# Family of 2x2 linear equations

This is my first question here, so please tell me if the question is too trivial for this location. I have read FAQ and I didn't find the rule against easy questions :)

I should construct a system of linear equation similar to this one:

\begin{align}x+2y & = 4 \\ 5x-y & = \frac15\end{align}

It is suggested that similarity should be in relation between coefficients. All that I see is that if we put it in form $y=ax+b$, it is true that $ab=-1$.

Any hints?

• Well, you could multiply both sides of the second equation with $5$... otherwise, I don't understand what "similarity" is supposed to mean here. – J. M. is a poor mathematician Sep 26 '11 at 18:15
• Okay, so any 2x2 system with unique solution can be considered similar to this one? – 1osmi Sep 26 '11 at 18:41
• You might want to introduce yourself to the notation $Ax=b$ with this example where $A$ is a matrix $x, b$ are vectors... It would turn out to be useful later if you will continue solving similar problems. The problem of yours now becomes $$\pmatrix{1 &2\\5 &-1}\pmatrix{x\\y} = \pmatrix{4\\ \frac{1}{5}}$$. with $x$ is a scalar not related to the previous $x$ I mentioned. You can then search for Gauss-Elimination online and easily solve this and many more problems. – user13838 Sep 26 '11 at 19:03

Your system of equations is of the form :

$a_1x+b_1y=c_1$

$a_2x-b_2y=c_2$

Note that $c_1=4a_1$ and $b_1=2a_1$ so coefficients of the first equation are of the $(a_1,2a_1,4a_1)$ form. For the second equation we can write $c_2=\frac{a_2}{25}$ and $b_2=\frac{a_2}{5}$ so coefficients of the second equation are of the $(a_2,\frac{a_2}{5},\frac{a_2}{25})$ form.

Now, if we take for example $a_1=2$ and $a_2=25$ we get the next system of equations which is similar to the first system :

$2x+4y=8$

$25x-5y=1$

Since $a_1$ and $a_2$ are arbitrary numbers you may write infinitely many similar systems.

Try subracting 5 times the first equation from the second. That will eliminate x, and you can solve for y.