How to solve linear congruence with two variable that is not system congruence? This problem from Number theory-James K. Strayer 
The congruence is 
$$2x+3y\equiv 4 \pmod{7}$$
since it is not system congruence the eliminate method does not work 
Can anyone help me?
 A: Note that $3\cdot 5\equiv 1\pmod{7}$, so your congruence is equivalent to
$$
5(2x+3y)\equiv 5\cdot4\pmod{7}
$$
that is $3x+y\equiv 6\pmod{7}$ and therefore to
$$
y\equiv 4x+6\pmod{7}
$$
Give $x$ any value between $0$ and $6$ and determine $y$ (up to congruence modulo $7$, of course).
A: $2x\equiv (4 -3y) mod 7$
$x\equiv (2-(3/2)y) mod 7$ (For $k=1$ since   $\frac{3+7k}{2}=5$ then $(3/2)\equiv 5 mod7 $
$x\equiv (2-5y) mod 7$.
Since $-5\equiv 2 mod 7$, we get $x\equiv (2+2y) mod 7$. Then $x\equiv (2(1+y)) mod 7$.
Then $y=\bar{-1}=\bar6$ for $x=\bar0$,  and $y=\bar{3}$ for $x=\bar1$ (Put $x=1$ in the $x=2(1+y)$ then we have $y=\frac{-1}{2}$, for $k=1$, $y=\frac{-1+7k}{2}=3 $), and  $y=\bar{0}$ for $x=\bar2$ (we can calculate by using same argument), and  $y=\bar{4}$ for $x=\bar3$, and $y=\bar{1}$ for $x=\bar4$, and $y=\bar{5}$ for $x=\bar5$, and $y=\bar{2}$ for $x=\bar6$. 
So the answer is the set of binaries $\{(\bar0,\bar6), (\bar1,\bar3),(\bar2,\bar0),(\bar3,\bar4),(\bar5,\bar5),(\bar6,\bar2) \}$
