## Direct Variation

 1.1 Direct Variation

 Definition of direct variation Direct variation explains the relationship between two variables, such that when variable $$y$$ increases, then variable $$x$$ also increases at the same rate and vice versa. This relation can be written as $$y$$ varies directly as $$x$$ . In general, for a direct variation, $$y$$ varies directly as $$x^n$$ can be written as \begin{aligned}x\propto x^n\end{aligned}\hspace{1mm}\text{(variation relation)} or \begin{aligned} x=kx^n \end{aligned} \hspace{1mm} \text{(equation relation)} where \begin{aligned} n=1,2,3,\frac{1}{2},\frac{1}{3} \end{aligned} and $$k$$ is a constant.
 Example 1 Given $$m=12$$ when $$n=3$$. Express $$m$$ in terms of $$n$$ if a) $$m$$ varies directly as $$n$$. b) $$m$$ varies directly as $$n^3$$. Solution: a) $$n\implies m = kn \dots (1)$$. Substitute $$m=12$$ and $$n=3$$ into $$(1)$$ $$12=k(3)\implies k=\dfrac{12}{3}=4$$. $$\therefore m=4n$$. b) $$m\propto n^3\implies m = ln^3 \dots (2).$$ Substitute $$m=12$$ and $$n=3$$ into $$(2)$$: $$12=l(3)^3\implies l=\dfrac{12}{27}=\dfrac{4}{9}$$ $$\therefore m=\dfrac{4}{9}n^3.$$

## Direct Variation

 1.1 Direct Variation

 Definition of direct variation Direct variation explains the relationship between two variables, such that when variable $$y$$ increases, then variable $$x$$ also increases at the same rate and vice versa. This relation can be written as $$y$$ varies directly as $$x$$ . In general, for a direct variation, $$y$$ varies directly as $$x^n$$ can be written as \begin{aligned}x\propto x^n\end{aligned}\hspace{1mm}\text{(variation relation)} or \begin{aligned} x=kx^n \end{aligned} \hspace{1mm} \text{(equation relation)} where \begin{aligned} n=1,2,3,\frac{1}{2},\frac{1}{3} \end{aligned} and $$k$$ is a constant.
 Example 1 Given $$m=12$$ when $$n=3$$. Express $$m$$ in terms of $$n$$ if a) $$m$$ varies directly as $$n$$. b) $$m$$ varies directly as $$n^3$$. Solution: a) $$n\implies m = kn \dots (1)$$. Substitute $$m=12$$ and $$n=3$$ into $$(1)$$ $$12=k(3)\implies k=\dfrac{12}{3}=4$$. $$\therefore m=4n$$. b) $$m\propto n^3\implies m = ln^3 \dots (2).$$ Substitute $$m=12$$ and $$n=3$$ into $$(2)$$: $$12=l(3)^3\implies l=\dfrac{12}{27}=\dfrac{4}{9}$$ $$\therefore m=\dfrac{4}{9}n^3.$$