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05 Jun 2021, 04:46
Kudos
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Difficulty:
95%
(hard)
Question Stats:
54% (03:05) correct
46%
(03:19)
wrong
based on 142
sessions
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Solve the following system of linear equations
3x−4y−z=5
3y−4z+2w=−5
2x+w=10
for positive integers x, y, z and w. Find the value of x+ y+ z + w.
A. 6
B. 8
C. 9
D. 10
E. 12
3x−4y−z=5
3y−4z+2w=−5
2x+w=10
for positive integers x, y, z and w. Find the value of x+ y+ z + w.
A. 6
B. 8
C. 9
D. 10
E. 12
05 Jun 2021, 08:25
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nick1816
Interesting. At first look, you might ask... how do I find values of 4 variables with 3 equations!
Note that x, y, z and w are all +ve integers.
Pick the last equation: 2x+w=10
There can be only 4 pair of values for (x,w) which can satisfy the equation and the +ve integer condition:
(1,8),(2,6),(3,4),(4,2)
Put these values in equations 1 and 2... and you will find that only one pair will satisfy the equations.
Start with equation 1:
Case 1 - Put x = 1
3x−4y−z=5
3−4y−z=5
-4y-z = 2
There is no value of +ve integers y and z which can satisfy the equation. Hence, (1,8) reject.
Case 2 - Put x = 2
3x−4y−z=5
6−4y−z=5
−4y−z=-1
4y+z=1
Same as case 1. (2,6) reject.
Case 3 - Put x = 3
3x−4y−z=5
9−4y−z=5
4y+z=4
Still a no. (3,4) reject.
Case 4 - Put x = 4
3x−4y−z=5
12−4y−z=5
4y+z=7 ---> To satisfy this equation, only one pair of (y,z) is possible.... (1,3).
Hence, x, y, z and w are 4,1,3,2.
Doubtful? To cross-check... put these values in equation 2. They satisfy!
Answer: 10.
General Discussion
