Bunuel wrote:

If x + z = 8, 2y + z =11, and 3x + y= 19, what is the average (arithmetic mean) of x, y, and z?

A. 2

B. 3

C. 4

D. 5

E. 6

1) Pick two equations, eliminate one variableWith three equations and two variables, usually, you can eliminate only one variable at first.

Look for a pair in which one variable has the

same coefficient (not always possible).

x + z = 8 (P)

2y + z = 11 (Q)

3x + y = 19 (R)

P and Q: z's coefficient is 1 in both. Use (Q - P)

2y + z = 11 (Q)

(-)x + z = 8 (P)___

2y - x = 3 (S)

2) Pair new equation with an original that has the same two remaining variables Pair (S) with (R). Both have x and y only.

3x + y = 19 (R), rewritten to line up x and y

y + 3x = 19 (R)

2y - x = 3 (S)

x- and y-coefficents do not match. Multiply bottom equation (S) by 3, then add (R + S)

y + 3x = 19 (R)

6y - 3x = 9__ (S\(_1\))

7y . . . = 28

y = 4 3) Find other variables. Substitute known value(s) into other equation(s)Find x: y + 3x = 19 (S)

4 + 3x = 19

3x = 15

x = 5Find z: x + z = 8 (P)

5 + z = 8

z = 34) Average of x + y + z?3, 4, 5 are consecutive integers. Median = mean

Mean = 4

OR

\(\frac{3 + 4 + 5}{3} =\frac{12}{3}=4\)Answer C

_________________

In the depths of winter, I finally learned

that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"