Step 1

Since 15 is not prime number .Prime factorization of 15 is \(\displaystyle{5}\times{3}\). Convert the given equation in \(\displaystyle{b}\text{mod}{5}\) and mod 3.

The solution of the equation \(\displaystyle{9}{x}\equiv{12}\pm{o}{d}{3}\) and

\(\displaystyle{9}{x}\equiv{12}\pm{o}{d}{5}\)

Now solve the above system that will be find out by finding common solution of above equations.

Step 2

Solve the equation \(\displaystyle{9}{x}\equiv{12}\pm{o}{d}{3}\).

The equivalent meaning of above equation is \(\displaystyle{\frac{{{3}}}{{{9}{x}}}}-{12}\) that is \(\displaystyle{\frac{{{3}}}{{{3}}}}{\left({3}{x}-{4}\right)}\). Since 3 divide 9x−12 always inn-respective of values of \(\displaystyle{x}\in{\mathbb{{{Z}}}}\). Therefore solution of equation is \(\displaystyle{\mathbb{{{Z}}}}\).

Now solve the equation \(\displaystyle{9}{x}\equiv{12}\pm{o}{d}{5}\).

\(\displaystyle{9}{x}\equiv{12}\pm{o}{d}{5}\)

\(\displaystyle{4}{x}\equiv{2}\pm{o}{d}{5}\)

\(\displaystyle-{x}\equiv{2}\pm{o}{d}{5}\)

\(\displaystyle{x}\equiv-{2}\pm{o}{d}{5}\)

\(\displaystyle{x}\equiv{3}\pm{o}{d}{5}\)

The solution of the equation \(\displaystyle{x}\equiv{3}\pm{o}{d}{5}\) is 5k+3 where k is an integer. The common solution of both the equation is 5k+3 where k is an integer.

The set form of the solution is \(\displaystyle\le{f}{t}{\left\lbrace{5}{k}+{3}.{k}\in{\mathbb{{{Z}}}}{r}{i}{g}{h}{t}\right\rbrace}\).

Since 15 is not prime number .Prime factorization of 15 is \(\displaystyle{5}\times{3}\). Convert the given equation in \(\displaystyle{b}\text{mod}{5}\) and mod 3.

The solution of the equation \(\displaystyle{9}{x}\equiv{12}\pm{o}{d}{3}\) and

\(\displaystyle{9}{x}\equiv{12}\pm{o}{d}{5}\)

Now solve the above system that will be find out by finding common solution of above equations.

Step 2

Solve the equation \(\displaystyle{9}{x}\equiv{12}\pm{o}{d}{3}\).

The equivalent meaning of above equation is \(\displaystyle{\frac{{{3}}}{{{9}{x}}}}-{12}\) that is \(\displaystyle{\frac{{{3}}}{{{3}}}}{\left({3}{x}-{4}\right)}\). Since 3 divide 9x−12 always inn-respective of values of \(\displaystyle{x}\in{\mathbb{{{Z}}}}\). Therefore solution of equation is \(\displaystyle{\mathbb{{{Z}}}}\).

Now solve the equation \(\displaystyle{9}{x}\equiv{12}\pm{o}{d}{5}\).

\(\displaystyle{9}{x}\equiv{12}\pm{o}{d}{5}\)

\(\displaystyle{4}{x}\equiv{2}\pm{o}{d}{5}\)

\(\displaystyle-{x}\equiv{2}\pm{o}{d}{5}\)

\(\displaystyle{x}\equiv-{2}\pm{o}{d}{5}\)

\(\displaystyle{x}\equiv{3}\pm{o}{d}{5}\)

The solution of the equation \(\displaystyle{x}\equiv{3}\pm{o}{d}{5}\) is 5k+3 where k is an integer. The common solution of both the equation is 5k+3 where k is an integer.

The set form of the solution is \(\displaystyle\le{f}{t}{\left\lbrace{5}{k}+{3}.{k}\in{\mathbb{{{Z}}}}{r}{i}{g}{h}{t}\right\rbrace}\).