∣ x ∣ + x + y − z = 1 9 x + ∣ y ∣ − y + z = − 1 8 x − y + ∣ z ∣ − z = 1 9
Let x , y , and z be integers that satisfy the equations above. What is the value of x z − y ?
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The full solution is pretty lengthy, so I'll only put up a partial solution. The way to do this problem is to assume signs of each of the x, y, and z variables. That means there are 8 possible assumptions:
x + + y + + z + − + − + − + − + + − − − − − + − + − −
An important thing to know is the s g n function: ∣ x ∣ = x ∗ s g n ( x )
So if we assume x to be positive, then: ∣ x ∣ = x , i f x > 0
If we assume x to be negative, then: ∣ x ∣ = − x , i f x < 0
After eliminating some assumptions either by contradiction or by checking by plugging in the value to the system, only - + - for x, y, and z respectively, works, and we arrive at
x = − 4 y = 5 z = − 1 4 x z − y = − 4 ( − 1 4 ) − 5 = 5 6 − 5 = 5 1
Let me know if I need to get into further detail.
Could you write out the whole solution?
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If we add the first two equations, we get
∣ x ∣ + ∣ y ∣ + 2 x = 1 .
Since we are looking for integer solutions, this means that we may conclude that x < 0 . Further manipulation of the equations does not allow us to conclude directly whether x or y must be positive or negative. The system of equations depends on whether the values of x and y are positive or negative.
If there is a solution, it must be among the following four cases:
a) x < 0 , y > 0 , z > 0
b) x < 0 , y > 0 , z < 0
c) x < 0 , y < 0 , z > 0
d) x < 0 , y < 0 , z < 0 .
Rewriting the equations for each case gives:
a) y − z = 1 9 ; x + z = − 1 8 ; x − y = 1 9 ⇒ ( x , y , z ) = ( 1 0 , − 9 , − 2 8 )
b) y − z = 1 9 ; x + z = − 1 8 ; x − y − 2 z = 1 9 ⇒ ( x , y , z ) = ( − 4 , 5 , − 1 4 )
c) y − z = 1 9 ; x − 2 y + z = − 1 8 ; x − y = 1 9 ⇒ not solvable
d) y − z = 1 9 ; x − 2 y + z = − 1 8 ; x − y − 2 z = 1 9 ⇒ ( x , y , z ) = ( 1 1 , 1 0 , − 9 ) .
We see that only case b) satisfies the conditions: x = − 4 , y = 5 , z = − 1 4 .
Thus, x z − y = ( − 4 ) ( − 1 4 ) − 5 = 5 1 .