

You isolate #x# on one side of the equation.

The thing to notice about this equation is that #sqrt(25)# is actually equal to

#sqrt(25) = sqrt(5""^2) = 5#

This means that your original equation becomes

#5 - 3x - 6 = 1#

You can easily solve this by first isolating #-3x# on one side of the equation first. Start from

#-1 - 3x = 1#

Add #+1# to both sides of the equation to get

#cancel(-1) + cancel(1) - 3x = 1 + 1#

#-3x = 2#

Now divide both sides of the equation by #-3# to get

#(cancel(-3) * x)/cancel(-3) = 2/-3#

#x = color(green)(-2/3)#