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Byju's Answer
Standard XII
Mathematics
Parts of a linear equation
Solve each of...
Question
Solve each of the following system of equations in R.
28. |x − 1| + |x − 2| + |x − 3| ≥ 6
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Solution
We have,
x
-
1
+
x
-
2
+
x
-
3
≥
6
The
LHS
of
the
inequation
contains
three
seperate
modulus
.
By
equating
these
expressions
to
zero
,
we
get
,
x
=
1
,
2
,
3
as
the
critical
points
.
These
points
divide
the
real
line
into
four
parts
,
i
.
e
.
(
-
∞
,
1
]
,
[
1
,
2
]
,
[
2
,
3
]
,
[
3
,
∞
)
So
,
let
us
consider
the
following
cases
.
CASE
1
:
-
∞
<
x
≤
1
Then
,
x
-
1
=
-
(
x
-
1
)
,
x
-
2
=
-
(
x
-
2
)
,
x
-
3
=
-
(
x
-
3
)
∴
x
-
1
+
x
-
2
+
x
-
3
≥
6
⇒
-
x
+
1
-
x
+
2
-
x
+
3
≥
6
⇒
-
3
x
≥
0
⇒
x
∈
(
-
∞
,
0
]
CASE
2
:
1
≤
x
≤
2
x
-
1
=
x
-
1
,
x
-
2
=
-
(
x
-
2
)
,
x
-
3
=
-
(
x
-
3
)
Then
,
x
-
1
+
x
-
2
+
x
-
3
≥
6
⇒
x
-
1
-
x
+
2
-
x
+
3
≥
6
⇒
x
≥
2
⇒
x
∈
(
-
∞
,
2
]
CASE
3
:
2
≤
x
≤
3
Then
,
x
-
1
=
x
-
1
,
x
-
2
=
x
-
2
,
x
-
3
=
-
(
x
-
3
)
∴
x
-
1
+
x
-
2
+
x
-
3
≥
6
⇒
x
-
1
+
x
-
2
-
x
+
3
≥
6
⇒
x
≥
6
⇒
x
∈
[
6
,
∞
)
CASE
4
:
When
3
≤
x
<
∞
Then
,
all
the
three
modulus
are
positive
.
∴
x
-
1
+
x
-
2
+
x
-
3
≥
6
⇒
x
-
1
+
x
-
2
+
x
-
3
≥
6
⇒
3
x
-
6
≥
6
⇒
3
x
≥
12
⇒
x
≥
4
⇒
x
∈
[
4
,
∞
)
Hence
,
the
solution
of
the
given
inequation
is
the
union
of
the
sets
formed
in
all
the
above
four
cases
.
∴
x
∈
(
-
∞
,
0
]
∪
(
-
∞
,
2
]
∪
[
6
,
∞
)
∪
[
4
,
∞
)
∴
x
(
-
∞
,
0
]
∪
[
4
,
∞
)
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