\begin{tabular}{l} Solve \\ \( \qquad \begin{array}{l}6 x-(2 x-3)=31 \\ x^{2}+2 x-15=0\end{array} \) \\ \hline\end{tabular}

Solve the equation \( 6x-(2x-3)=31 \). Solve the equation by following steps: - step0: Solve for \(x\): \(6x-\left(2x-3\right)=31\) - step1: Subtract the terms: \(4x+3=31\) - step2: Move the constant to the right side: \(4x=31-3\) - step3: Subtract the numbers: \(4x=28\) - step4: Divide both sides: \(\frac{4x}{4}=\frac{28}{4}\) - step5: Divide the numbers: \(x=7\) Solve the equation \( x^{2}+2x-15=0 \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(x^{2}+2x-15=0\) - step1: Factor the expression: \(\left(x-3\right)\left(x+5\right)=0\) - step2: Separate into possible cases: \(\begin{align}&x-3=0\\&x+5=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=3\\&x=-5\end{align}\) - step4: Rewrite: \(x_{1}=-5,x_{2}=3\) The solutions to the system of equations are \( x = 7 \) and \( x = -5 \) or \( x = 3 \).