Question

The answer is 5,-2,-3 but i need help with steps that make sense

What is the solution to the system of equations?
x+3y−z=2
4x+2y+5z=1
3x+z=12




(−1, 2, 3)

(3, 0, 3)

(5, −2, −3)

(−4, 1, −3)

Answer 1

`\left\{\begin{array}{ccc}x+3y-z=2\\4x+2y+5z=1\\3x+z=12&\to z=12-3x\end{array}\right\\\\\text{substitute the value of z to the first and second equation}\\\\\left\{\begin{array}{ccc}x+3y-(12-3x)=2\\4x+2y+5(12-3x)=1\end{array}\right\\\\\left\{\begin{array}{ccc}x+3y-12+3x=2&|+12\\4x+2y+60-15x=1&|-60\end{array}\right\\\\\left\{\begin{array}{ccc}4x+3y=14&|\cdot2\\-11x+2y=-59&|\cdot(-3)\end{array}\right`

`\underline{+\left\{\begin{array}{ccc}8x+6y=28\\33x-6y=177\end{array}\right}\ \ \ \ |\text{add both sides of the equations}\\.\ \ \ \ \ \ \ 41x=205\ \ \ \ |:41\\.\ \ \ \ \ \ \ x=5\\\\\text{substitute the value of x to the first equation}\\\\4(5)+3y=14\\20+3y=14\ \ \ \ |-20\\3y=-6\ \ \ \ \ |:3\\y=-2\\\\\text{substitute the value of x to the equation:}\ z=12-3x\\\\z=12-3(5)=12-15=-3\\\\x=5,\ y=-2,\ z=-3\to(5,\ -2,\ -3)`