$\displaystyle 4m+ 3n= 400$
Express all possible positive integer values of m, n in terms of a third variable, k. (Use modular arithmetic)
Hello, DivideBy0!
We can solve this with "normal" algebra . . .
$\displaystyle 4m+ 3n\:=\: 400$
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 $\displaystyle k.$
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
Hello, DivideBy0!
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.