The solution to the system of equations using elimination method is (1, 1, 0)
Given the system of equations;
-2x+2y+3z=0 ............. 1
-2x-y+z=-3 .............2
2x+3y+3z=5 ...............3
Subtracting 1 from 2 will give;
2y+y + 3z-z = 0-(-3)
3y + 2z = 3 ...................... 4
Add equation 2 and 3
-y + 3y + z + 3z = -3 + 5
2y+4z = 2
y+2z = 1 .............. 5
Solve 4 and 5 simultaneously:
3y + 2z = 3 .......... 4
y+2z = 1 .............. 5
Subtract 4 from 5 to have:
3y - y = 3 - 1
2y = 2
y = 1
Since y + 2z = 1
1 + 2z = 1
2z = 0
z = 0
Substitute y = 1 and z = 0 into the equation 2:
Recall from 2;
-2x - y + z = -3
-2x - 1 = -3
-2x = -3 + 1
-2x = -2
x = 1
Hence the solution to the system of equations is (1, 1, 0)
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