How to solve linear congruence with two variable that is not system congruence?

Question

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?

Answer 1

$2x\equiv (4 -3y) mod 7$

$x\equiv (2-(3/2)y) mod 7$ (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 $\{(\bar0,\bar6), (\bar1,\bar3),(\bar2,\bar0),(\bar3,\bar4),(\bar5,\bar5),(\bar6,\bar2) \}$

Answer 2

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).