Albarellak

2021-04-21

Solve the linear congruence $7x+3y\equiv 10\left(b\text{mod}16\right)$.

AGRFTr

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

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