Solving Systems Part 2

Question

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

Answer 1

Because the second two equations have coefficents for two variables which are the equal but of opposite signs, the best start would be to add these two equations together and solve for x.

 

x - 2y - z = 2

3x + 2y + z = 2

 

By adding these two equations together, the y and z terms cancel and we get

x + 3x = 2 + 2

4x = 4

x = 1

 

We can now plug this value of x into the top two equations to give us two equations in two unknowns.

 

1 + y + z =2

1 - 2y - z = 2

 

We can now add these two equations together which causes the z terms to cancel out

 

1 + 1 + y - 2y = 2 + 2

2 - y = 4

y = -2

 

We can now plug these values for x and y in any of the equations to get the z value. Plugging into the first equation, we get: 

 

x+y+z = 2

1-2+z = 2

-1+z = 2

z = 3

 

Therefore, x = 1, y = -2, and z = 3