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

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