# Simaltaneous Equations

• January 5th 2007, 01:42 AM
gramit108
Simaltaneous Equations
I need a little help... just can't figure this question out.

Quote:

For what values of a and b does the system

ax + bz = 2
ax + by + 4z = 4
ay + 2z = b

have:
• (i) no solution

• (ii) a unique solution

• (iii) a one parameter solution

• (iv) a two parameter solution

Using gaussian elimination I got (2-b)z = 2 - b.

Substituting a=0 gives z=1 and b=2 but nothing for x or y so at a guess that would be the 1 parameter solution.

Substituting b=0 gives ax=2, ay+4z=4 and ay+2z = 0. Rearranging and subsituting gives 2z=4 so z=2, and -ay=4. So -ay=2ax which means -y=2x

That is all I could think to do.
• January 5th 2007, 02:10 AM
earboth
Quote:

Originally Posted by gramit108
I need a little help...

Using gaussian elimination I got (2-b)z = 2 - b.

Substituting a=0 gives z=1 and b=2 but nothing for x or y so at a guess that would be the 1 parameter solution.

Substituting b=0 gives ax=2, ay+4z=4 and ay+2z = 0. Rearranging and subsituting gives 2z=4 so z=2, and -ay=4. So -ay=2ax which means -y=2x

That is all I could think to do.

Hello,

first: Are you certain that there isn't a typo in the first line of your system of equations?

I used determinants and got the solution:

$x=\frac{2}{a+b}\ \wedge \ y=\frac{2(a+2b)}{(a+b)(b-2a)}\ \wedge \ z=\frac{a(a+2b)}{(a+b)(b-2a)}$

Now you can determine the values of a and b so that there are solutions according to the items (i) to (iv) of your problem.

EB
• January 5th 2007, 04:48 AM
gramit108
yes, the first equation had a typo, has been edited. Thanks a lot this helps. Would it be possible for you to describe how you got the results?