Systems of equations.

• Feb 18th 2010, 07:42 AM
MathBlaster47
Systems of equations.
Here is a problem I don't quite know what to do with.

Solve the system of equations.
$3x+4y+z=7$
$2y+z=3$
$-5x+3y+8z=-31$

should I just solve for z by isolating it on one side of the equation like I would normally or is there a better method?
• Feb 18th 2010, 08:04 AM
Pulock2009
Quote:

Originally Posted by MathBlaster47
Here is a problem I don't quite know what to do with.

Solve the system of equations.
$3x+4y+z=7$
$2y+z=3$
$-5x+3y+8z=-31$

should I just solve for z by isolating it on one side of the equation like I would normally or is there a better method?

i would prefer the matrix method. X=A^(-1).B where X=(x,y,z),
A=(3,4,1)
(2,0,1)
(-5,3,8)
B=(7,3,-31) where A is a 3 by 3 matrix and X and B are 3 by 1 matrices.
• Feb 18th 2010, 08:33 AM
masters
Quote:

Originally Posted by MathBlaster47
Here is a problem I don't quite know what to do with.

Solve the system of equations.
$3x+4y+z=7$
$2y+z=3$
$-5x+3y+8z=-31$

should I just solve for z by isolating it on one side of the equation like I would normally or is there a better method?

Hi MathBlaster47,

There are several ways to solve a system like this, but your way seems to be a good start.

(1) $3x+4y+z=7$

(2) $2y+z=3$

(3) $-5x+3y+8z=-31$

Solve (2) for z: $z=3-2y$

Substitute this for y in (1) and (3).

(1) $3x+4y+3-2y=7$

(3) $-5x+3y+8(3-2y)=-31$

Solve these two equations for x and y by using substitution or elimination method. Then, substitute your value for y into (2) to find z.

(1) $3x+2y=4$

(3) $-5x-13y=-55$

Multiply (1) by 5. Multiply (2) by 3.

(1) $15x+10y=20$

(3) $-15x-39y=-165$

$-29y=-145$
$\boxed{y=5}$