Question

Can someone please thoroughly explain to me how to solve systems of equations with THREE variables?
Here is an example to make things easier:
2x + 4y + 3z = 6

3x + 5y + 6z =3

2x + 3y + 4z = 8

Answer 1

use two of the equations to reduce the number of variables to two

Answer 2

then use another two equations and do the same

Answer 3

make x the subject for all 3 equations.
equate (1) and (3) and (2) and (3) to remove the x
solve the two equations for y and z
thats the method in general

Answer 4

solve for the first two then for the second two:
first use elimination to get rid of one variable, ex:\[~~~~2x + 4y + 3z = 6\]\[\underline{-~2x + 3y + 4z =8}\]\[~~~~~0~+~y~-z~=-2\]solve for \(y\):\[y-z=-2~~~\implies~~~y=z-2\]use elimination again to get rid of the same variable:\[~~~~3~(2x+3y+4z=8)~~~~\implies~~~~~6x+~9y+12z=24\]\[\underline{-~2~(3x+5y+6z=3)}~~~\implies~~\underline{-~6x+10y+12z=9}\]\[~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~0~-~y~~+~0~~=15\]\(y=-15\), so solve for \(z\):\[y=z-2~~~~~~\implies~~~~~-15=z-2~~~~~\implies~~~~~z=-13\]now given that \(y=-15\) and \(z=-13\) we can solve for \(x\) using any equation...I'll choose the first one:\[2x+4y+3x=6~~~~\implies~~~~2x+4(-15)+3(-13)=6~~~~\implies~~2x-60-39=6\]solve for \(x\):\[2x-60-36=6~~~\implies~~~2x=105~~~\implies~~~x=52.5\]
so your solutions are:\[x=~~52.5\]\[y=-15\]\[z=-13\]

Note: this is only a sample way of HOW to do this type of a problem

Answer 5

Thank you both so much.