# 3 Simultaneous equations help please.

• May 31st 2008, 12:38 PM
AshleyT
Find the value of k which the simultaneous equations have solutions:
2y - z = k
2x + 3y - z = 2
2x - y + z = 0

The answer is k = 1 and its done by eliminating x from equations 2 and 3 and getting 4y - 2z = 2....Then comparing that with the first equation.

BUT... I don't get how you can do that? Because if you eliminate x from one equation, you can't then use that equation to eliminate x in another equation...because it won't have an x?

• May 31st 2008, 01:32 PM
bobak
Quote:

Originally Posted by AshleyT
Find the value of k which the simultaneous equations have solutions:
2y - z = k
2x + 3y - z = 2
2x - y + z = 0

The answer is k = 1 and its done by eliminating x from equations 2 and 3 and getting 4y - 2z = 2....Then comparing that with the first equation.

BUT... I don't get how you can do that? Because if you eliminate x from one equation, you can't then use that equation to eliminate x in another equation...because it won't have an x?

Your eliminating x from two equations to get another equation without x and comparing it to the first, I don't really understand where your confusion is coming form.

Bobak
• May 31st 2008, 01:54 PM
AshleyT
Quote:

Originally Posted by bobak
Your eliminating x from two equations to get another equation without x and comparing it to the first, I don't really understand where your confusion is coming form.

Bobak

Uh, i'm confused as to how you can eliminate x from the two equation.
Because if you eliminate it from one(giving a total of two equations without x) how would you eliminate it from the 3rd?

Thanks :)
• May 31st 2008, 02:07 PM
Jonboy
this is what you should do.

You have the equations:

\$\displaystyle 2x + 3y - z = 2\$
\$\displaystyle 2x - y + z = 0\$

Subtract both equations. That means subtract like terms.

That means you'll have: \$\displaystyle 4y -2z = 2\$

\$\displaystyle 4y -2z = 2\$

Divide both sides by 2: \$\displaystyle 2y - z = 1\$

Look here:
\$\displaystyle 2y - z = 1\$
\$\displaystyle 2y - z = k\$

k must be one.
• Jun 1st 2008, 06:31 AM
AshleyT
Okey,
I think i just managed to confuse myself which was why.
Yea that sounds right, thanks jon and bobak :)