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solve the system by elimination

-2x+2y+3z=0
-2x-y+z=-3
2x+3y+3z=5

User Zucker
by
6.1k points

1 Answer

6 votes

Answer: x = 1, y = 1, z = 0

Explanation:

EQ 1: -2x + 2y + 3z = 0 EQ 2: -2x - y + z = -3

EQ 3: 2x + 3y + 3z = 5 EQ 3: 2x + 3y + 3z = 5

5y + 6z = 5 2y + 4z = 2

Eliminate one of the variables for the new equations:

2(5y + 6z = 5) --> 10y + 12z = 10

-3(2y + 4z = 2) --> -6y - 12z = -6

4y = 4

y = 1

Substitute y = 1 into one of the new equations to solve for z:

5(1) + 6z = 5

5 + 6z = 5

6z = 0

z = 0

Substitute y = 1 and z = 0 into one of the original equations to solve for x:

2x + 3(1) + 3(0) = 5

2x + 3 + 0 = 5

2x = 2

x = 1

User Kalyan Raghu
by
5.9k points