Solve the system of linear equation in three variables X+y+z=5 ,2x-y+z =9, x-2y+ 3z =16
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Answer:
Given system of linear equations-
x
+
y
+
z
=
5
2
x
−
y
+
z
=
9
x
−
2
y
+
3
z
=
16
(Eq.1)
(Eq.2)
(Eq.3)
Now equation (1) + equation (2)-
x
+
y
+
z
+
(
2
x
−
y
+
z
)
=
14
3
x
+
2
z
=
14
(Eq.4)
Now Equation (3)+2 (Equation (1))-
x
−
2
y
+
3
z
+
2
(
x
+
y
+
z
)
=
16
+
2
(
5
)
3
x
+
5
z
=
26
(Eq.5)
Now subtract the equation (4) from equation (5)-
3
x
+
5
z
−
(
3
x
+
2
z
)
=
26
−
14
3
z
=
12
z
=
4
From equation (4)-
3
x
+
2
(
4
)
=
14
3
x
=
14
−
8
3
x
=
6
x
=
2
From equation (1)-
2
+
y
+
4
=
5
y
=
5
−
6
y
=
−
1
So the
x
=
2
,
y
=
−
1
and
z
=
4
are the required solutions to the given system of linear equations.
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