Equations and inequalities as parameters: proving that an equation holds. I have $x-y=3$ and $y\le1$ and $x\ge\frac12$.
I proved that $\sqrt{(2x-1)^2}+\sqrt{(2y-2)^2}=7$ and that $-\frac52\le y\le 1$ and $\frac12\le x\le4$.
How can I prove that $|x+y-5|+|x+y+2|=7$?

ok let us  consider following cases :
first side is  just  $-x-y+5$,second case is   $x+y+2$,so sum is $5+2=7$
to be more deeply,let us take such  situations


*

*$y=1$


2.$x=4$
we have
$|x+y-5]+|x+y+2|=x+y-5+x+y+2=2*x+2*y-3=2*(x+y)-3
=2*(5)-3=7$
can you continue from this?
just consider this situation when $x<0$  then $|x|=-x$,else  if $x>=0$ then $|x|=x$.ok let us consider following case,when
$y=-5/2$   and $x=1$  ,we have that $-5/2+1-5<0$,that why we will have
$-x-y+5$,
for second we have  $-5/2+1+2>0$ ,
that why we have second as $x+y+2$,together is is still $7$
