The denominator of y cannot be zero as this would make y
undefined
.
Equating the denominator to zero and solving gives the value that x cannot be.
solve
3
x
−
2
=
0
⇒
x
=
2
3
←
excluded value
⇒
domain is
x
∈
R
,
x
≠
2
3
Rearrange to make x the subject
y
(
3
x
−
2
)
=
1
⇒
3
x
y
−
2
y
=
1
⇒
3
x
y
=
1
+
2
y
⇒
x
=
1
+
2
y
3
y
solve
3
y
=
0
⇒
y
=
0
←
excluded value
⇒
range is
y
∈
R
,
y
≠
0
graph{1/(3x-2) [-10, 10, -5, 5]}
