MATH SOLVE

2 months ago

Q:
# Solve the system by elimination-2x+2y+3z=0, -2x-y+z=-3, 2x+3y+3z=5

Accepted Solution

A:

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 ...............3Subtracting 1 from 2 will give;2y+y + 3z-z = 0-(-3)3y + 2z = 3 ...................... 4Add equation 2 and 3-y + 3y + z + 3z = -3 + 52y+4z = 2y+2z = 1 .............. 5Solve 4 and 5 simultaneously:3y + 2z = 3 .......... 4y+2z = 1 .............. 5Subtract 4 from 5 to have:3y - y = 3 - 12y = 2y = 1Since y + 2z = 11 + 2z = 12z = 0z = 0Substitute y = 1 and z = 0 into the equation 2:Recall from 2;-2x - y + z = -3-2x - 1 = -3-2x = -3 + 1-2x = -2x = 1Hence the solution to the system of equations is (1, 1, 0)Learn more here: