Get the intersection of $f(x)$ and $f(y)$

Question

I'm a bit of a maths noob but I have the problem that I have two functions of the following form:

$$f(x) = 0.1 x + 3 \quad \text{ and } \quad f(y) = 0.3 y + 400 $$

meaning that I have a graph where I have a "normal" line in my plot and a vertical line which is dependent on $y$ and not $x$.

I am now trying to get the intersection of these two functions but I do not really know how to do this. I am helpful for any answers! Thank you. Here is an example. How can I get the intersection of the blue line with the orange or the green one?

Answer 1

If you are looking for the intersection of the lines

$y=0.1x+3$

and

$x=0.3y+400$

then you can just substitute $0.1x+3$ for $y$ in the second equation to get

$x=0.3(0.1x+3) +400 = 0.03x + 400.9$

and then solve this equation to find the value of $x$.

Answer 2

If we have a line written as a function of $x$, such as $y=ax+b$, then we can rewrite it as a function of $y$ by "solving for x", that is $x=\frac{1}{a}(y-b)$, and similarly we can go the other way $$x = ay+b \Rightarrow y = \frac{1}{a}(x-b) \quad \quad (\text{provided that }a\neq0)$$ So in order to solve your problem, we can convert $g(y)=0.3y+400$ into a function of $x$ using the method above $$x = \frac{3}{10}y+400 \Rightarrow y = \frac{10}{3}(x-400)=\frac{10}3 x - \frac{4000}{3}$$ Now we can find the $x$-coordinate by solving $$\frac{x}{10} + 3 = \frac{10}3 x - \frac{4000}{3}$$