# Solve the system of linear equations using the elimination or substitution method 3x + 2y - z= 1 (i) 2x - 2y + 4z= -2 (ii) -x + 1/2y - z = 0 (iii)Show complete solution and explain the answer.

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### 1 Answer

3x+2 -z = 1...............(i)

2x-2y +4z =-2 ...........(ii)

-x+(1/2)y -z = 0 .............(iii)

First we will add (i) and (ii)

==> 5x +3z = -1 ..................(IV)

Now we will multiply (iii) by 4and add tp (ii)

=> 2*(iii)==> -4x +2y -4z = 0

==> (ii)==> 2x-2y +4z=-2

==> -2x = -2

==> **x= 1**

Now we will find the value of z using the equation (IV)

==> 5x+3z = -1

==> 5(1) + 3z = -1

==> 3z = -6 ==> **z = -2**

Now we will substitute in (iii) to find y.

==> -x + (1/2)y -z == 0

==> -1 + (1/2)y +2 = 0

==> 1+(1/2)y= 0

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

==> **y= -2**

**Then, the answer is : (x,y,z)= (1,-2,-2) **