3 equations and the area of a triangle

Solve the triple-system of equations. By simplification, $x+2y=z+1$ , $x+2y=-z-1$ , $2x-z=4y+28$ , and $4x^2-5y+7z=1700$ . The $2$ possible expressions for $x$ and $z$ resulted in the fact that an equation with an absolute value in both sides will split into $2$ equations (the $1$ st and $2$ nd equation, shown above). The third equation resulted in from the equation $2^{2x-z} = 16^{y+7}$ , where $2x-z=4(y+7) = 4y+28$ . I multiplied by $4$ because of the fact that $16=2^4$ . The $4$ th equation (the quadratic) is given in the problem.

A good variable to solve for, in my opinion (you could solve for any) is $y$ : In terms of $y$ , $x=27+6y$ (where $z=8y+26$ ), and $x= \frac{2y}{3}\ +9$ (where $z=-10 -$ $\frac{8y}{3}$ ).

You should get $4$ possible values for $(x+y+z)$ : $-23.484$ , $35.172$ , $35.165$ , and $-25.495$ . Then, Sum all of the values to get $21.358$ . Since the area of an equiangular, or equilateral triangle is $(s^2 * 0.25\sqrt{3})$ , where $s=k=21.358$ , the area is $197.52$ units $^2$ .