Solve this system of linear equations This is a system from Shilov's Linear Algebra, page 7
He gives a system of linear equation
$a_{11}x_{1}+a_{12}x_{2}=b_{1}$
$a_{21}x_{1}+a_{22}x_{2}=b_{2}$
He says that "eliminating one of the unknowns in the usual way, we can easily obtain the formulas":
$x_{1}=\dfrac{b_{1}a_{22}-b_{2}a_{12}}{a_{11}a_{22}-a_{21}a_{12}}$
$x_{2}=\dfrac{a_{11}b_{2}-a_{21}b_{1}}{a_{11}a_{22}-a_{21}a_{12}}$
So far, I have only been able to come up with:
$$x_{1}=\dfrac{b_{1}-a_{12}x_{2}}{a_{11}}$$
$$x_{1}=\dfrac{b_{2}-a_{22}x_{2}}{a_{21}}$$
How do you successfully eliminate the unknowns to obtain his result?
 A: Given 
$a_{11}x_{1}+a_{12}x_{2}=b_{1}$ and 
$a_{21}x_{1}+a_{22}x_{2}=b_{2}$,
to eliminate $x_2$, multiply the first equation by $a_{22}$ and the second by $a_{12}$ to get
$a_{22}a_{11}x_1+a_{22}a_{12}x_2=a_{22}b_1$ and 
$a_{12}a_{21}x_1+a_{12}a_{22}x_2=a_{12}b_2$ 
and then subtract the second equation from the first.  Can you take it from here?
A similar strategy could be used to eliminate $x_1$ and solve for $x_2$.
A: Applying the answer of Tanner, I tried to multiply $a_{21} (a_{11}x_{1}+a_{12}x_{2}=b_{1})$ and $a_{11} (a_{21}x_{1}+a_{22}x_{2}=b_{2})$ to get two equations:
$a_{21}a_{11}x_{1}+a_{21}a_{12}x_{2}=a_{21}b_{1}$
$a_{11}a_{21}x_{1}+a_{11}a_{22}x_{2}=a_{11}b_{2}$
The term $a_{21}a_{11}x_{1}$ cancels after substraction, the equation is $a_{21}a_{12}x_{2}-a_{11}a_{22}x_{2}=a_{21}b_{1}-a_{11}b_{2}$
Therefore $x_{2}=\dfrac{a_{21}b_{1}-a_{11}b_{2}}{a_{21}a_{12}-a_{11}a_{22}}$
Something is not right since the answer in the book is
$x_{2}=\dfrac{a_{11}b_{2}-a_{21}b_{1}}{a_{11}a_{22}-a_{21}a_{12}}$
My answer is incorrect for some reasons, can you explain where is my mistake?
Should I switch the equation or what?
A: My suggestion, you could follow the multiplications indicated in these factors with $b_{1}$ and $b_{2}$. Observe: $a_{11}b_{2} - a_{21}b_{2}$ indicates that the first equation was multiplied by $- a_{21}$ and the second equation was multiplied by $a_{11}$, then he added the results and isolated the value of $x_{2}$. Try to do the same to get the value of $x_{1}$.
A: With "the usual way" I would understand adding, subtracting, dividing and multiplying with constants or even whole terms.
The way I would do it is this one:


*

*Divide each equation with their $x_1$ coefficient.

*Subtract 2nd from 1st equation.

*Repeat the procedure for $x_2$.


After the first step I get:
$$x_1 + \frac{a_{12}}{a_{11}}x_2 = \frac{b_1}{a_{11}}$$
$$x_1 + \frac{a_{22}}{a_{21}}x_2 = \frac{b_2}{a_{21}}$$
After 2nd step:
$$\big(\frac{a_{12}}{a_{11}} - \frac{a_{22}}{a_{21}}\big)x_2 = \frac{b_1}{a_{11}} - \frac{b_2}{a_{21}}$$
If I divide by the constant factor before $x_2$, it just looks like the formula which is mentioned in the question.
I guess, this is exactly what the Gauß algorithm would do, but my solution was just created intuitively.
