where is the 2x coming from here - Consider the congruence: [ 2x \equiv 14 \pmod{26} ] This congruence has multiple solutions. For instance: [ x = 7 ] [ 2 \cdot 7 = 14 \equiv 14 \pmod{26} ] And: [ x = 20 ] [ 2 \cdot 20 = 40 \equiv 14 \pmod{26} ] Both ( x = 7 ) and ( x = 20 ) satisfy the congruence, showing that there are multiple solutions