Solve for x and y simultaneously, where: 6y = x2 − 2x − 6 and sum_n=34 x(n−2)2 = 10y

The summation runs for $n=3$ and $n=4$.
For $n=3$: $x(3-2)^2 = x(1)^2 = x$
For $n=4$: $x(4-2)^2 = x(2)^2 = 4x$
Adding these terms: $x + 4x = 5x$.
So, the second equation simplifies to:

$$5x = 10y$$

Divide both sides by 10:

$$y = \frac{5x}{10}$$ $$y = \frac{1}{2}x$$

Substitute $y = \frac{1}{2}x$ into $6y = x^2 - 2x - 6$:

$$6\left(\frac{1}{2}x\right) = x^2 - 2x - 6$$ $$3x = x^2 - 2x - 6$$

Subtract $3x$ from both sides:

$$0 = x^2 - 2x - 3x - 6$$ $$0 = x^2 - 5x - 6$$

We need two numbers that multiply to $-6$ and add to $-5$. These numbers are $-6$ and 1.

$$(x - 6)(x + 1) = 0$$

This gives two possible values for $x$:

$$x - 6 = 0 \implies x = 6$$ $$x + 1 = 0 \implies x = -1$$

For $x = 6$:

$$y = \frac{1}{2}(6)$$ $$y = 3$$

For $x = -1$:

$$y = \frac{1}{2}(-1)$$ $$y = -\frac{1}{2}$$

The solutions are the pairs $(x, y)$:

$$(6, 3) \text{ and } \left(-1, -\frac{1}{2}\right)$$