Answer
(a) yes
(b) no
Work Step by Step
(a) Using $x=2$, we get:
Left side:
$1-[2-(3-2)]=1-[2-1]=1-1=0$ and RHS $=4(2)-(6+(2))=8-8=0$
Right side:
$4(2)-(6+2)=8-8=0$
The sides match, so $x=2$ is a solution.
(b) Using $x=4$, we get:
Left side:
$1-[2-(3-4)]=1-[2-(-1)]=1-3=-2$
Right side:
$4(4)-(6+4)=16-10=6$
The sides do not match, so $x=4$ is not a solution.
