Solution of simple algebraic equations

Question

Suppose $a=\alpha+2\beta$ and $b=3\alpha+5\beta$. Then by solving this, we get $\alpha=2b-5a$ and $\beta=3a-b$.

I tried substitution method to solve this, but i couldn't get the above solutions viz., $\alpha$ and $\beta$.

I am afraid if this is too basic and it might not be appropriate for this site.

Answer 1

Multiply $a=\alpha+2\beta$ by $3$ to get $3a=3\alpha+6\beta.$

Now subtract this from $b=3\alpha+5\beta$ to get $\beta=3a-b$.

Similarly, if you multiply $a=\alpha+2\beta$ by $5$ and $b=3\alpha+5\beta$ by $2$,

you can subtract the equations to eliminate $\beta $ and solve for $\alpha$.

Or you could say $a=\alpha+2(3a-b)$ and solve for $\alpha$ that way.

Answer 2

Since $a=\alpha+2\beta$ and $b=3\alpha+5\beta$,$$b-3a=(3\alpha+5\beta)-3(\alpha+2\beta)=-\beta.$$So, $\beta=3a-b$ indeed. And now it is easy to get $\alpha$.

Answer 3

\begin{align} a&=\alpha+2\beta\tag{$\color{blue}{1}$}\\ b&=3\alpha+5\beta\tag{$\color{blue}{2}$}\\ 3\times(\color{blue}1)-(\color{blue}2):3a-b&=\beta\tag{$\color{blue}{3}$}\\ \text{sub}\ (\color{blue}3)\to(\color{blue}1):a&=\alpha+2(3a-b)\implies\alpha=2b-5a \end{align}

So we have the solution $\boxed{(\alpha,\beta)=(2b-5a,2a-b)}$.