# Systems in 3 variables #2

• Dec 2nd 2009, 05:44 PM
kelsikels
Systems in 3 variables #2
Have to find x,y,and z.

1.)
x - y - 5z = -2
y + 3z = 9
x + y + z = 16

• Dec 2nd 2009, 06:49 PM
Soroban
Hello, kelsikels!

Quote:

1) Solve: .$\displaystyle \begin{array}{ccc} x - y - 5z &=& \text{-}2 \\ \qquad y + 3z &=& 9 \\ x + y + z &=& 16 \end{array}$

There is an infinite number of solutions:

. . $\displaystyle \begin{Bmatrix}x &=& 7 + 2t \\ y &=& 9 - 3t \\ z &=& t \end{Bmatrix}$

• Dec 3rd 2009, 02:13 AM
HallsofIvy
kelsikels, you should have at least tried yourself! For example, if you add the first and second equations, the "y" and "-y" cancel leaving you an equation in only x and z. If you add the first and third equations, again "y" and "-y", giving a second equation in only x and y. If those are "independent" (neither equation is just a multiple of the other) you can solve those two equations for x and z, then go back and solve for y. If they are not, as Soroban implies, you can solve for, say, x, in terms of z and then solve for y in terms of z. That gives a different solution for each value of z.