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If x + z = 8, 2y + z =11, and 3x + y= 19, what is the average (arithme
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28 Dec 2017, 21:38
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Re: If x + z = 8, 2y + z =11, and 3x + y= 19, what is the average (arithme
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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 = 8x = 112y x2y = 3 3x+y = 19 6x+2y = 38 x2y = 3 Adding 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
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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 = 5Using value of x=5 in equation (3), we get y = 4Similarly using the value of x in equation (1), we get z = 3Therefore, 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
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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
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30 Dec 2017, 10:09
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 ycoefficents 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
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Re: If x + z = 8, 2y + z =11, and 3x + y= 19, what is the average (arithme
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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. Answer: C
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Re: If x + z = 8, 2y + z =11, and 3x + y= 19, what is the average (arithme &nbs
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