The last problem in this group illustrates the concept of an extraneous solution. The equation involves absolute values, but to illustrate the concept we solve it in an unusual way: by squaring instead of considering two cases for the absolute value.

Enter (without the quotation marks) "$==>$" if the left equation implies the right, "$<==$" if the right equation implies the left, "$<==>$" if either equation implies the other, and "$><$" if neither equation implies the other.

$|x+1| = 2x -1$ $(x+1)^2 = (2x-1)^2$.
(Square on both sides. The square of the absolute value is the square of what is inside that absolute value.)

$(x+1)^2 = (2x-1)^2$ $x^2+2x+1 = 4x^2-4x+1$.
(Apply the Distributive Law, or, more specifically, the binomial formulas.)

$x^2+2x+1 = 4x^2-4x+1$ $3x^2-6x = 0$.
(Subtract $(x^2+2x+1)$ on both sides and switch sides.)

$3x^2-6x = 0$ $x(x-2) = 0$.
(Factor, and divide by 3 on both sides.)

$x(x-2) = 0$ $x=0$.
(one possible solution.)

$x(x-2) = 0$ $x=2$.
(the other possible solution.)

Only one of the two proposed solutions satisfies the original equation $|x+1| = 2x-1$. It is $x=$.