$\displaystyle 4m+ 3n= 400$

Express all possible positive integer values of m, n in terms of a third variable, k. (Use modular arithmetic)

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- Sep 14th 2007, 04:29 AMDivideBy0possible values for simple equation
$\displaystyle 4m+ 3n= 400$

Express all possible positive integer values of m, n in terms of a third variable, k. (Use modular arithmetic) - Sep 14th 2007, 02:51 PMSoroban
Hello, DivideBy0!

We can solve this with "normal" algebra . . .

Quote:

$\displaystyle 4m+ 3n\:=\: 400$

Express all possible positive integer values of $\displaystyle m,\, n$ in terms of a third variable, $\displaystyle k.$

Solve for $\displaystyle m\!:\;\;m \:=\;100-\frac{3}{4}n$ .**[1]**

Since $\displaystyle m$ is an integer, $\displaystyle n$ must be a multiple of 4.

. . That is: .$\displaystyle n \:=\:4k$ for some integer $\displaystyle k.$

Substitue into [1]: .$\displaystyle m \:=\:100-\frac{3}{4}(4k)\:=\:100-3k$

And we have parametric equations for all solutions:

. . $\displaystyle \begin{array}{ccc}m & = & 100-3k \\ n & = & 4k\end{array}$ . for any integer $\displaystyle k.$

- Sep 15th 2007, 03:31 AMDivideBy0
Thanks, I have two question though

How would you express them if you could only have them as natural numbers, would you have 1 =< k =< 33?

and What are parametric equations? thanks again - Sep 15th 2007, 05:32 AMCaptainBlack
As

$\displaystyle

4m+3n=400

$

$\displaystyle 4|n$, so $\displaystyle n=4k$ for some $\displaystyle k \in \bold{Z}$

Then we have:

$\displaystyle

m+k=100

$,

or:

$\displaystyle

m=100-k

$.

Now if you choose any $\displaystyle k \in \bold{Z}$ then:

$\displaystyle m=100-k,\ n=4k$

is a solution to the original equation.

RonL - Sep 15th 2007, 07:41 AMSoroban
Hello, DivideBy0!

Quote:

How would you express them if you could only have them as natural numbers?

Would you have: .$\displaystyle 1 \:\leq\: k \:\leq \:33$? . . . . Yes!

What are parametric equations?

A parameter is an "extra variable".

Instead of having $\displaystyle y$ as a function of $\displaystyle x\!:\;y \:=\:f(x)$,

. . we can have: .$\displaystyle \begin{array}{cc}x\text{ as a function of }t\!: & x \:=\:g(t) \\

y\text{ as a function of }t\!: & y \:=\:h(t)\end{array}$

This opens the door to an entire*universe*of fascinating functions and graphs

. . . . . curves with loops, that intersect itself, etc.