Answer:
x = 3
Step-by-step explanation:
Given the system of equations:
[tex]\begin{cases} xy = 6 - 2x - 3y\\ yz = 6 - 4y - 2z\\ xz = 30 - 4x - 3z \end{cases}[/tex]
To find the positive solution of x, begin by rearranging the first equation to isolate y:
[tex]xy=6-2x-3y \\\\xy+3y=6-2x \\\\y(x+3)=6-2x \\\\y=\dfrac{6-2x}{x+3},\quad x\neq -3[/tex]
Now, rearrange the second equation and solve for y:
[tex]yz = 6 - 4y - 2z \\\\ yz + 4y= 6 - 2z \\\\ y(z + 4)= 6 - 2z \\\\ y=\dfrac{6 - 2z}{z + 4},\quad z\neq -4[/tex]
Set the equations equal to each other and solve for z:
[tex]\dfrac{6-2x}{x+3}=\dfrac{6 - 2z}{z + 4} \\\\ (z+4)(6-2x)=(6-2z)(x+3) \\\\ 6z-2xz+24-8x=6x+18-2xz-6z \\\\ 6z-2xz+2xz+6z=6x+18-24 +8x\\\\ 12z=14x-6 \\\\14x=12z+6 \\\\ 7x=6z+3\\\\6z=7x-3\\\\z=\dfrac{7x-3}{6},\quad x\neq -3[/tex]
Rearrange the third equation to isolate z:
[tex]xz = 30 - 4x - 3z \\\\ xz +3z= 30 - 4x \\\\ z(x+3)=30-4x\\\\z=\dfrac{30-4x}{x+3}, \quad x\neq 3[/tex]
Now, set the two equations for z equal to each other and solve for x:
[tex]\dfrac{7x-3}{6}=\dfrac{30-4x}{x+3} \\\\ (7x-3)(x+3)=6(30-4x) \\\\ 7x^2+21x-3x-9=180-24x \\\\7x^2+18x-9=180-24x \\\\7x^2+42x-189=0 \\\\7(x^2+6x-27)=0\\\\x^2+6x-27=0\\\\x^2+9x-3x-27=0\\\\x(x+9)-3(x+9)=0\\\\(x-3)(x-9)=0\\\\\\x-3=0 \implies x=3\\\\x+9=0 \implies x=-9[/tex]
Therefore, the positive solution of x is:
[tex]\Large\boxed{\boxed{x=3}}[/tex]
