Thread: 3 Simultaneous equations help please.

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?

Thanks for any help you can offer .

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?

Thanks for any help you can offer .

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

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

this is what you should do.

You have the equations:




Subtract both equations. That means subtract like terms.

That means you'll have:



Divide both sides by 2:

Look here:



k must be one.

Okey,
I think i just managed to confuse myself which was why.
Yea that sounds right, thanks jon and bobak