What is the average (arithmetic mean) of x, y, and z?

Statement 1
$$-x-4y+3z=14$$ =
Values of x, y and z are unknown, so we can't arrive at a definite answer hence, Statement 1 is INSUFFICIENT.

Statement 2
$$5x+8y+z=26$$
Value of x, y and z are unknown, so we can't arrive at a definite answer. Hence, Statement 2 is INSUFFICIENT.

Combining statements 1 and 2 together
Adding Statement 1 and 2 together
$$-x-4y+3z=14$$ ..............................1
$$5x+8y+z=26$$ ...............................2

We obtain
$$4x+4y+4z=40$$
Dividing both sides of the equation by 4, we obtain
$$\frac{\left(4x+4y+4z\right)}{4}=\frac{40}{4}$$
$$\left(x+y+z\right)=10$$
$$Average\ arithemetic\ mean=\frac{EFX}{EF}=\frac{10}{3}$$
both statement together is SUFFICIENT.

$$answer\ is\ Option\ C$$

DivyaD wrote:What is the average (arithmetic mean) of x, y, and z?

(1) —x − 4y + 3z = 14

(2) 5x + 8y + z = 26

$$? = \frac{{x + y + z}}{3}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\boxed{\,? = x + y + z\,}$$
$$\left( 1 \right)\,\, - x - 4y + 3z = 14\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y,z} \right) = \left( { - 14,0,0} \right)\,\,\,\, \Rightarrow \,\,\,? = - 14 \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y,z} \right) = \left( { - 10, - 1,0} \right)\,\,\,\, \Rightarrow \,\,\,? = - 11 \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,{\rm{INSUFF}}.$$
$$\left( 2 \right)\,\,5x + 8y + z = 26\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y,z} \right) = \left( {0,0,26} \right)\,\,\,\, \Rightarrow \,\,\,? = 26 \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y,z} \right) = \left( {4,0,6} \right)\,\,\,\, \Rightarrow \,\,\,? = 10 \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,{\rm{INSUFF}}{\rm{.}}$$
$$\left( {1 + 2} \right)\,\,\,\left\{ \matrix{
\, - x - 4y + 3z = 14 \hfill \cr
\,5x + 8y + z = 26 \hfill \cr} \right.\,\,\,\,\,\mathop \Rightarrow \limits^{\left( + \right)} \,\,\,\,\,4\left( {x + y + z} \right) = 40\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,{\rm{SUFF}}{\rm{.}}$$


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.