Could someone please help
2 answers:
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Assuming and
, as well as
and
, we have
Since both the real and imaginary parts are positive, belongs to the first quadrant.
Answer:
Given: The following system:
4x + y − 2z=18 ......[1]
2x-3y+3z = 21 ......[2]
x-3y=6 ......[3]
we can write equation [3] as;
3y = x-6 ......[4]
Multiply by 3 in equation [1] to both sides of an equation we get,
or
12x+3y-6z=54 ......[5]
Substituting the equation [4] in [2] and [5] we get;
2x-(x-6)+3z=21 or
2x-x+6+3z=21
Simplify:
x+3z=15 .....[6] [combine like terms]
12x+x-6-6z =54
Simplify:
13x-6z=60 ......[7] [Combine like terms]
On Solving equation [6] and [7] simultaneously,
x+3z=15
13x-6z=60
we get the value of x
i.e, x=6
Substitute the value of x in equation x+3z=15 we get
6+3z=15 or
3z=9
Simplify:
z=3
Also, substitute the value of x=6 in equation [3] we get the value of y;
x-3y=6
6-3y=6 or
-3y = 0
Simplify:
y = 0
Therefore, the solution to the system of three linear equation is, (6, 0 , 3)