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# If x + z = 8, 2y + z =11, and 3x + y= 19, what is the average (arithme

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If x + z = 8, 2y + z =11, and 3x + y= 19, what is the average (arithme  [#permalink]

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28 Dec 2017, 21:38
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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

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Re: If x + z = 8, 2y + z =11, and 3x + y= 19, what is the average (arithme  [#permalink]

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29 Dec 2017, 06:07
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

z = 8-x = 11-2y
x-2y = -3
3x+y = 19

6x+2y = 38
x-2y = -3
7x = 35
x = 5
y=4
z=3
Mean of x,y,z (5,4,3) = 4
C
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If x + z = 8, 2y + z =11, and 3x + y= 19, what is the average (arithme  [#permalink]

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29 Dec 2017, 07:14
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

Since x + z = 8, z = 8 - x -> (1)

2y + 8 -x = 11 -> -x + 2y = 3 -> (2)

3x + y = 19
Multiplying this equation by 2, 6x + 2y = 38 -> (3)

Solving equation (2) and (3), -7x = - 35 -> x = 5
Using value of x=5 in equation (3), we get y = 4
Similarly using the value of x in equation (1), we get z = 3

Therefore, the average of x,y, and z is $$\frac{3+4+5}{3}$$ = 4(Option C)
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Re: If x + z = 8, 2y + z =11, and 3x + y= 19, what is the average (arithme  [#permalink]

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29 Dec 2017, 10:38
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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If x + z = 8, 2y + z =11, and 3x + y= 19, what is the average (arithme  [#permalink]

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30 Dec 2017, 10:09
1
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 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 $$\frac{3 + 4 + 5}{3} =\frac{12}{3}=4$$

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Re: If x + z = 8, 2y + z =11, and 3x + y= 19, what is the average (arithme  [#permalink]

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03 May 2018, 16:28
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

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.

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Re: If x + z = 8, 2y + z =11, and 3x + y= 19, what is the average (arithme &nbs [#permalink] 03 May 2018, 16:28
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