JOMO 5, Short 2
x , y and z are non-negative integers where x y + 3 x + 2 y = 2 and y z + 4 y + 3 z = 5 2 . Find the value of x + y + z + x y + y z + x z + x y z
The answer is 25.
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3 solutions
For those who didn't notice: the answer when the values of x , y , z are known could've been easily found out by substituting them into ( x + 1 ) ( y + 1 ) ( z + 1 ) − 1 , which is a bit faster for getting the final answer.
yup......It's quite clear that this is the only solution.. for any positive value except these the solution will not hold for first equation as then x y + 3 x + 2 y > 2
From the equations we can get x = ( 2 − 2 y ) / ( y + 3 ) . . . . 1 and also y = ( 5 2 − 3 z ) / ( z + 4 ) . . . . . 2 . Using these two equations we can substitute in equation 2 into equation 1 to get x in terms of z . This gives us x = ( 8 z − 9 6 ) / ( 5 2 − 3 z ) . . . . 3 . Now equation 1 = 3 . Doing this gives us a z in terms of y but a slightly different equation compared to 2 . This is z = ( 1 9 6 − 4 y ) / ( 1 5 + y ) . . . . 4 . Plug equation 4 into the given equations in the question to get y = 1 . Use y = 1 to find x and z which are x = 0 and z = 1 2 . Hence x + y + z + x y + y z + x z + x y z = 2 5 .
Factorise:
( x + 2 ) ( y + 3 ) = 8
y + 3 ) ( z + 4 ) = 6 4
( y + 3 ) is a factor of 8 and 64 and x , y , z > = 0
Therefore y = 1
Therefore x = 0 , z = 1 2
( x + 1 ) ( y + 1 ) ( z + 1 ) − 1 = 2 5
We have { x y + 3 x + 2 y = 2 y z + 4 y + 3 z = 5 2 ⇒ ⎩ ⎪ ⎨ ⎪ ⎧ x = 0 y = 1 z = 1 2 Hence x + y + z + x y + y z + x z + x y z = 2 5