Algebra Question

Your question is kind of vague, but I will try to clarify and answer it. We have
\[z=\frac{x-y}{x}=1-\frac{y}{x}\]
Suppose the values of $x$ and $y$ change from $(x_1,y_1)$ to $(x_2,y_2)$. Then the change in value of $z$ is
\[z_2-z_1=(1-\frac{y_2}{x_1})- (1-\frac{y_1}{x_1})=\frac{y_1}{x_1}-\frac{y_2}{x_2}\]
\[=\frac{y_1 x_2-y_2x_1}{x_1 x_2}=\frac{y_1 x_2-y_1x_1+y_1x_1-y_2x_1}{x_1 x_2}\]
\[=\frac{y_1 (x_2-x_1)-x_1(y_2-y_1)}{x_1 x_2}=\frac{y_1}{x_1x_2}(x_2-x_1)-\frac{1}{x_2}(y_2-y_1).\]
So
\[z_2-z_1=\frac{y_1}{x_1x_2}(x_2-x_1)-\frac{1}{x_2}(y_2-y_1).\]
You can measure the change in $z$ due to change in $x$
\[\Delta x=\frac{y_1}{x_1x_2}(x_2-x_1)\]
and the change in $z$ due to change in $y$
\[\Delta y=-\frac{1}{x_2}(y_2-y_1).\]
In your example we have
\[\Delta x=\frac{y_1}{x_1x_2}(x_2-x_1)=\frac{2}{4\cdot 5}(5-4)=\frac{1}{10}=10 \%,\]
and
\[\Delta y=-\frac{1}{x_2}(y_2-y_1)= -\frac{1}{5}(1-2)=\frac{1}{5}=20 \%.\]

This took me about 30 minutes to answer. I would appreciate a tip!