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Subtract 2y + z = 11 from x + z = 8 to get 2y - x = 3 (equation 1). This eliminates one equation already.

Multiply 3x + y = 19 by 2 to get 6x + 2y = 38 (equation 2).

6x + 2y = 38 (equation 2) minus 2y - x = 3 (equation 1) is 7x = 35 because y cancels out. Hence, x = 5.

If x = 5, then z = 3 and y = 4. The average of x, y and z is 4, which is C.
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Bunuel
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 variable

With 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 = 5

Find z: x + z = 8 (P)
5 + z = 8
z = 3

4) 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
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Bunuel
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

Subtracting the first equation from the second, we have:

2y - x = 3

Multiplying the above equation by 3 and adding that to the third equation, we have:

7y = 28

So y = 4. Substituting y = 4 into the third equation, we have:

3x + 4 = 19

3x = 15

So x = 5. Substituting x = 5 into the first equation, we have:

5 + z = 8

z = 3

Thus, the average of x, y and z is (5 + 4 + 3)/3 = 4.

Answer: C
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