\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}

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Solve the equation $$ 6x-(2x-3)=31 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$6x-\left(2x-3ight)=31$$
- 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:
  $$\left(x-3ight)\left(x+5ight)=0$$
- 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 $$ x = 3 $$.
The solutions are $$ x = 7 $$, $$ x = -5 $$, and $$ x = 3 $$.

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