$$ 3 y+4 x=-1 \quad $$ and $$ 5 y-x=2 $$

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Solve the system of equations $$ 3y+4x=-1;5y-x=2 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}3y+4x=-1\5y-x=2\end{array}ight.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}3y+4x=-1\x=-2+5y\end{array}ight.$$
- step2: Substitute the value of $$x:$$
  $$3y+4\left(-2+5yight)=-1$$
- step3: Simplify:
  $$23y-8=-1$$
- step4: Move the constant to the right side:
  $$23y=-1+8$$
- step5: Add the numbers:
  $$23y=7$$
- step6: Divide both sides:
  $$\frac{23y}{23}=\frac{7}{23}$$
- step7: Divide the numbers:
  $$y=\frac{7}{23}$$
- step8: Substitute the value of $$y:$$
  $$x=-2+5	imes \frac{7}{23}$$
- step9: Calculate:
  $$x=-\frac{11}{23}$$
- step10: Calculate:
  $$\left\{ \begin{array}{l}x=-\frac{11}{23}\y=\frac{7}{23}\end{array}ight.$$
- step11: Check the solution:
  $$\left\{ \begin{array}{l}x=-\frac{11}{23}\y=\frac{7}{23}\end{array}ight.$$
- step12: Rewrite:
  $$\left(x,yight) = \left(-\frac{11}{23},\frac{7}{23}ight)$$
The solution to the system of equations $$3y+4x=-1$$ and $$5y-x=2$$ is $$x=-\frac{11}{23}$$ and $$y=\frac{7}{23}$$.
$$ x=-\frac{11}{23} $$ and $$ y=\frac{7}{23} $$.

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