3 variable equation

Question

30x+10y+20z=130

15x+20y+10z=95

10x+25y+20z=140

Answer 1

first reduce the coefficients where possible by dividing the whole equation by the greatest common factor

 

3x + y + 2z = 13

3x + 4y + 2z = 19

2x + 5y + 4z = 28

 

Now you pick two pairs of equations and eliminate the SAME variable in each pair.

 

Here are good choices for pairs.

 

3x + 4y + 2z = 19
2x + 5y + 4z = 28

(eliminate z by multiplying the first equation by 2 and then subtracting the equations)

 

and

 

3x + y + 2z = 13
2x + 5y + 4z = 28

(eliminate z by multiplying the first equation by 2 and then subtracting the equations)

 

Then take your two new equation that have only x and y and solve for x and y

 

Finally use the values for x and y in one of the original equations to find z

 

Answer 2

I would simply the equations first by dividing all coefficients by 5:

6x+2y+4z=26

3x+4y+2z=19

2x+5y+4z=28

 

Next, I would eliminate z by subtracting the third equation from the first: 4x-3y=-2

 

Solve for x: x=(3y-2)/4

 

Now multiple the 3nd equation in the original 3 by 3: 6x+15y+12z=84

 

Subtract the 1st equation from the above: 13y+8z=58

 

Solve for z: z=(58-13y)/8

 

Now plug the equations for x and z in terms of y into the 2nd of the original equation and solve for y:

3*[(3y-2)/4]+4y+2*[(58-13y)/8]=19

24y=48

y=2

 

Now solve for x and z:

x=(3y-2)/4=(3*2-2)/4=(6-2)/4=4/4=1

z=(58-13y)/8=(58-13*2)/8=(58-26)/8=32/8=4

 

Now check your answers by plugging in x,y,z into the 3 equations:

 

30(1)+10(2)+20(4)=30+20+80=130

15(1)+20(2)+10(4)=15+40+40=95

10(1)+25(2)+20(4)=10+50+80=140

 

 

Answer 3

30x+10y+20z=130

3x+y+2z=13

 

15x+20y+10z=95

3x+4y+2z=19

 

10x+25y+20z=140

2x+5y+4z=28

 

3x+y+2z=13

3x+4y+2z=19

2x+5y+4z=28

 

subtract first new equation from second new equation

3x+4y+2z=19

3x+y+2z=13

3y=6

y=2

 

substitute y=2 into second and third new equations

3x+8+2z=19

2x+10+4z=28

3x+2z=11

2x+4z=18

multiply first equation here by 2

6x+4z=22

2x+4z=18

subtract

4x=4

x=1

substitute x=1 into first equation before you multiplied it by 2

3x+2z=11

3+2z=11

2z=8

z=4

(1,2,4) is the solution