Solve the linear congruence 7x+3y≡10(bmod16).

Step 1
Consider the linear congruence ${\displaystyle 7x+3y\equiv 10\left(b\text{mod}16\right)}$.
Since gcd (7, 3) = 1 we know at least one solution exists.
However, the difference between a linear congruence in one variable and a linear congruence in two variables becomes clear when we see that the congruence ${\displaystyle 7x+3y\equiv 10\left(b\text{mod}16\right)}$ has multiple solutions.
The existence of one solution comes to fruition upon converting the aforementioned linear congruence to the form ${\displaystyle 7x\equiv 10-3yb\text{mod}16}$ and setting ${\displaystyle y\equiv 0b\text{mod}16}$.
This leads us to the linear congruence ${\displaystyle 7x\equiv 10b\text{mod}16}$.
After multiplying both sides of our congruence by 7, we find ${\displaystyle x\equiv 6b\text{mod}16}$.
Therefore, one solution to the linear congruence ${\displaystyle 7x+3y\equiv 10\left(b\text{mod}16\right)}$ is given by
${\displaystyle x\equiv 6b\text{mod}16}$
${\displaystyle y\equiv 0b\text{mod}16}$
Step 2
Our difference maker comes into play when we let ${\displaystyle y\equiv 1b\text{mod}16}$.
This gives rise to the congruence ${\displaystyle 7x\equiv 7b\text{mod}16}$.
In this case we have ${\displaystyle x\equiv 1b\text{mod}16}$.
As a result, we find another solution of ${\displaystyle 7x+3y\equiv 10b\text{mod}16}$ is
${\displaystyle x\equiv 1b\text{mod}16}$
${\displaystyle y\equiv 1b\text{mod}16}$