Simplify: (a) |6 − x| + |x − 10| if |4 − x| = 4 − x and |2 − x| = x − 2.

(b) Simplify 2 − |2 − |x − 2|| if x < −2.

(c) Evaluate a^2− 2ab if |a − 2| + |3 + b| = 0

(d) Find all possible values of y for which |2y + 7| = |3y − 1|.

(a) 16-2x

(b)-x-2 or x+6

(c)

Step-by-step explanation:

(a)

|6-x|+|x-10|,

|4-x|=4-x,so x<4

|2-x|=x-2=-(2-x)

2-x<0

2<x

or x>2

2<x<4

so |6-x|=6-x,in 2<x<4

|x-10|=-(x-10),in 2<x<4

|6-x|+|x-10|=6-x-(x-10)=6-x-x+10=16-2x

(b)

2-|2-|x-2||,if x<-2

|x-2|=-(x-2) if x<-2

=2-|2-{-(x-2)}|

=2-|2+x+2|

x<-2

x+2<0

=2-|x+4|

=2-(x+4),if x>-4,or -4<x<-2

=2-x-4

=-x-2

if x<-4

then |x+4|=-(x+4)

2-|x+4|=2-{-(x+4)}

=2+x+4

=x+6