#"We could square both sides : "#
#(x+2)^2 = (2x+1)^2#
#=> x^2 + cancel(4x) + 4 = 4x^2 + cancel(4x) + 1#
#=> 3x^2 - 3 = 0#
#=> x^2 = 1#
#=> x = pm 1#
#"The absolute value is > 0 and squaring yields also values > 0."#
#"So we will have the same solutions."#
#"We could also work with the definition of |x| : "#
#={(x " , "x>=0),(-x " , "x<=0):}#
#"But here we have 4 cases, 2 for the LHS (left hand side of"#
#"the equation) and 2 for the RHS, so it is a lot of work to deal"#
#"with 4 cases, squaring is easier."#