# SOLUTION: If x, y, and z are positive intergers such that the value of x+y is even and the value of {{{(x+y)^2}}}+x+z is odd, which of the following must be true? (A) x is odd (B) x is e

Algebra ->  Algebra  -> Expressions -> SOLUTION: If x, y, and z are positive intergers such that the value of x+y is even and the value of {{{(x+y)^2}}}+x+z is odd, which of the following must be true? (A) x is odd (B) x is e      Log On

 Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

 Algebra: Evaluation of expressions, parentheses Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on expressions Question 92788: If x, y, and z are positive intergers such that the value of x+y is even and the value of +x+z is odd, which of the following must be true? (A) x is odd (B) x is even (C) If z is even, then x is odd (D) If z is even, then xy is even (E) xy is evenAnswer by Edwin McCravy(8879)   (Show Source): You can put this solution on YOUR website!``` If x, y, and z are positive integers such that the value of x+y is even and the value of (x+y)²+x+z is odd, which of the following must be true? (A) x is odd (B) x is even (C) If z is even, then x is odd (D) If z is even, then xy is even (E) xy is even There are 8 possible cases: We can use any even and odd integer. The easiest even positive integer is 2 and the easiest odd integer is 1. So let's use these in those 8 cases x y z x+z=? (x+y)²+x+z=? ------------------------------------------------------------ 1. 2 2 2 2+2=4=EVEN (2+2)²+2+2=4²+2+2=16+2+2=20=EVEN 2. 2 2 1 2+1=3=ODD (2+2)²+2+1=4²+2+1=16+2+1=19=ODD 3. 2 1 2 2+2=4=EVEN (2+1)²+2+2=3²+2+2= 9+2+2=13=EVEN 4. 2 1 1 2+1=3=ODD (2+1)²+2+1=3²+2+1= 9+2+1=12=EVEN 5. 1 2 2 1+2=3=ODD (1+2)²+1+2=3²+1+2= 9+1+2=12=EVEN 6. 1 2 1 1+2=3=ODD (1+2)²+1+1=3²+1+1= 9+1+1=11=ODD 7. 1 1 2 1+1=2=EVEN (1+1)²+1+2=2²+1+2= 4+1+2= 7=ODD 8. 1 1 1 1+1=2=EVEN (1+1)²+1+1=2²+1+1= 4+1+1= 6=EVEN There are only one case when x+y is EVEN and (x+y)²+x+z is ODD, That is case 7, where x=1, y=1 and z=2 So x is odd, y is odd, and z is even. The only one of the choices that must be true is A. Edwin```