How do you solve and check for extraneous solutions in $abs(x-1) = 5x + 10$ ?

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Answer 1 · 2015-08-02T19:42:33

Solution: $x = -3/2$
Extraneous solution: $x = -11/4$

If you take into account the fact that the absolute value of a number is always positive regardless if said number is positive or negative

$color(blue)( |n| = {(n",", "if "n>=0), (-n",", "if "n<0) :})$

then you can say that the solutions to this equation must satisfy the condition

$5x+10 >0 <=> x> -2$

Now, your absolute value equation will produce two solutions, depending on which condition is true

$|x-1| = x-1$

This will get you

$x-1 = 5x+10 => x = color(red)(-11/4)$

$|x-1| = -(x-1) = -x+1$

The solution to the equation will be

$-x+1 = 5x + 10 => x = -9/6 = color(green)(-3/2)$

As you can see, $x = -11/4$ does not satisfy the condition $x> -2$ , which means that this solution will be extraneous.

The only solution to this equation will thus be $x = -3/2$ .

How do you solve and check for extraneous solutions in abs(x-1) = 5x + 10abs(x-1) = 5x + 10?