Question

Solve for x.
And solve for Z

Answer 1

From Tales theorem:
`(8x-2)/(2x+1)=(3z+z+22)/(3z)`

Solving for x:

`(8x-2)/(2x+1)=(4z+22)/(3z)\\\\(8x-2)(3z)=(2x+1)(4z+22)\qquad\qquad\{divide\ both\ sides\ by\ 2\}\\\\(4x-1)(3z)=(2x+1)(2z+11)\\\\12xz-3z=2x(2z+11)+1(2z+11)\\\\12xz-2x(2z+11)=3z+2z+11\\\\x(12z-4z-22)=5z+11\\\\x(8z-22)=5z+11\qquad\qquad\{divide\ both\ sides\ by\ (8z-22)\}\\\\x=(5z+11)/(8z-22)`

Solving for z:

`(8x-2)/(2x+1)=(4z+22)/(3z)\\\\(8x-2)(3z)=(2x+1)(4z+22)\qquad\qquad\{divide\ both\ sides\ by\ 2\}\\\\(4x-1)(3z)=(2x+1)(2z+11)\\\\(12x-3)(z)=(2x+1)(2z)+(2x+1)(1)\\\\(12x-3)(z)-(4x+2)(z)=(2x+1)\\\\\ [(12x-3)-(4x+2)](z)=(2x+1)\\\\(12x-3-4x-2)(z)=(2x+1)\\\\(8x-5)(z)=(2x+1)\qquad\qquad\{divide\ both\ sides\ by\ (8x-5)\}\\\\z=(2x+1)/(8x-5)`