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

##### 1 Answer

#### Answer:

Solution:

Extraneous solution:

#### Explanation:

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

*If*#(x-1)>0# ,*then*

#|x-1| = x-1#

This will get you

*If*#(x-1)<0# ,*then*

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

The solution to the equation will be

As you can see, *extraneous*.

The only solution to this equation will thus be