Dạng 3: Tìm số chưa biết trong một đẳng thức

$x=64:8=8$

$x=60:\left(-15\right)=-4$

$\begin{array}{l}12x=30-6\\ 12x=24\\ x=24:12\\ x=2\end{array}$

$\begin{array}{l}\left(-11\right).x=130-9\\ \left(-11\right).x=121\\ x=121:\left(-11\right)\\ x=-11\end{array}$

$\begin{array}{l}\left(-12\right).x=\left(-15\right).\left(-4\right)-12\\ \left(-12\right).x=60-12\\ \left(-12\right).x=48\\ x=48:\left(-12\right)\\ x=-4\end{array}$

$\begin{array}{l}\left(-9\right).x+3=\left(-2\right).\left(-7\right)+16\\ \left(-9\right).x+3=14+16\\ \left(-9\right).x+3=30\\ \left(-9\right).x=30-3\\ \left(-9\right).x=27\\ x=27:\left(-9\right)\\ x=-3\end{array}$

$\begin{array}{l}\left(-12\right).x-34=2\\ \left(-12\right).x=2+34\\ \left(-12\right).x=36\\ x=36:\left(-12\right)\\ x=-3\end{array}$

x.y = 13 nên x$\in$U(13) và y$\in$U(13)

Do đó ta có:

$\left(x;y\right)\in \left\{\left(1;13\right);\left(13;1\right);\left(-1;-13\right);\left(-13;-1\right)\right\}$