# Thread: Application of Linear Law to Non-linear Relations

1. ## Application of Linear Law to Non-linear Relations

Given x and y are related by the equation xy=px+q/x where p and q are constant.A straight line graph is obtained by plotting y against 1/(x^2),it passes through the points (0,0.5) and (4,-2.5).calculate the value of p and q.

2. ## hi

$xy = px +q/x$ /divide by x

then $y = p + q/(x^2)$ rearange terms

$y = q/x^2 +p$

we know that when x = 0 then y = 0.5

$0.5 = (q/o^2) +p$ that gives $p = 0.5$

Likewise plug in the values from (4,-2.5)

$y = q/x^2 + 0.5$ and there it is.

you get $-2.5 = q/4 +0.5$ and solve for q