Solve the system: x+2y -z=3 2x-y+2z=8 x+y+z=3
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x+2y -z=3......(1)
2x-y+2z=8......(2)
x+y+z= 3......(3)
First we will add(3) and (2):
==> 3x +3z = 11........(4)
Now multiply (2) by 2 and then add to (1):
==> 5x +3z = 19.....(5)
Now subtract (4) from (5):
==> 2x = 8
==> x= 4
Substitute in (5) :
5(4) + 3z = 19
==> 3z= -1
==> z= -1/3
Now substitute in (3):
4+ y -1/3 = 3
==> y= -1+1/3
==> y= -2/3
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Being a system of linear equations, we'll calculate the determinant of the matrix of the system, which is formed from the coefficients of x, y and z.
We'll identify a = 1, b = 2, c = -1, d = 2, e = -1, f = 2, g = 1, h = 1, i = 1
det (A)=2(-1*1 - 1*2)-2(2*1 + 1)+1*(2*2-1)
det (A) = -6-6+6= -6
Since det A has a value different from zero, that means that the system is determined and it has a single solution.
The system will be solved using Cramer formula.
With this formula, we'll calculate each unknown in this way:
x = det x/ det A
y = det y/ detA
z = det z/ det A
To calculate det x, we'll substitute in the determinant A, the column of the coefficient of the unknown x. So, instead of a, d, g, we'll have 3,8,3.
det x = 3(-1*1 - 1*2) - 8(2*1 + 1)+ 3*(2*2-1)
det x = -9-24+9
det x = -24
x = -24/-6
x = 4
Now, we can substitute x in the first and the third equation:
2y - z = 3-4
y + z = 3-4
We'll add these equations and we'll get:
3y = -2
y = -2/3
-2/3 + z = -1
z = -1 + 2/3
z = -1/3
The solution of the system is: (4 , -2/3 , -1/3).
Given:
x + 2y - z = 3 ... (1)
2x - y + 2z = 8 ... (2)
x + y + z = 3 ... (3)
Adding equations (1) and (3):
x + x + 2y + y - z + z = 3 + 3
2x + 3y = 6 ... (4)
Multiplying equation (3) by 2:
2x + 2y + 2z = 6 ... (5)
Subtracting equation (2) from (5):
2x - 2x + 2y + y + 2z - 2z = 6 - 8
3y = -2
y = -2/3
Substituting this value of y in equation (4):
2x + 3*(-2/3) = 6
2x - 2 = 6
2x = 6 + 2 = 8
x = 8/2 = 4
Substituting these values of x and y in equation (3):
4 - 2/3 + z = 3
10/3 + z = 3
z = 3 - 10/3 = -1/3
Answer:
x = 4
y = -2/3
z = -1/3
To solve:
x+2y-z =3.........(1)
2x-y+2z=8........(2)
x+y+z=3...........(3)
Elimination of z:
Eq(2)+Eq(1)*2 gives: 4x+3y = 14.........(4)
Eq(1)+Eq(3): gives: 2x+3y = 6..........(5)
Elimination of y:
Eq(4)-(5) gives: 2x=8, x = 8/2 = 4. From (5) , 3y = 6-2x =6-2(4) = -2, y=-2/3,
From (3), x+y+z =3, 4 -2/3+z = 3 , z = 3-4+2/3 = -1/3
(x,y,z) = (4,-2/3,-1/3)
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