Lagrange multiplier with 2 constraints

Question

I can't find X, Y, and Z in the following equation. it's dificult $$ F=5x^2+3y^2-2z+2xy+1+{\alpha}(3x+z-5)+{\beta}(4y-3z) $$\begin{cases} F_x=10x+2y+3{\alpha}=0 \\ F_y=6y+2x+4{\beta}=0 \\ F_z=-2+{\alpha}-3{\beta}=0 \\ g_1=3x+z=5 \\ g_2=4y-3z=0 \\ \end{cases}

Answer 1

Taking $12F_z-4F_x+9F_y$ eliminates $\alpha,\beta$ and gives you $$-22x+46y=24\ .$$ Together with the equations from $g_1$ and $g_2$ this is a system of three linear equations in three unknowns and I'm sure you can solve it even though the arithmetic is a bit messy.