4. Ilocrporre rpacur ypaBHens (22 + y? + 6y)(x - y) = 0
To solve the equation \((22 + y^2 + 6y)(x - y) = 0\), we can set each factor equal to zero.
First Factor: \(22 + y^2 + 6y = 0\)
This is a quadratic equation in terms of \(y\). We can rearrange it to standard form: \[ y^2 + 6y + 22 = 0 \]
Next, we can use the quadratic formula, which is: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \(a = 1\), \(b = 6\), and \(c = 22\).
Calculate the discriminant: \[ b^2 - 4ac = 6^2 - 4(1)(22) = 36 - 88 = -52 \]
Since the discriminant is negative, there are no real solutions for \(y\). The solutions will be complex numbers.
Now, calculating the complex solutions: \[ y = \frac{-6 \pm \sqrt{-52}}{2(1)} = \frac{-6 \pm \sqrt{52}i}{2} = \frac{-6 \pm 2\sqrt{13}i}{2} = -3 \pm \sqrt{13}i \]
Therefore, the solutions for \(y\) are: \[ y = -3 + \sqrt{13}i \quad \text{and} \quad y = -3 - \sqrt{13}i \]
Second Factor: \(x - y = 0\)
This gives us: \[ x = y \]
In conclusion, the overall solutions for the system are:
- \(y = -3 + \sqrt{13}i\) and \(x = -3 + \sqrt{13}i\)
- \(y = -3 - \sqrt{13}i\) and \(x = -3 - \sqrt{13}i\)
So, the pairs of solutions \((x, y)\) are: \[ (x, y) = (-3 + \sqrt{13}i, -3 + \sqrt{13}i) \quad \text{and} \quad (x, y) = (-3 - \sqrt{13}i, -3 - \sqrt{13}i) \]
