Question : For what values of $\displaystyle \lambda$ and $\displaystyle \mu$ , the simultaneous equations

x + y + z = 6

x + 2y + 3z = 10

x + 2y + $\displaystyle \lambda$z = $\displaystyle \mu$

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- Oct 25th 2009, 02:32 AMzorroFind the value of 2 terms in the simultaneous equation
Question : For what values of $\displaystyle \lambda$ and $\displaystyle \mu$ , the simultaneous equations

x + y + z = 6

x + 2y + 3z = 10

x + 2y + $\displaystyle \lambda$z = $\displaystyle \mu$ - Oct 25th 2009, 02:46 AMMush

Note that the first two terms of the last two equations are identical. We can rewrite these equations as:

$\displaystyle x + 2y = 10 - 3z $

$\displaystyle x + 2y = \mu - \lambda z $

So we can conclude that $\displaystyle 10 - 3z = \mu -\lambda z $.

You can solve for lambda and mu by comparing coefficients of $\displaystyle z^1 \, \, \text{and} z^0 $ on each side of this equation. - Oct 31st 2009, 09:56 PMzorroPlease elaborate on ur solution
- Oct 31st 2009, 10:29 PMharbottle
$\displaystyle z^1=z;\quad z^0=1$

- Nov 1st 2009, 12:44 AMzorroI am still unable to understand
- Nov 1st 2009, 12:50 AMmr fantastic
Forget the $\displaystyle z^0$$\displaystyle z^1$ business.

If $\displaystyle 10 - 3z = \mu -\lambda z$ for all values of z then it should be quite plain that you need to equate the constant term on each side of the equation and you need to equate the coefficient of z on each side of the equation. - Nov 1st 2009, 01:09 AMzorro
- Nov 1st 2009, 01:40 AMmr fantastic
I will not.

Are you honestly saying that you do not know what the constant term is in $\displaystyle 10 - 3z$ and $\displaystyle \mu - \lambda z$? If that's the case then sorry but there's nothing educational to be gained by someone writing a solution for you to simply copy.

The cold hard fact (based on this thead and others) is that you need to go back and thoroughly revise basic material (such as polynomials) because the questions you are asking (and no doubt other questions you have yet to meet or ask) assume you are competent with that material.

Then read threads like this one again. - Nov 1st 2009, 01:46 AMzorro
- Nov 1st 2009, 01:50 AMmr fantastic
- Nov 1st 2009, 02:08 AMzorro
- Nov 1st 2009, 02:14 AMmr fantastic
- Nov 1st 2009, 02:40 AMzorro
- Nov 1st 2009, 03:19 AMRaoh
(Happy)yes,it is correct ,$\displaystyle \lambda =3,\mu =10$.

- Nov 1st 2009, 03:40 AMmr fantastic
These are the values for the system to have infinite solutions. But having reviewed the original post, I find that that the OP hasn't said what's meant to be happening with the system ....

This question is incomplete. What is meant to happen with these equations? Do you want:

Infinite solutions? Unique solution? No solution?