Answer:
See Explanation
Step-by-step explanation:
Required
Proportion problems
An example is:
y is directly proportional to x such that: y=4 when x = 2;
Derive the equation
For direct proportions, we have:
[tex]y\ \alpha\ x[/tex]
This gives:
[tex]y = kx[/tex]
Make k the subject
[tex]k = y/x[/tex]
So:
[tex]k = 4/2 =2[/tex]
So, the equation is:
[tex]y = kx[/tex]
[tex]y = 2x[/tex]
Assume the above question is for inverse proportion
The variation will be:
[tex]y\ \alpha\ \frac{1}{x}[/tex]
This gives:
[tex]y\ = \frac{k}{x}[/tex]
Make k the subject
[tex]k =x*y[/tex]
[tex]k =2* 4 = 8[/tex]
So, the equation is:
[tex]y\ = \frac{k}{x}[/tex]
[tex]y = \frac{8}{x}[/tex]
