$x+ay=4, ax+9y=b$, Find the values of $a$ and $b$ for which the system has more than one set of solutions
I've tried $2$ different methods that my teacher taught me to get it, but I keep getting the wrong answer. Can someone please show me the steps as to how to get the correct answer?
Solution 1:
$x+ay=4\Rightarrow x = 4-ay\Rightarrow a(4-ay)+9y=b\Rightarrow 4a-a^2y+9y=b\Rightarrow (9-a^2)y=b-4a$. From this you can see that in order to have more than $1$ set of solution, it must be true that: $9-a^2=0=b-4a$. This yields: $(a,b) = (\pm 3,\pm 12)$