Question

How do you solve the system of equations algebraically `-x-3y+z=54, 4x+2y-3z=-32, 2y+8z=78`?

Answer 1

Solution : `x =11 , y = -17 , z =14`

Explanation:

`-x-3y+z = 54 (1) ,4x +2y -3z = -32 (2)`

`2y+8z= 78(3) : y = (78-8z)/2= 39-4z` Putting `y = 39-4z`

In equation (1) and (2) we get `-x -3(39-4z)+ z =54` or

`-x +13z =171 (4) and 4x +2(39-4z) -3z = -32 or 4x-11z = -110 (5)`

`(4)*4: -4x +52z = 684 (6)` Adding (5) and (6) we get

`41 z = 574 :. z = 1 4 :. 2y + 8*14 =78 :. 2y = 78-112`

or`2y = -34 :. y = -17 ; -x - 3* -17 + 14 =54 or x = 51+14-54=11`

Solution : `x =11 , y = -17 , z =14` [Ans]