# System of equations ?

Could you please explain or show me how to solve this system of equations problem

4x-2y+z=5

2x+y-2z=4

x+3y-2z=6

### 11 Answers

- la consoleLv 71 month ago
(1) : 4x - 2y + z = 5

(2) : 2x + y - 2z = 4

(3) : x + 3y - 2z = 6

You calculate (2) - [2 * (3)] and you obtain the equation (4)

(4) : (2x + y - 2z) - 2.(x + 3y - 2z) = 4 - (2 * 6)

(4) : 2x + y - 2z - 2x - 6y + 4z = 4 - 12

(4) : - 5y + 2z = - 8

(4) : 2z = 5y - 8

(4) : z = (5y - 8)/2

You calculate (1) - [4 * (3)]

(4x - 2y + z) - 4.(x + 3y - 2z) = 5 - (4 * 6)

4x - 2y + z - 4x - 12y + 8z = 5 - 24

- 14y + 9z = - 19

9z = 14y - 19

z = (14y - 19)/9 → recall (4): z = (5y - 8)/2

(14y - 19)/9 = (5y - 8)/2

2.(14y - 19) = 9.(5y - 8)

28y - 38 = 45y - 72

28y - 45y = - 72 + 38

- 17y = - 34

→ y = 2

Recall (4): z = (5y - 8)/2

z = (10 - 8)/2

→ z = 1

Recall (3) : x + 3y - 2z = 6

x = 6 - 3y + 2z

x = 6 - 6 + 2

→ x = 2

- PinkgreenLv 71 month ago
4x-2y+z=5------(1)

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

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

The answer to this problem is

x=2

y=2

z=1

Method:

(i) (3)-(2), you can obtain a new equation

in x, y only by eliminating z, call it (4).

(ii) 2*(1)+(2) or 2*(1)+(3), You obtain another

new equation in x, y only, call it (5).

(iii) from (4) & (5), you can solve for the value

of x, & y. Then use any of the (1), (2) & (3),

you can find the solution of z with the help

of the solutions of x & y.

- formengLv 61 month ago
First rearrange the equations listing only the coefficients and according to the least leading coefficient, i.e,

1+3-2=6

2 +1 -2 =4

4-2+1=5

Now use the Gauss-Jordan method to reduce the system to Reduced Row Echelon form;

It starts out like this

1+3-2=6

0 -5 2 =-8 ---> -2R1 +R2 Replaces R2

0-14+9=-19-->-4R1 +R3 replaces R3

Continue in this manner until you get the following rref form.

1 0 0 2

0 1 0 2

0 0 1 1

Which means x=2, y=2, and z = 1. If you don't understand what I'm doing look up Gauss-Jordan method in any algebra book.

If you have a TI-graphing calculator you can check your answer by entering the coefficients as a matrix, going to Matrix, Math and seledting rref( Entering the matrix and press Enter

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- PhilipLv 61 month ago
4x-2y+z=5...(1).;

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

x+3y-2z=6...(3).;

Step 1. We pick any 2 eqns and eliminate 1 variable.

[2(2)-(1)]---> 4y-5z =3...(4).

Step 2. We pick another 2 eqns and eliminate the same variable as in step 1.

[2(3)-(2)]---> 5y-2z = 8...(5).

Now we have 2 eqns in 2 unknowns. Using (4) & (5) we eliminate one variable.

[4(5)-5(4)]---> -8z+25z = 32-15, ie., 17z = 17, ie., z = 1. Then (5)---> y = 2 and

(1)---> 4x = 5+2y-z = 5+4-1 = 8, ie., x = 2. Then (x,y,z) = (2,2,1).

Note that in multiplying and subtracting eqns we concentrate term by term which

helps us avoid making mistakes.

- KrishnamurthyLv 71 month ago
4x - 2y + z = 5

2x + y - 2z = 4

x + 3y - 2z = 6

10x - 3y = 14

x - 2y = -2

20x - 6y = 28

3x - 6y = -6

17x = 34

x = 2

y = 2

z = 1

- RRLv 71 month ago
It's a simultaneous equation. You have to eliminate one unknown at a time and substitute. It's a bit long-winded but quite straight-forward. I will number the equations to help you follow:

(i) 4x-2y+z=5

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

(iii) x+3y-2z=6

___________________

Eliminate z

(ii) - (iii)

(iv) x - 2y = -2

------

(i) x 2

(iv) 8x-4y+2z =10

(iv) + (iii)

(vi) 9x-y = 16

_______________

You now have a simultaneous equation with 2 unkowns

(iv) x - 2y = -2

(vi) 9x - y = 16

____________________

Eliminate y and find x

(vi) x 2

(viii)18x - 2y = 32

(viii) - (iv)

17x = 34

x = 2

___________________

Substitute x = 2 into (iv)

2 - 2y = -2

2y = -4

y = -2

__________________

Substitute x = 2 and y = -2 into (i)

(4 x 2) - (2 x -2) + z = 5

8 - 4 + z = 5

z= 5 - 8 + 4

z = 1

_______________

ANS x = 2, y = -2, z = 1 Phew!

- Ian HLv 71 month ago
2z = x + 3y – 6 = 2x + y – 4 = 4y - 8x + 10

x – 2y = -2

10x – 3y = 14

6y = 3x + 6 = 20x – 28

17x = 34, x = 2, y = (x + 2)/2 = 2

z = 5 – 4x + 2y = 1

- AmyLv 71 month ago
Add or subtract two equations in the right ratio to eliminate one of the variables.

2(4x-2y+z) + (2x+y-2z) = 2(5) + 4

10x - 3y = 14

Add or subtract a different two equations in the right ratio to eliminate the same variable.

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

x - 2y = -2

Add or subtract your resulting equations in the right ratio to eliminate one of the variables.

(10x - 3y) - 10(x - 2y) = 14 - 10(-2)

17y = 34

This gives you the value of the one remaining variable.

y = 2

Plug that into one of your two-variable equations to solve for the other.

x - 2(2) = -2

x = 2

And finally plug both values into one of the original equations to solve for the last variable.

4(2) - 2(2) + z = 5

z = 1

Verify your solution by plugging all the values into all the original equations.

4(2) - 2(2) + 1 = 5

2(2) + 2 - 2(1) = 4

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