<s>[INST] Solve the following system of equations by Cross Multiplication Method Solve the following System of equations by Cross Multiplication Method
$$2x-2y+3z=20$$
$$x+3y-z=12$$
$$3x-y+4z=22$$
My Attempt:
Let us solve first two equations:
$$\dfrac {x}{2-9}=\dfrac {y}{3-(-2)}=\dfrac {z}{6-(-2)} = k(let)$$
$$\dfrac {x}{-7}=\dfrac {y}{5}=\dfrac {z}{8}=k$$
So we have: $x=-7k$, $y=5k$ and $z=8k$
Now, putting these values of $x$,$y$ and $z$ in third equation,
$$3(-7k)-5k+4(8k)=22$$
$$k=\dfrac {11}{3}$$
Thus,
$x=\dfrac {-77}{3}$,
$y=\dfrac {55}{3}$ and 
$z=\dfrac {88}{3}$.
However the actual solution is $(x,y,z)=(21,-7,-12)$. Where has the error occured in this solution?
 [/INST] We want to use the Cross Multiplication Method (CMM) to solve
$$\begin{align} 2x-2y+3z&=20 \tag 1 \\ x+3y-z&=12 \tag 2 \\ 3x-y+4z&=22 \tag 3 \end{align}$$
We can write $(1)$ and $(2)$ as
$$\begin{align} 2x-2y+ (3z - 20)&=0  \\ x+3y-(z+12)&=0  \end{align}$$
By the CMM
$$\dfrac{x}{-2(-z-12)-3(3z - 20)} = \dfrac{y}{(3z-20)-2(-z-12)} = \dfrac{1}{2(3) - 1(-2)}$$
Simplifying
$$\dfrac{x}{84-7z} = \dfrac{y}{4 + 5z} = \dfrac{1}{8}$$
Solving
$$x = \dfrac{84-7z}{8}, y = \dfrac{4+5z}{8}$$
We now substitute these into $(3)$
$$3\left(\dfrac{84-7z}{8}\right) - \left(\dfrac{4+5z}{8}\right) + 4 z = 22$$
This gives $z = -12$, hence
$$(x, y, z) = (21, -7, -12)$$
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