System with $x + ay = 3 , 2x - y = b$

Question

Quesstion about this system: $x + ay = 3 , 2x - y = b$

Where $a, b \in\mathbb R$. The system admits as solution $(1, 2)$. I'm struggling with the logic of getting a final result. Is there an efficient way of calculating this

Answer 1

If $a=-\frac{1}{2}$, then the first equation becomes $x-\frac{1}{2}y=3$ or $2x-y=6$. So if $b\neq6$, we have no solutions and if $b=6$ we have infinitely many solutions.

Suppose that $a\neq -\frac{1}{2}$. Then from the first we have $x=3-ay$ and substituting into the second gives $$2x-y=6-2ay-y=6-y(2a+1)=b$$ or $$y=\frac{6-b}{2a+1}$$

and then we have $$x=3-ay=3-\frac{a(6-b)}{2a+1}=\frac{ab+3}{2a+1}$$ So for $a\neq -\frac{1}{2}$ the set of all solutions is given by $$(x,y)=\big(\frac{ab+3}{2a+1},\frac{6-b}{2a+1}\big)$$

Setting $a=1$ and $b=0$ we obtain one solution $(x,y)=(1,2)$.

If $a=0$ we have $(x,y)=(3,6-b),$ etc.

Answer 2

The point $(1,2)$ is a solution to the system iff $1+a\cdot2=3$ and $2\cdot1-2=b$ are both true statements. From here $a=1$ and $b=0$.