# Mason's question via Facebook about solving a system of equations (2)

#### Prove It

##### Well-known member
MHB Math Helper
Solve the following system for \displaystyle \begin{align*} x, y, z \end{align*}:

\displaystyle \begin{align*} z &= 12 - x + 4\,y \\ z &= 4 + 5\,x + 3\,y \\ z &= 5 - 12\,x - 5\,y \end{align*}
As all the z coefficients are the same, it's a good idea to eliminate the z values in the second and third equations, so apply R2 - R1 to R2 and R3 - R1 to R3...

\displaystyle \begin{align*} z &= 12 - x + 4\,y \\ 0 &= -8 + 6\,x - y \\ 0 &= -7 - 11\,x - 9\,y \end{align*}

Now we can multiply the second equation by 9 in order to eliminate the y terms...

\displaystyle \begin{align*} z &= 12 - x + 4\,y \\ 0 &= -72 + 54\,x - 9\,y \\ 0 &= -7 - 11\,x - 9\,y \end{align*}

Now apply R3 - R2 to R3

\displaystyle \begin{align*} z &= 12 - x + 4\,y \\ 0 &= -72 + 54\,x - 9\,y \\ 0 &= 65 - 65\,x \end{align*}

Thus \displaystyle \begin{align*} 65\,x = 65 \implies x = 1 \end{align*}, giving

\displaystyle \begin{align*} 54 \, \left( 1 \right) - 9\,y &= 72 \\ -9\,y &= 18 \\ y &= -2 \end{align*}

and

\displaystyle \begin{align*} z &= 12 - 1 + 4\,\left( -2 \right) \\ z &= 3 \end{align*}

Thus the solution is \displaystyle \begin{align*} \left( x , y, z \right) = \left( 1, -2, 3 \right) \end{align*}.

#### HallsofIvy

##### Well-known member
MHB Math Helper
Equivalently, since z is equal to each of 12−x+4y, −8+6x−y, and −7−11x−9y, they are all equal to each other:
12- x+ 4y= -8+ 6x- y and
-8+ 6x- y= -7- 11x- 9y.

From there, do the same as Prove It.