SOLUTION: if w,x,y,and z are consecutive integers and if w>x>y>z, then (w-z)(y-8)+x-z=?
I keep getting -5 for the answer nit matter what numbers I plug in. The answer is -1.
Thank you
Algebra.Com
Question 631811:  if w,x,y,and z are consecutive integers and if w>x>y>z, then (w-z)(y-8)+x-z=?
I keep getting -5 for the answer nit matter what numbers I plug in. The answer is -1.
Thank you 
Answer by KMST(5328)   (Show Source): You can put this solution on YOUR website!
 I believe the problem was meant to be
"if w,x,y,and z are consecutive integers and if w>x>y>z, then (w-z)(y-x)+x-z=?"
In the posting, the expression came up as "(w-z)(y-8)+x-z" and that will not work.
 
The reason that you are getting -5 instead of -1 is that you are using increasing consecutive numbers as w,x,y,and z.
The numbers are supposed to be decreasing from w to z.
In the problem, w is the largest number, x is one less than w, and so on.
It says w>x which means w is greater than x.
Similarly x is one more than y (consecutive and greater than), and y is one more than z.
 
Expressing all numbers as a function of x:
w=x+1
x=x
y=x-1 and
z=x-2
w-z=(x+1)-(x-2)=x+1-x+2=3
y-x=x-1-x=-1
x-z=x-(x-2)=x-x+2=2
Putting it all together:
(w-z)(y-x)+x-z=(3)(-1)+2=-3+2=-1
for any x, meaning
for any four consecutive integers called w, x, y, and z
listed in order from greatest to smallest.
 
If you make w, x, y, and z be consecutive integers listed from smallest to greatest (as w
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