# solving linear systems in three variables?

use elimination to solve each system of equations, i dont know how to solve for three variables, can you guys help me out? these are the two systems i need help with. can you please show the steps too so i get an understanding of how to solve more of these, thank you for the help

1.

8x+3y-6z=4

x-2y-z=2

4x+y-2z=-4

2.

2x-y+3z= 7

5x-4y-2z= 3

3x+3y+2z= -8

### 2 Answers

- PramodLv 78 years agoBest Answer
Q1.

8 x + 3 y - 6 z = 4 .............. (1)

1 x - 2 y - 1 z = 2 .............. (2)

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

Understanding steps : Step (i) Consider eqns (1) and (2). ASSUME all the numerals and z as constants. Re-write these eqns. as follows

8 x + 3 y = 6 z + 4 .............. ( modified 1)

1 x - 2 y = - 1 z + 2 .............. ( modified 2)

Step (ii) use elimination method to find value of x and y in terms of z . We get these values as follows ( I am not showing the intermediate steps )

.............(3) (z+2) - [ - 2 ( 6z+4) ]. . . . 15 z + 14

=> x = ----------------------------- ------ . = ------------ ...................... (4)

. . . . . . . . . . . (3+16). . . . . . . . . . . . . .19

Substitute this value of x in eqn (2). We get y = - (12+2z)/19 .............. (5)

Step (iii) Follow the same method and again find value of x and y considering eqns (2) and (3)

We get -----

=> x = (5 z - 6) / 9 and y = - (2z + 12) / 9 ................... (6)

From eqns (4) and (6) we see that ---

(15z+14) / 19 = x = (5 z - 6) / 9

-------------------------------------------

=> z = -- 6 ......... Answer

--------------------------------------------

Now if we substitute value of z ( ie -- 6 ) in the given eqns. we get --

First eqn => 8 x + 3 y = -- 32

Sec. eqn => . . x - 2 y = -- 4

Third eqn => 4x + y = -- 16

We can now find x and y easily by taking any two eqns ( say eqn1 and 2 ) and solving for x and y

by elimination method.

Answer x = -- 4 , y = 0 and z = -- 6 ........ Answer Answer

Q2. First step ---

2 x - y + 3 z = 7 ................ (1)

5 x - 4 y - 2 z = 3 ....................(2)

3 x + 3 y + 2 z = - 8 ...............(3)

Multiply eqn (1) by 2 and eqn (2) by 3 ---

4 x - 2 y + 6 z = 14 ........modified (1)

15 x - 12 y - 6 z = 9 ..............modified.(2)

----------------------------------------- Add the two eqns

19 x - 14 y = 23 ........... (4)

Second step : Considering eqns (2) and (3)

5 x - 4 y - 2 z = 3

3 x + 3 y + 2 z = - 8

------------------------------------- Add the two eqns

8 x - y = - 5 ............... (5)

Final step: Subtract 14 times the eqn (5) from eqn (4)

..19 x - 14 y = 23 ........... (4)

±112 x ∓ 14 y = ∓ 70 ............... (5)

-------------------------------------------------------------------- Subtract ( dont forget to change sign )

-- 93 x + 0 (y) = 93

=> x = -- 1 ............... Answer

Again multiply eqn (4) by 8 and eqn (5) by 19

..152 x - 112 y = 184 ........... (4)

±152 x ∓ 19 y = ∓ 95 ............... (5)

---------------------------------------- Subtract

...... . . . . . -- 93 y = 279

=> y = - (279/93) = - 3 .......... Answer

----------------- -------------------------------------

Ans: x = -- 1 , y = - 3 and z = + 2

----------------- -------------------------------------

>>>>>>>>>>>>>>>>>>

.

- Seamus OLv 78 years ago
They want you to use the elimination method:

1. 8x + 3y - 6z = 4 ... [eqn 1]

x - 2y - z = 2 ... [eqn 2]

4x + y - 2z = -4 ... [eqn 3]

eliminating z to get an equation in terms of x and y:

[eqn 2] * 6 → 6x - 12y - 6z = 12 ... [eqn 4]

[eqn 1] - [eqn 4] → 2x + 15y = -8 ......................... [eqn 5] ... [be careful with signs]

eliminating z to get a second equation in x and y:

[eqn 2] * 2 → 2x - 4y - 2z = 4 ... [eqn 6]

[eqn 3] - [eqn 6] → 2x + 5y = -8 ...........................[eqn 7]

using the two equations in x and y to eliminate x:

[eqn 5] - [eqn 7] → 10y = 0

y = 0

solving for x:

subs y = 0 into [eqn 7] → 2x = -8

x = -4

solving for z:

subs x = -4, y = 0 into [eqn 2] → -4 - 0 - z = 2

z = -6

so the solution is (-4, 0, -6)

Second one is done the same way ... except I'd start by eliminating y:

2x - y + 3z = 7 ... [eqn 1]

5x - 4y - 2z = 3 ... [eqn 2]

3x + 3y + 2z = -8 ... [eqn 3]

[eqn 1] * 4 → 8x - 4y + 12z = 28 ... [eqn 4]

[eqn 2] - [eqn 4] → -3x -14z = -25 ............................. [eqn 5]

[eqn 1] * 3 → 6x - 3y + 9z = 21 ... [eqn 6]

[eqn 3] + [eqn 6] → 9x + 11z = 13 ............................. [eqn 7] ... [NOTE: The equations had to be added to eliminate the y terms]

[eqn 5] * 3 → -9x - 42z = -75 ... [eqn 8]

[eqn 7] + [eqn 8] → -31z = -62

z = 2

subs z = 2 into [eqn 7] → 9x + 22 = 13

9x = -9

x = -1

subs x = -1, z = 2 into [eqn 1] → -2 - y + 6 = 7

y = -3

so the solution is (-1, -3, 2)