How can I solve a modular equation with no modular inverse (but there's a solution)?

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While reading a book about cryptography, I stumbled across the following calculations:

$k_1\equiv \overline{(5\times24-22\times5)}(24\times21-22\times5)$ modulo $26$,

where $\overline{x}$ denotes a multiplicative inverse.

The book skipped the calculations and gave the correct solution which you can check

$k_1\equiv3$ modulo $26$

The problem is that, if you simplify the first congruence, you get:

$k_1\equiv \overline{10} \times 394 \equiv \overline{10} \times 4$ modulo $26$

And, since $GCD(10, 26)=2\ne1$, there is no multiplicative inverse for $\overline{10}$ modulo $26$.

So, how can I solve such equation and what am I missing here involving the multiplicative inverses?

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To solve $10k_1\equiv4\pmod{26}, $ though $10$ does not have an inverse mod $26$,

divide through by $2$ to get $5k_1\equiv2\pmod{13}$. $5$ has an inverse mod $13$ (it's $8$),

so $5k_1\equiv2\pmod{13}$ means $k_1\equiv3\pmod{13}$.

That translates to $k_1\equiv3$ or $16\pmod{26}$.