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

1 Answer
Sep 10, 2015

Answer:

#x=0" "# or #" "x = 4#

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 #0 in (-oo, 1)#, #x=0# will be a valid solution to the original equation.

  • #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 #4 in (1,5)#, so #x=4# will be a valid solution.

  • #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, #14/3 !in (5, + oo)#, so this solution will be extraneouis.

The only two valid solutions will be

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