#### Question

Solve the following systems of equations:

x − y + z = 4

x + y + z = 2

2x + y − 3z = 0

#### Solution

We have,

x − y + z = 4 ...(i)

x + y + z = 2 ....(ii)

2x + y − 3z = 0 ....(iii)

From equation (i), we get

z = 4 - x + y

z = -x + y + 4

Substituting z = -x + y + 4 in equation (ii), we get

x + y + (-x + y + 4) = 2

=> x + y - x + y + 4 = 2

=> 2y + 4 = 2

`=> 2y = 2 - 4 = -2`

=> 2y = -2

`=> y = (-2)/2 = -1`

Substituting the value of z in equation (iii), we get

2x + y -3(-x + y + 4) = 0

=> 2x + y + 3x - 3y - 12 = 0

=> 5x - 2y - 12 = 0

=> 5x - 2y = 12 ....(iv)

Putting y = -1 in equation (iv), we get

`5x - 2xx (-1) = 12`

=> 5x + 2 = 12

=> 5x = 12 - 2 = 10

`=> x = 10/5 = 2`

Putting x = 2 and y = -1 in z = -x + y + 4 we get

z = -2 + (-1) + 4

=-2 - 1 + 4

= -3 + 4

= 1

Hence, solution of the giving system of equation is x = 2, y = -1, z = 1