ons $$ -4 x+y=5 $$ and $$ -x+4 y=12 $$

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Solve the system of equations $$ -4x+y=5;-x+4y=12 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}-4x+y=5\-x+4y=12\end{array}ight.$$
- step1: Solve the equation:
  $$\left\{ \begin{array}{l}y=5+4x\-x+4y=12\end{array}ight.$$
- step2: Substitute the value of $$y:$$
  $$-x+4\left(5+4xight)=12$$
- step3: Simplify:
  $$15x+20=12$$
- step4: Move the constant to the right side:
  $$15x=12-20$$
- step5: Subtract the numbers:
  $$15x=-8$$
- step6: Divide both sides:
  $$\frac{15x}{15}=\frac{-8}{15}$$
- step7: Divide the numbers:
  $$x=-\frac{8}{15}$$
- step8: Substitute the value of $$x:$$
  $$y=5+4\left(-\frac{8}{15}ight)$$
- step9: Simplify:
  $$y=\frac{43}{15}$$
- step10: Calculate:
  $$\left\{ \begin{array}{l}x=-\frac{8}{15}\y=\frac{43}{15}\end{array}ight.$$
- step11: Check the solution:
  $$\left\{ \begin{array}{l}x=-\frac{8}{15}\y=\frac{43}{15}\end{array}ight.$$
- step12: Rewrite:
  $$\left(x,yight) = \left(-\frac{8}{15},\frac{43}{15}ight)$$
The solution to the system of equations $$ -4x+y=5 $$ and $$ -x+4y=12 $$ is $$ (x,y) = (-\frac{8}{15},\frac{43}{15}) $$ or $$ (x,y) = (-0.5\dot{3},2.8\dot{6}) $$.
The solution to the system of equations is $$ x = -\frac{8}{15} $$ and $$ y = \frac{43}{15} $$.

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