# Proportionality

just need help with a quick proportionality question I may have lost my bearings on.

I am trying to write a single expression for $z$ in terms of $x$ and $y$ that corrsponds to this statement.

A unit increase in $x$ gives an increase $k$ in $z$ and a unit increase in $y$ gives an increase $l$ in $z$ where $k$ and $l$ are independent of $x$ and $y$.

I think the answer is $x+1=z(k+1)$ and $y+1=l(k+1)$ but I don't understand the statemnt '$k$ in $z$' and '$l$ in $z$' so am not sure if I'm correct.

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Try to solve the simpler problem "A unit increase in $x$ gives an increase $7$ in $y$." By making things concrete, you will see what's going on. – André Nicolas Oct 4 '12 at 22:11

I would read this as $z$ being a function of $x$ and $y$ where $$z(x+1,y)=z(x,y)+k$$ $$z(x,y+1)=z(x,y)+l$$ and more generally $$z(x,y)=z_0 +kx+ly$$ where $z_0=z(0,0)$.

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You want $z$ to be some function of both $x$ and $y$ that has the following two properties:

1. every time $x$ increases by $1$ while $y$ remains fixed, $z$ increases by $k$; and
2. every time $y$ increases by $1$ while $x$ remains fixed, $z$ increases by $\ell$.

Let’s ignore $y$ for a moment. Suppose that we let $z=kx$. Then when $x$ increases by $1$ unit to $x+1$, $z$ increases to $k(x+1)=kx+k$, so $z$ really does increase by $k$ units. This $z$ satisfies the first of the two conditions. Similarly, $z=\ell y$ satisfies the second condition. What happens if you set $z=kx+\ell y$?

(You can actually use any function of the form $z=kx+\ell y+c$, where $c$ is any constant.)

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