If $\,\rm\color{#90f}{prime}$ $\,ax\!+\!c\mid ay\!+\!d\,$ then $\,\color{#90f}{(a,c)=1}\,$ (except degenerate cases), thus by $\rm\color{#c00}{Bezout}$

$$ (a,c)=1\Rightarrow \exists\, j,k\!:\ \color{#c00}{ja\!-\!kc=-d},\,\ {\rm so}\, \bmod ax\!+\!c\!:\,\ y \equiv \dfrac{\color{#c00}{-d}}a\equiv \dfrac{\color{#c00}{ja\! - \!kc}}a \equiv \bbox[5px,border:1px solid #c00]{j + k x}\qquad $$

e.g. your $\,15x\!-\!7\mid 15y\!+\!2\,$ has $\,ja\!-\!jc = -d\,$ $\leadsto \,15j\!+\!7k = -2,\,$ so $\!{\bmod 7\!:\ j \equiv -2},\,$ so $\,\color{#0a0}{j = -2\!+\!7m},\,$ so $\,k = (-2\!-\!15\:\!\color{#0a0}j)/7 = (-2\!-\!15(\color{#0a0}{-2\!+\!7m)})/7 = 4-15m,\,$ so $\,j+kx \equiv -2\!+\!4x +(7\!-\!15x)m,\,$ which agrees with your sought solution for $\,m=0$.

Note $\,(a,ax\!+\!c)\! =\! (a,c)\!=\!1\,$ so $\bmod ax\!+\!c\!:\ a^{-1}$ exists, so $\,\frac{n}a := na^{-1}\,$ uniquely exists too.

If $\,\rm\color{#90f}{prime}$ $\,ax\!+\!c\mid ay\!+\!d\,$ then $\,\color{#90f}{(a,c)=1}\,$ (except degenerate cases), thus by $\rm\color{#c00}{Bezout}$

$$ (a,c)=1\Rightarrow \exists\, j,k\!:\ \color{#c00}{ja\!-\!kc=-d},\,\ {\rm so}\, \bmod ax\!+\!c\!:\,\ y \equiv \dfrac{\color{#c00}{-d}}a\equiv \dfrac{\color{#c00}{ja\! - \!kc}}a \equiv \bbox[5px,border:1px solid #c00]{j + k x}\qquad $$

Explicitly: $\,\ {(ax\!+\!\color{#c00}c)\color{#c00}k = \color{#c00}a(kx\!+\!\color{#c00}j)\color{#c00}{+d}}$

e.g. your $\,15x\!-\!7\mid 15y\!+\!2\,$ has $\,ja\!-\!jc = -d\,$ $\leadsto \,15j\!+\!7k = -2,\,$ so $\!{\bmod 7\!:\ j \equiv -2},\,$ so $\,\color{#0a0}{j = -2\!+\!7m},\,$ so $\,k = (-2\!-\!15\:\!\color{#0a0}j)/7 = (-2\!-\!15(\color{#0a0}{-2\!+\!7m)})/7 = 4-15m,\,$ so $\,j+kx \equiv -2\!+\!4x +(7\!-\!15x)m,\,$ which agrees with your sought solution for $\,m=0$.

Note $\,(a,ax\!+\!c)\! =\! (a,c)\!=\!1\,$ so $\bmod ax\!+\!c\!:\ a^{-1}$ exists, so $\,\frac{n}a := na^{-1}\,$ uniquely exists too.

If $\,\rm\color{#90f}{prime}$ $\,ax\!+\!c\mid ay\!+\!d\,$ then $\,\color{#90f}{(a,c)=1}\,$ (except degenerate cases), thus by $\rm\color{#c00}{Bezout}$

$$ (a,c)=1\Rightarrow \exists\, j,k\!:\ \color{#c00}{ja\!-\!kc=-d},\,\ {\rm so}\, \bmod ax\!+\!c\!:\,\ y \equiv \dfrac{\color{#c00}{-d}}a\equiv \dfrac{\color{#c00}{ja\! - \!kc}}a \equiv \bbox[5px,border:1px solid #c00]{j + k x}\qquad $$

e.g. your $\,15x\!-\!7\mid 15y\!+\!2\,$ has $\,ja\!-\!jc = -d\,$ $\leadsto \,15j\!+\!7k = -2,\,$ so $\!{\bmod 7\!:\ j \equiv -2},\,$ so $\,\color{#0a0}{j = -2\!+\!7m},\,$ so $\,k = (-2\!-\!15j)/7 = (-2\!-\!15(\color{#0a0}{-2\!+\!7m)})/7 = 4-15m,\,$ so $\,j+kx \equiv -2\!+\!4x +(7\!-\!15x)m,\,$ which agrees with your sought solution for $\,m=0$.

Note $\,(a,ax\!+\!c)\! =\! (a,c)\!=\!1\,$ so $\bmod ax\!+\!c\!:\ a^{-1}$ exists, so $\,\frac{n}a := na^{-1}\,$ uniquely exists too.

If $\,\rm\color{#90f}{prime}$ $\,ax\!+\!c\mid ay\!+\!d\,$ then $\,\color{#90f}{(a,c)=1}\,$ (except degenerate cases), thus by $\rm\color{#c00}{Bezout}$

$$ (a,c)=1\Rightarrow \exists\, j,k\!:\ \color{#c00}{ja\!-\!kc=-d},\,\ {\rm so}\, \bmod ax\!+\!c\!:\,\ y \equiv \dfrac{\color{#c00}{-d}}a\equiv \dfrac{\color{#c00}{ja\! - \!kc}}a \equiv \bbox[5px,border:1px solid #c00]{j + k x}\qquad $$

e.g. your $\,15x\!-\!7\mid 15y\!+\!2\,$ has $\,ja\!-\!jc = -d\,$ $\leadsto \,15j\!+\!7k = -2,\,$ so $\!{\bmod 7\!:\ j \equiv -2},\,$ so $\,\color{#0a0}{j = -2\!+\!7m},\,$ so $\,k = (-2\!-\!15\:\!\color{#0a0}j)/7 = (-2\!-\!15(\color{#0a0}{-2\!+\!7m)})/7 = 4-15m,\,$ so $\,j+kx \equiv -2\!+\!4x +(7\!-\!15x)m,\,$ which agrees with your sought solution for $\,m=0$.

Note $\,(a,ax\!+\!c)\! =\! (a,c)\!=\!1\,$ so $\bmod ax\!+\!c\!:\ a^{-1}$ exists, so $\,\frac{n}a := na^{-1}\,$ uniquely exists too.