Absolute Confusion

Algebra Level 4

x + x + y z = 19 x + y y + z = 18 x y + z z = 19 \left| x \right| \, +\, x\, +\, y\, -\, z\, =\, 19\\ x\, +\, \left| y \right| \, -\, y\, +\, z\, =\, -18\\ x\, -\, y\, +\, \left| z \right| \, -\, z\, =\, 19

Let x x , y y , and z z be integers that satisfy the equations above. What is the value of x z y xz-y ?


The answer is 51.

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2 solutions

Martin Falk
Mar 10, 2015

If we add the first two equations, we get

x + y + 2 x = 1 |x| + |y| + 2x =1 .

Since we are looking for integer solutions, this means that we may conclude that x < 0 x<0 . Further manipulation of the equations does not allow us to conclude directly whether x x or y y must be positive or negative. The system of equations depends on whether the values of x x and y 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 x<0, y>0, z>0

b) x < 0 , y > 0 , z < 0 x<0, y>0, z<0

c) x < 0 , y < 0 , z > 0 x<0, y<0, z>0

d) x < 0 , y < 0 , z < 0 x<0, y<0, z<0 .


Rewriting the equations for each case gives:

a) y z = 19 ; x + z = 18 ; x y = 19 y-z=19; x+z=-18; x-y=19 ( x , y , z ) = ( 10 , 9 , 28 ) \Rightarrow (x,y,z)=(10,-9,-28)

b) y z = 19 ; x + z = 18 ; x y 2 z = 19 y-z=19; x+z=-18; x-y-2z=19 ( x , y , z ) = ( 4 , 5 , 14 ) \Rightarrow (x,y,z)=(-4,5,-14)

c) y z = 19 ; x 2 y + z = 18 ; x y = 19 y-z=19; x-2y+z=-18; x-y=19 \Rightarrow not solvable

d) y z = 19 ; x 2 y + z = 18 ; x y 2 z = 19 y-z=19; x-2y+z=-18; x-y-2z=19 ( x , y , z ) = ( 11 , 10 , 9 ) \Rightarrow (x,y,z)= (11,10,-9) .

We see that only case b) satisfies the conditions: x = 4 , y = 5 , z = 14 x=-4, y=5,z=-14 .

Thus, x z y = ( 4 ) ( 14 ) 5 = 51 xz-y= (-4)(-14)-5= \boxed{51} .

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 + + + + + + + + + + + + \begin{matrix} \underline { x } & \underline { y } & \underline { z } \\ + & + & + \\ + & + & - \end{matrix}\\ \begin{matrix} + & - & + \\ - & + & + \\ + & - & - \end{matrix}\\ \begin{matrix} - & - & + \\ - & + & - \\ - & - & - \end{matrix}

An important thing to know is the s g n sgn function: x = x s g n ( x ) \quad \left| x \right| \quad =\quad x*sgn\left( x \right)

So if we assume x to be positive, then: x = x , i f x > 0 \quad \left| x \right| \quad =\quad x,\quad if\quad x\quad >\quad 0

If we assume x to be negative, then: x = x , i f x < 0 \quad \left| x \right| \quad =\quad -x,\quad if\quad x\quad <\quad 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 = 14 x z y = 4 ( 14 ) 5 = 56 5 = 51 x\, =\, -4\\ y\, =\, 5\\ z\, =\, -14\\ xz\, -\, y\, =\, -4(-14)\, -\, 5\, =\, 56\, -\, 5\, =\, \boxed { 51 }

Let me know if I need to get into further detail.

Could you write out the whole solution?

Tor Andreas Sandve Kringeland - 6 years, 3 months ago

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