# Solving the equations .

Say , I have two equations :

$$y_1=a+bx_{1}+e_1$$ $$y_2=a+bx_{2}+e_2$$

Say , $a=.5$ , $b=2.1$ , $x_1=2$ , $x_2=2.2$ .

Now if $e_1=e_2$ , I have to find the relationship between $y_1$ and $y_2$ .

So i started as :

$$e_1=e_2$$ $$\Rightarrow y_1-(a+bx_{1})=y_2-(a+bx_{2})$$ $$\Rightarrow y_1-[.5+(2.1)(2)]=y_2-[.5+(2.1)(2.2)]$$ $$\Rightarrow y_1-4.7=y_2-5.12$$ $$\Rightarrow y_1=y_2-5.12+4.7$$ $$\Rightarrow y_1=y_2-.42$$ But when i started to cross-check , that is ,

Holding $a=.5$ and $b=2.1$ and $x_{1}=2$ , $x_{2}=2.2$ and the relationship $y_1=y_2-.42$ , when $y_1=5$ , $y_2=4.58$ , then

$$e_1=y_1-(a+bx_{1})=5-[.5+(2.1)(2)]=5-4.7=0.3$$

and

$$e_2=y_2-(a+bx_{2})=4.58-[.5+(2.1)(2.2)]=4.58-5.12=-.54$$

That is , $$e_1\ne e_2\quad\text{!!! }$$

But for this example I established the relationship if $$e_1=e_2\quad\text{then}\quad y_1=y_2-.42$$ . Then why is the converse not true , i.e., if $$y_1=y_2-.42\quad\text{then}\quad e_1=e_2$$

???

## 1 Answer

If $y_1 = y_2 - 0.42$, then if $y_1= 5$, the value of $y_2$ should not be $4.58$, because $5\neq 4.58-0.42!$