How can I find the formula to calculate the solution of a system of linear equations in relation to a third variable?

Question

So let's say you had the equation $y=100x$ and $y=1000(x-D)$. How would you be able to find a formula to calculate the solution in relation to $D$? I'm looking for something like $x=5D-2$. Basically I want to find how the solution $x$ and $y$ would change when I change $D$.

Answer 1

The two equations $y= 100x$ and $y=1000(x-D) $ define two lines on the plane, that cross at exactly one point, since they are not parallel. This point has two coordinates $(x,y)$ and since they lie on each line, $100x =y = 1000(x-D) $ has to be satisfied. Solving this for x we obtain $$ x= (10/9) D$$ and as for y $$ y=( 1000/9) D. $$

Answer 2

Caution: my first solution relates to the particular way the parameter $D$ appears in the equations.

As the equations are linear in all variables, you can exploit this linearity.

• solve the system with $D=0$,

• solve the system with $D=1$,

and the general solution is the combination of the above.

• $y=100x$ and $y=1000x\to x,y=0$.

• $y=100x$ and $y=1000(x-1)\to x=\dfrac{10}9,y=\dfrac{1000}9$.

Then

$$x=0+\dfrac{10}9D,y=0+\dfrac{1000}9D.$$

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The normal way is to solve as if all coefficients were constants, but keep the parameter in the expressions.

We first show the solution for $D=9$, then for arbitrary $D$:

$$\begin{cases}y-100x=0,\\y-1000x=-9000\end{cases}$$

and by elimination (first minus second)

$$900x=9000$$

then

$$x=10$$

and by backsubstitution

$$y=100\cdot10.$$

Next $$\begin{cases}y-100x=0,\\y-1000x=-1000D\end{cases}$$

and by elimination (first minus second)

$$900x=1000D$$

then

$$x=\frac{10D}9$$

and by backsubstitution

$$y=100\frac{10D}9.$$