Solve the linear congruence x+2y -= 1(mod 5) 2x+y -= 1(mod 5)

Question
Congruence
asked 2021-02-05
Solve the linear congruence
\(\displaystyle{x}+{2}{y}\equiv{1}{\left(\text{mod}{5}\right)}\)
\(\displaystyle{2}{x}+{y}\equiv{1}{\left(\text{mod}{5}\right)}\)

Answers (1)

2021-02-06
Step 1
Given a system of linear congruences
\(\displaystyle{x}+{2}{y}\equiv{1}{\left(\text{mod}{5}\right)}\)
\(\displaystyle{2}{x}+{y}\equiv{1}{\left(\text{mod}{5}\right)}\)
Solve it.
Step 2
Multiply the first congruence by 2.
\(\displaystyle{2}{x}+{4}{y}\equiv{2}{\left(\text{mod}{5}\right)}\)
Add the preceding one with the second congruence in the system.
\(\displaystyle{4}{x}+{5}{y}\equiv{3}{\left(\text{mod}{5}\right)}\)
\(\displaystyle\Rightarrow{4}{x}\equiv{3}{\left(\text{mod}{5}\right)}\)
Step 3
Multiply the congruence by \(\displaystyle{4}^{{-{1}}}{\left(\text{mod}{5}\right)}={4}\) to get
\(\displaystyle{16}{x}\equiv{12}{\left(\text{mod}{5}\right)}\)
\(\displaystyle\Rightarrow{x}\equiv{2}{\left(\text{mod}{5}\right)}\)
Step 4
Plugging back in x for the second equation.
\(\displaystyle{4}+{y}\equiv{1}{\left(\text{mod}{5}\right)}\)
\(\displaystyle\Rightarrow{y}\equiv-{3}{\left(\text{mod}{5}\right)}{\quad\text{or}\quad}{y}\equiv{2}{\left(\text{mod}{5}\right)}\)
Thus the solution of the system of congruences is
\(\displaystyle{x}\equiv{2}{\left(\text{mod}{5}\right)},{y}\equiv{2}{\left(\text{mod}{5}\right)}\)
0

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