$$24,-11$$
So we begin with $gcd(24,-11)$
$$24=-2*-11+2=gcd(-11,2)$$
$$-11=-6*2+1=gcd(2,1)=1$$
So $gcd(24,-11)=1$ now how do I find x,y such that $1=24x-11y$?
$$24,-11$$
So we begin with $gcd(24,-11)$
$$24=-2*-11+2=gcd(-11,2)$$
$$-11=-6*2+1=gcd(2,1)=1$$
So $gcd(24,-11)=1$ now how do I find x,y such that $1=24x-11y$?
Easier than back-substitution is too keep track of the linear relations from the start
$$\begin{align} 24 &= \ \ 1(24)+0(11)\\ 11 &=\ \ 0(24)+1(11)\\ 2 &=\ \ 1(24)-2(11)\ \ =\ \rm row_1 - 2\cdot row_2\\ 1 &=\! -5(24)\!+\!11(11)\, = \ \rm row_2 - 5\cdot row_3 \end{align}$$
i.e. we perform the Euclidean algorithm on the LHS and in parallel on the RHS.
See this answer for much more on this version of the extended Euclidean algorithm.
The standard trick is to substitute backward from the equations you already have. For example, you have $-11=-6\cdot2+1$, so that means $1=-11+6\cdot2$. Using your first equation, we can re-write $2$ to yield the new equation $1=-11+6\cdot(24+2\cdot-11)$, which implies $\boxed{1=6\cdot24-11\cdot13}$ by distributing the $6$ and combining like terms.