How would one solve this system

Question

$$12x^2=6z\\2y=-z\\6x-y=7$$

It's been many years since I've dealt with system equations, and now find myself in need to solve them. I am not quite sure what to do; I am interested in finding $x$ and $y$, so assuming that $z \neq 0$ (can I do this?), I isolate it and get $$2x^2 = z = -2y \\ -x^2 = y \\ 6x - (-x^2) = 7 \\ 6x + x^2 = 7 $$ which gives $$x^2 + 6x - 7 = 0 \\ (x+7)(x-1)=0$$ so $$ x = -7 \\or \\x = 1$$ which gives $$y = -(-7)^2 = -49 \\or \\ y = -(1)^2= -1 $$ But the answer is supposed to be $(1,-1)$ and $(7,35)$ ?? Where's my mistake?

Answer 1

eliminating $y,z$ we get $x^2+6x-7=0$ with $x-1=1,x_2=-7$ therefore $6-y=7$ thus $y_1=-1$ or $-42-y=7$ thus $y_2=-49$ now you can calculate $z$

Answer 2

The only real solutions to $12x^2=6z$ satisfy $z\ge 0$, so the second equation implies that $y\le 0$. This rules out $(7,35)$ as a solution to your problem.

Answer 3

Given is $f(x,y)=4x^3+y^2$ with the constraint $y=6x-7$. The easiest way to solve this, is plugging in $y$ into the function, which gives $$f(x)=4x^3+(6x-7)^2=4x^3+36x^2-84x+49$$ and we are asked to find the extrema. Differentiating: $$f'(x)=12x^2+72x-84=0$$ Dividing by $12$ gives $f'(x)=x^2+6x-7=(x+7)(x-1)=0$. From this follow the solutions $x=-7$ and $x=1$.

Another method is with Lagrange multipliers. The gradient of the constraint is $\left(\begin{array}{c}6\\-1\end{array}\right)$. The gradient from the function $f(x,y)$ is $2\left(\begin{array}{c}6x^2\\y\end{array}\right)$. These two gradients must be parallel to have an extremum at a given point $(x,y)$.

From this follows that $\frac{6x^2}{y}=\frac{6}{-1}$ and $x^2=-y$. Plugging this expression for $y$ into the constraint $6x-y=7$ gives $x^2+6=-7$, which is the same quadratic equation as before.

To summarize; there is a local maximum at $f(-7,-49)=1029$ and a local minimum at $f(1,-1)=5$.

To answer the question, your approach is correct, so maybe the answer model is incorrect.