# How do you solve #abs(x-5)+ abs(2-2x) = 7#?

##### 1 Answer

#### Explanation:

The idea here is that you need to looks at the intervals for which the two expressions inside the moduli *change signs*.

Since the absolute value of a real number is **always positive**, regardless of the sign of the number, you need to take into account the fact that these expressions can be negative as well.

So, you know that

#x - 5 > 0 implies x > 5" "# and#" "1-x>0 implies x < 1#

This will produce **three intervals** which need to be considered

#x in (-oo, 1)#

In this case, you have

#x - 5 < 0 implies |x-5| = -(x-5)#

and

#2(1-x) > 0 implies |2-2x| = 2 - 2x#

The equation becomes

#-(x-5) + 2 - 2x = 7#

#-x + 5 + 2 - 2x = 7#

#-3x = 0 implies x = 0#

Since you have

#x in (1,5)#

This time, you have

#x-5 < 0 implies |x-5| = -(x-5)#

and

#2(1-x) < 0 implies |2-2x| = -(2-2x)#

The equation will be

#-x + 5 - 2 + 2x = 7#

#x = 7 - 3 = 4#

Once again, you have

#x in (5, + oo)#

This time, you have

#x - 5 >0 implies |x-5| = x-5#

and

#2(1-x)<0 implies |2-2x| = -(2-2x)#

The equation will be

#x - 5 -2 + 2x = 7#

#3x = 14 implies x = 14/3#

However, *extraneouis*.

The only two valid solutions will be

#x = color(green)(0)" "# or#" "x = color(green)(4)#