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