(2) $$ 12 x-4 y=28 $$ $$ 12 x-9 y-33=0 $$

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Solve the system of equations $$ 12x-4y=28; 12x-9y-33=0 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}12x-4y=28\12x-9y-33=0\end{array}ight.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}x=\frac{7+y}{3}\12x-9y-33=0\end{array}ight.$$
- step2: Substitute the value of $$x:$$
  $$12	imes \frac{7+y}{3}-9y-33=0$$
- step3: Simplify:
  $$-5-5y=0$$
- step4: Move the constant to the right side:
  $$-5y=0+5$$
- step5: Remove 0:
  $$-5y=5$$
- step6: Change the signs:
  $$5y=-5$$
- step7: Divide both sides:
  $$\frac{5y}{5}=\frac{-5}{5}$$
- step8: Divide the numbers:
  $$y=-1$$
- step9: Substitute the value of $$y:$$
  $$x=\frac{7-1}{3}$$
- step10: Simplify:
  $$x=2$$
- step11: Calculate:
  $$\left\{ \begin{array}{l}x=2\y=-1\end{array}ight.$$
- step12: Check the solution:
  $$\left\{ \begin{array}{l}x=2\y=-1\end{array}ight.$$
- step13: Rewrite:
  $$\left(x,yight) = \left(2,-1ight)$$
The solution to the system of equations is $$ x = 2 $$ and $$ y = -1 $$.
$$ x = 2 $$, $$ y = -1 $$

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