Questions: Solve the following system of equations. x^2 + y^2 = 19 x^2 - y = 7 If there is more than one solution, enter additional solutions with the "or" button. If there is no real solution, use the "No solution" button. (x, y) = ( , )

Solution Steps

To solve the given system of equations, we can use substitution or elimination methods. First, express $y$ from the second equation in terms of $x$. Substitute this expression into the first equation to form a single equation in terms of $x$. Solve for $x$ and then use the value(s) of $x$ to find the corresponding $y$ value(s).

We start with the system of equations: $$\begin{cases} x^{2} + y^{2} = 19 \\ x^{2} - y = 7 \end{cases}$$ From the second equation, we can express $y$ in terms of $x$: $$y = x^{2} - 7$$

Substituting $y$ into the first equation gives: $$x^{2} + (x^{2} - 7)^{2} = 19$$ Expanding this leads to a single equation in $x$: $$x^{2} + (x^{4} - 14x^{2} + 49) = 19$$ This simplifies to: $$x^{4} - 13x^{2} + 30 = 0$$

Let $z = x^{2}$. The equation becomes: $$z^{2} - 13z + 30 = 0$$ Using the quadratic formula, we find: $$z = \frac{13 \pm \sqrt{(13)^{2} - 4 \cdot 1 \cdot 30}}{2 \cdot 1} = \frac{13 \pm \sqrt{49}}{2} = \frac{13 \pm 7}{2}$$ This gives us: $$z = 10 \quad \text{or} \quad z = 3$$ Thus, $x^{2} = 10$ or $x^{2} = 3$.

1. For $x^{2} = 10$:

   • $x = \pm \sqrt{10}$
   • $y = 10 - 7 = 3$

2. For $x^{2} = 3$:

   • $x = \pm \sqrt{3}$
   • $y = 3 - 7 = -4$

Final Answer