$$ \begin{array}{l}3 x-4 y=7 \ 3 x+2 y=-5\end{array} $$ When the second equation is subtracted from the first, the resulting equation is $$ -6 y=-12 $$ $$ -2 y=-12 $$ $$ -6 y=12 $$ $$ 2 y=12 $$

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Solve the system of equations $$ 3x-4y=7;3x+2y=-5 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}3x-4y=7\3x+2y=-5\end{array}ight.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}x=\frac{7+4y}{3}\3x+2y=-5\end{array}ight.$$
- step2: Substitute the value of $$x:$$
  $$3	imes \frac{7+4y}{3}+2y=-5$$
- step3: Simplify:
  $$7+6y=-5$$
- step4: Move the constant to the right side:
  $$6y=-5-7$$
- step5: Subtract the numbers:
  $$6y=-12$$
- step6: Divide both sides:
  $$\frac{6y}{6}=\frac{-12}{6}$$
- step7: Divide the numbers:
  $$y=-2$$
- step8: Substitute the value of $$y:$$
  $$x=\frac{7+4\left(-2ight)}{3}$$
- step9: Calculate:
  $$x=-\frac{1}{3}$$
- step10: Calculate:
  $$\left\{ \begin{array}{l}x=-\frac{1}{3}\y=-2\end{array}ight.$$
- step11: Check the solution:
  $$\left\{ \begin{array}{l}x=-\frac{1}{3}\y=-2\end{array}ight.$$
- step12: Rewrite:
  $$\left(x,yight) = \left(-\frac{1}{3},-2ight)$$
The solution to the system of equations $$3x-4y=7$$ and $$3x+2y=-5$$ is $$x = -\frac{1}{3}$$ and $$y = -2$$.

Now, let's subtract the second equation from the first equation to find the resulting equation:

$$3x - 4y = 7$$

$$3x + 2y = -5$$

Subtracting the second equation from the first equation:

$$(3x - 4y) - (3x + 2y) = 7 - (-5)$$

$$-6y = 12$$

Therefore, the resulting equation is $$-6y = 12$$.
The resulting equation is $$-6y = 12$$.

\( \begin{array}{l}3 x-4 y=7 \\ 3 x+2 y=-5\end{array} \) When the second equation is subtracted from the first, the resulting equation is \( -6 y=-12 \) \( -2 y=-12 \) \( -6 y=12 \) \( 2 y=12 \)

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