how do you solve systems of equations with 3 variables?

Question

Example:
X+Y=6
X+Z=-2
Y+Z=2

Answer 1

One of the methods is:
elimination of x, then elimination of y,
solve z, find y en find x. (It may be another order.)

Elimination of X
X = 6 - Y and X = -2 - Z
So 6 - Y = -2 - Z

Now we have
6 - Y = -2 - Z or Y = Z + 8 and Y+Z=2

Elimination of Y gives (Z + 8) + Z = 2

Solve Z
2Z = -6 or Z = -3

Find Y
Y = Z + 8 = -3 + 8 = 5

Find X
X+Y=6. So X + 5 =6 and X=1

Conclusion: (X,Y,Z) = (1 , 5 , -3)
The check you can do yourself.

Answer 2

U can find the value by using Matrices & detrminants
By using Cramar's law (determinants):
system of equation must have the 3 variables in each equation
(if there is a variable doesn't exist in an equation, we put it & multiply it by zero)

therfore the system of equation:
X + Y + (0*Z) = 6
X + Z + (0*Y) = -2
Y + Z + (0*X) = 2

then apply Cramar's law of determinants

i hope u got ur point from my explanation

Answer 3

For your example, Let the above equations equal "1", "2", and "3", as follows:

1. X+Y = 6
2. X+Z = -2
3. Y+Z = 2

Add equation 1 to equation 2 and subtract by equation 3. You now have 2X = 2, so X = 1. From this, you can find ouy that Z=-3, and Y=5.

Answer 4

Two equations at a time:

X+Z=-2
Y+Z=2

Z = -2 - X

Y + (-2 - X) = 2
Y - X = 4

Use this result with the other given equation:

Y - X = 4
X + Y = 6

Add them together:

2Y - X + X = 10
2Y = 10

Y = 5

X + (5)=6, X = 1
(1) + Z=-2, Z = -3

Answer 5

you have 3 variables, and 3 equations. Try expressing 'x' in terms of " y and z" or 'y' in terms of 'x and z' or 'z' in terms of 'x and y'.
For example:

since equation 1 is:
x + y = 6, therefore x = 6 - y.

equation 2 can now be written as: (6 - y)+ Z= -2

which is: - y+z = -2-6 = -8

hence equation 2 is now : z-y = -8
equation 3 is already in terms of 'y and z'.
z + y = 2
z - y = -8
you can now solve them simultaneously.

Answer 6

first solve x in equation1 ( x=6-y ), substitute that in for x of eq2 ( 6-y+z=-2 ) and solve for y ( y=8+z ), sub that into eq3 ( 8+z+z=2 ) and solve for z ( z=-3 ) sub back into the other eqs for z=-3, y=5, and x=1.

Answer 7

3x-2y+z=-4 ----------------- (a million) 5x+2y-2z=-9 ----------------- (2) -2x+y+3z = 20 ----------------- (3) upload equation (a million) with (2): (this might cancel the y variable) 3x+5x+z-2z = -4- 9 8x-z = -13 z = 8x+13 ---------- (4) replace (4) in (3): -2x+y+3(8x+13) = 20 -2x+y+24x+39 = 20 y+22x = -19 y = -22x-19 ---------- (5) replace (4) and (5) in (a million): 3x-2(-22x-19)+8x+13 = -4 3x+44x+38+8x+13 = -4 55x = -fifty 5 x = -a million we've x = -a million, hence y = 3 and z = 5 (via substituting x = -a million in (4) and (5))

Answer 8

also solve and sub works

Solve 3. for Z
Z=2-Y

Substitute Z into 2.
X+2-Y=-2
and solve for X
X=-4+Y

Sub X into 1.
-4+Y+Y=6
2Y=10

Y=5
Find X
X=1
Find Z
Z=-3

Some systems work better with different methods

Answer 9

u should try to figyre it u r self
its a simple thing looking at last equation which is terms of y n z make another eqn in trms of y n z from 1 & 2
and then solve it

Answer 10

make a triangle matrix of it
:..aX + bY + cZ = d
:..........eY + fZ = g
:..................hZ = i
next solve Z = i/h;
then Y, then X

1X + 1Y +0Z = 6
........1Y + 1Z = 2
1X + 0Y +1Z = -2

subtract from last row the first row :
1X + 1Y +0Z = 6
........1Y + 1Z = 2
........-Y +1Z = -8

add second row to last row
1X + 1Y +0Z = 6
........1Y + 1Z = 2
........ +2Z = -6

now solve Z = -6/2, then Y = 2 - Z = 2+3 = 5 etc etc