# 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.
\$\displaystyle 3x+4y+z=7\$
\$\displaystyle 2y+z=3\$
\$\displaystyle -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.
\$\displaystyle 3x+4y+z=7\$
\$\displaystyle 2y+z=3\$
\$\displaystyle -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.
\$\displaystyle 3x+4y+z=7\$
\$\displaystyle 2y+z=3\$
\$\displaystyle -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) \$\displaystyle 3x+4y+z=7\$

(2) \$\displaystyle 2y+z=3\$

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

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

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

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

(3) \$\displaystyle -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) \$\displaystyle 3x+2y=4\$

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

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

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

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