# Using Parameters

• July 9th 2009, 09:39 PM
Stroodle
Using Parameters
A question from my text is:

The simultaneous equations $x+2y+3z=13,\;-x-3y+2z=2$ and $-x-4y+7z=17$ have infinitely many solutions. Describe these solutions through the use of a parameter.

I understand how to do this. I just don't understand the purpose of using a parameter?

• July 9th 2009, 09:47 PM
Chris L T521
Quote:

Originally Posted by Stroodle
A question from my text is:

The simultaneous equations $x+2y+3z=13,\;-x-3y+2z=2$ and $-x-4y+7z=17$ have infinitely many solutions. Describe these solutions through the use of a parameter.

I understand how to do this. I just don't understand the purpose of using a parameter?

The parameter (let's say $t$) is used to define $x$, $y$, and $z$ solutions. Since $t\in\mathbb{R}$, that implies that $t$ can take on any value (hence, the system has infinite solution). Its just a nice way of writing the solution, because you can't possibly list all the solutions (you can even call it a general solution for any $t$)!
• July 9th 2009, 11:35 PM
Stroodle
Actually. I don't quite understand how to do these. Can someone show working for:

Using a parameter find all the solutions for:

$2x-y+z=0$ and $y+2z=2$

Thanks
• July 10th 2009, 05:53 AM
mr fantastic
Quote:

Originally Posted by Stroodle
Actually. I don't quite understand how to do these. Can someone show working for:

Using a parameter find all the solutions for:

$2x-y+z=0$ and $y+2z=2$

Thanks

Let $x = t$ where t is any real number.

Then:

$-y + z = -2t$ .... (1)

$y + 2z = 2$ .... (2)

Now solve (1) and (2) simultaneously for y and z in terms of t.
• July 10th 2009, 06:57 AM
Soroban
Hello, Stroodle!

Quote:

Using a parameter find all the solutions for: . $\begin{array}{cc}2x-y+z\:=\:0 & [1] \\ y+2z\:=\:2& [2]\end{array}$
Here's one solution . . . [There are many others.]

Solve one of the equations for one of its variables.
. . Solve [2] for $y\!:\;\;{\color{blue}y \:=\:2-2z}$

Substitute into the other equation.
. . $2x - (2 - 2z) + z \:=\:0$
And solve for the third variable.
. . ${\color{blue}x \:=\:1 - \tfrac{3}{2}z}$

So we have: . $\begin{Bmatrix}x &=& 1 - \frac{3}{2}z \\ y &=& 2 - 2z \\ z &=& z \end{Bmatrix}$

On the right side, replace $z$ with a parameter $t.$

Therefore, we have: . $\boxed{\begin{array}{ccc} x &=& 1 - \frac{3}{2}t \\ y &=& 2 - 2t \\ z &=& t \end{array}}$

• July 10th 2009, 08:11 PM
Stroodle
Awesome. Thanks for your help guys!!