Question

Solve the following systems of equations:

x − y + z = 4
x − 2y − 2z = 9
2x + y + 3z = 1

Answer 1

We have,

x − y + z = 4  .....(i)

x − 2y − 2z = 9 ....(ii)

2x + y + 3z = 1 .....(iii)

From equation (i), we get

z = 4 - x + y

=> z = -x + y + 4

Subtracting the value of z in equation (ii), we get

x - 2y - 2(-x + y + 4) = 9

`=> x - 2y + 2x - 2y - 8 = 8`

=> 3x - 4x = 9 + 8

=> 3x - 4y = 17 ......(iv)

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

2x + y + 3(-x + y + 4) = 1

=> 2x + y + 3x + 3y + 12 = 1

=> -x + 4y =1 - 12

=> -x + 4y = -11 .....(v)

Adding equations (iv) and (v), we get

3x - x - 4y + 4y = 17 - 11

=> 2x = 6

=> x = 6/2 = 3

Putting x = 3 in equation (iv), we get

`3 xx 2  - 4y = 17`

=> 9 - 4y = 17

=> -4y = 17 - 9

=> -4y = 8

`=> y = 8/(-4) = -2`

Putting x = 3 and y = -2 in z = -x + y + 4 we get

z = -3 -2 + 4

=> x = -5 + 4

=> z= -1

Hence, solution of the giving system of equation is x = 3, y = -2, z = -1