Consider the provided system of equation

\(x+3y+4z=3\)

\(2z+6y+9z=5\)

\(3x+y-2z=7\)

The matrix equation of the linear system is given by,

\(AX=b\)

Where, \(A[(1,3,4),(2,6,9),(3,1,-2)]\)

given, \(|A|=8\)

Find the value of x by using cramer's rule

\(x=\frac{|D_x|}{|D|}\)

here, \(D_x=[(3,3,4),(5,6,9),(7,1,-2)]\)

and, \(|D|\) is the determinant of the matrix A

Since,when system of equation as,

\(a_1x+b_1y+c_1z=d_1\)

\(a_2x+b_2y+c_2z=d_2\)

\(a_3+b_3+c_3z=d_3\)

Then, \(D_x=[(d_1,b_1,c_1),(d_2,b_2,c_2),(d_3,b_3,c_3)]\)

Now, the value of x find as,

\(x=\frac{|D_x|}{|D|}\)

\(=|(3,3,4),(5,6,9),\frac{7,1,-2}{|A|}\)

\(=\frac{(-36+20+189)-(168+27-30)}{8}\)

\(=\frac{173-165}{8}\)

\(=\frac{8}{8}\)

\(=1\)

Thus, \(x=1\)