Question 868819
TIP: Use "abs(2x-6)" to get {{{abs(2x-6)}}} .
 
{{{abs(2x-6)=abs(x+1)}}}
{{{2x-6>=0}}}<--->{{{x>=3}}} and {{{x+1>=0}}}<--->{{{x>=-1}}}
 
For the cases {{{x>=3}}} and {{{x<-1}}} both expressions have the same sign,
and we look for solutions to {{{2x-6=x+1}}}
{{{2x-6=x+1}}}
{{{2x-x=1+6}}}
{{{x=7}}}
As {{{x=7}}} is within one of the cases above.
{{{highlight(x=7)}}} is a solution.
 
For the case {{{-1<=x<3}}} ,
{{{x+1>=0}}} so {{{abs(x+1)=x+1}}},
but {{{2x-6<0}}} so {{abs(2x-6)=-(2x-6=-2x+6}}} ,
and we look for solutions to
{{{x+1=-2x+6}}}
{{{x+2x=6-1}}}
{{{3x=5}}}
{{{x=5/3}}}
Since {{{x=5/3=1&1/3}}} is within the case {{{-1<=x<3}}} ,
{{{highlight(x=5/3)}}} is a solution.
 
Verification:
For {{{x=7}}}:
{{{abs(2x-6)=abs(2*7-6)=abs(14-6)=abs(8)=8}}} and
{{{abs(x+1)=abs(7+1)=abs(8)=8}}}
 
For {{{x=5/3}}}:
{{{abs(2x-6)=abs(2*(5/3)-6)=abs(10/3-6)=abs(10/3-18/3)=abs(-8/3)=8/3}}} and
{{{abs(x+1)=abs(5/3+1)=abs(5/3+3/3)=abs(8/3)=8/3}}}