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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...Solve the following system for $\displaystyle \begin{align*} x, y, z \end{align*}$:

$\displaystyle \begin{align*} z &= 12 - x + 4\,y \\ z &= 4 + 5\,x + 3\,y \\ z &= 5 - 12\,x - 5\,y \end{align*}$

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

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