# Solve this system of equations.  {2x-2y+z=3 {5y-z=-31 {x+3y+2z=-21 (All in one bracket)

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Solve the system `{[2x-2y+z=3],[5y-z=-31],[x+3y+2z=-21]}`

We can use linear combinations to solve. We choose to eliminate z:

Add the first equation to the second equation: 2x+3y=-28

Add the third equation to twice the second equation: x+13y=-83

Solve the new system:

2x+3y=-28
x+13y=-83    Subtract twice the second equation from the first:

-23y=138 ==>y=-6

y=-6==>2x-18=-28 ==> 2x=-10 ==>x=-5

x=-5,y=-6==>-30-z=-31==>-z=-1==>z=1

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The solution is x=-5,y=-6,z=1

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Given:

2x-2y+z=3 -----(1),  5y-z= -31 -----(2), x+3y+2z= -21 ----(3)

multiplying Eq.(3) by 2 : (x+3y+2z)*2= -21*2

=> 2x+6y+4z= -42 ----(4)

Subtracting eqn 1 from eqn. 4  i.e.  eqn.[(4)- (1)]

2x+6y+4z)-(2x-2y+z) = -42-3

=> 2x+6y+4z-2x+2y-z= -45

=> 8y+3z= -45 -------(5)

Now we multiply Eq.(2) by 3 i.e  Eq.(2)*3

(5y-z)*3=-31*3

=> 15y- 3z= -93   ------(6)

Adding Eq.(5) and Eq. (6)  We get

(8y+3z)+(15y-3z)= -45+(-93)

=> 8y+3z+15y-3z= -138

=> 23y= -108

=>  y = (-138)/23 = -6

=> y = -6

Now substituting value of y= -6 in equation (5) :- 8y+3z= -45

8(-6)+3z = -45

=> -48+3z= -45

=> 3z= -45 + 48 = 3

=> 3z= 3

z= 1

Again substituting value of x = -6 and z =1 in equation (1)

2x-2y+z=3

=> 2x-2(-6)+1 = 3

=> 2x +12+1= 3

=> 2x+ 13=3

=>2x = 3 -13

=> 2x = -10

x= (-10)/2 = -5

x = -5

Hence,   x= -5, y= -6, z=1  ---> Answer