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23 May 2021, 21:02
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A
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Difficulty:
55%
(hard)
Question Stats:
67% (02:21) correct
33%
(02:35)
wrong
based on 183
sessions
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Not Attempted Yet
If we solve the following system of equations
x+y+2z=10
3x−y+z=5,
where x, y, z are non-negative integers, what is the sum of all possible values of x?
A. 3
B. 5
C. 7
D. 9
E. 11
x+y+2z=10
3x−y+z=5,
where x, y, z are non-negative integers, what is the sum of all possible values of x?
A. 3
B. 5
C. 7
D. 9
E. 11
24 May 2021, 10:57
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nick1816
Let us remove y by adding the two equations
x+y+2z=10
3x−y+z=5
4x+3z=15......so z is odd and x is a multiple of 3 as 4x=3(5-z)
So z=1 and only possibility of x is 3.
OR
Let us remove z by multiplying second equation by 2 and subtracting i from ii.
x+y+2z=10
6x−2y+2z=10
5x-3y=0 or 5x=3y......so x is a multiple of 3.
Now 4x+3z=15, and x is a multiple of 3, that is 0, 3, 6..
4*3+3z=15......z=1......possible
4*6+3z=15......z=-3....No
Only possible value is 3.
A
24 May 2021, 22:26
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chetan2u
As they say "non-negative integers" 0 could be another possible value of x, couldn't it? (x=y=0, z=5). The answer does not change because it asks for the "sum of all possible values of x".
