Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

2 Answers

up vote 1 down vote accepted

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 $z_0=z(0,0)$.

share|improve this answer
add comment

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.)

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.