can't solve $a-2b=3$ & $5b=4a+1$ $$a-2b=3\tag1\label1$$
$$5b=4a+1\tag2\label2$$
Coming to \eqref{1}, $a=2b+3$. Substituting that in \eqref{2}, we get
$5b=4(2b+3)+1$.
$5b=8b+12+1$.
$3b=-13$.
$b=-13/3$.
Substituting value of $b$ in \eqref{1}.
$a-2(-13/3)=3$.
$a-(-26/3)=3$.
Adding $26/3$ to both sides.
$a-0=3+26/3$.
$a=35/3$.
Substituting $a$ and $b$ in \eqref{2}.
$5b=4a+1$.
$5(-13/3)=4(35/3)+1$.
$-65/3=4(35/3)+1$.
$-65/3=140/3+1$.
$-65/3=143/3$.
$-65=143$.
But this is not true.

No, you messed up where you said "Adding 26/3 to both sides." If you check again, you will see that this should actually be "subtracting 26/3 to both sides" since $$a-\left(-\frac{26}{3}\right)=a+\frac{26}{3}=3\implies a=-\frac{17}{3}$$.
Your answer and algebra for $b$ are correct, giving $a=-\frac{17}{3}, b=-\frac{13}{3}$.
