\( \begin{array}{l}1.2 .1 \quad\left(\frac{125 x^{7}}{x}\right)^{\frac{2}{3}} \\ 1.2 .2 \quad(\sqrt{3}+3)^{2}-2 \sqrt{27} \\ \text { Solve for } x \text { and } y \text { simultancously: } \\ y=x+2 \\ x y+y^{2}-10(x+1)=0\end{array} \)

To solve the given system of equations simultaneously, we need to find the values of \( x \) and \( y \) that satisfy both equations. Given equations: 1. \( y = x + 2 \) 2. \( xy + y^2 - 10(x + 1) = 0 \) We can substitute the value of \( y \) from the first equation into the second equation to solve for \( x \). Substitute \( y = x + 2 \) into the second equation: \( x(x + 2) + (x + 2)^2 - 10(x + 1) = 0 \) Simplify the equation: \( x^2 + 2x + x^2 + 4x + 4 - 10x - 10 = 0 \) \( 2x^2 - 4x - 6 = 0 \) Now, we have a quadratic equation in terms of \( x \). We can solve this equation to find the values of \( x \). Let's solve the quadratic equation \( 2x^2 - 4x - 6 = 0 \) to find the values of \( x \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(2x^{2}-4x-6=0\) - step1: Factor the expression: \(2\left(x-3\right)\left(x+1\right)=0\) - step2: Divide the terms: \(\left(x-3\right)\left(x+1\right)=0\) - step3: Separate into possible cases: \(\begin{align}&x-3=0\\&x+1=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=3\\&x=-1\end{align}\) - step5: Rewrite: \(x_{1}=-1,x_{2}=3\) The solutions to the quadratic equation \( 2x^2 - 4x - 6 = 0 \) are \( x = -1 \) and \( x = 3 \). Now, we can substitute these values of \( x \) back into the first equation \( y = x + 2 \) to find the corresponding values of \( y \). For \( x = -1 \): \( y = -1 + 2 = 1 \) For \( x = 3 \): \( y = 3 + 2 = 5 \) Therefore, the solutions to the system of equations are \( x = -1, y = 1 \) and \( x = 3, y = 5 \).