Hi to everyone,

I'm new here and this my 1st post. can anyone help me out?

Here's the problem:

If w^2 +x^2+y^2+z^2 = 4 and w(x+y+z) + x(y+z) + yz = 16, then solve for

w + x + y + z = _________.:confused:

Thanks and glad to be a member.:)

hansu

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- Nov 27th 2007, 08:07 PMhansuAlgebra 4 unknowns
Hi to everyone,

I'm new here and this my 1st post. can anyone help me out?

Here's the problem:

If w^2 +x^2+y^2+z^2 = 4 and w(x+y+z) + x(y+z) + yz = 16, then solve for

w + x + y + z = _________.:confused:

Thanks and glad to be a member.:)

hansu - Nov 27th 2007, 09:47 PMTruthbetold
My first instinct was to say all variables are 1 because that fits the first equation. But it doedsn't work.

This problem is actually easy to do, though hard to see.

The second equation is unnecessary and just there to throw you off.

Look at closely.

Is there anything you can do to simplify this equation?

Moving things around won't help, it will just complicate things.

You have to do something to every term in that equation.

Still don't know?

Find the square root of every term and there''s your answer. - Nov 28th 2007, 05:45 PMhansuAlgebra 4 unknowns reply
Hi there Truthbetold,

Thanks for the urgent reply to my problem. But I'm still at a lost. Can you expand on your solution to the problem?

Thanks again.:) - Nov 28th 2007, 06:34 PMTruthbetold
..original equation

.....square root

You just got to look at the equation and think about what you can do.

You might be able to do a very complex substitution and find the value of each variable (what each letter equals), but that is unnecessary.

It just wants you to find w + x +y + z = ?.

Take the square root of both sides since whatever you do to one side you to do the other, and you're done.

The second equation is unnecessary and is just there to complicate matters. - Nov 28th 2007, 08:46 PMDivideBy0
You can't do that!