anonymous
  • anonymous
The sum of two numbers is 110. Their difference is 28. What is the smallest of the two numbers?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
A. 56 B. 41 C. 69 D. 55
anonymous
  • anonymous
x + y = 110 x - y = 28. what do you think? assume y
anonymous
  • anonymous
btw. x = 69,y=41. but don't look at the answer. find out why :) first.!

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anonymous
  • anonymous
Well, it says which is the smallest of the two numbers, so I'd have to go with 41 because that is smaller.
anonymous
  • anonymous
Is tha
anonymous
  • anonymous
Is that right or am I wrong about that one?
anonymous
  • anonymous
41+69=110 looks good to me
mathmate
  • mathmate
Here's the trick: bigger number = (sum+difference)/2 smaller number = (sum-difference)/2
anonymous
  • anonymous
I am so confused on which one it is. :/
anonymous
  • anonymous
So it must be 41 because that is the smaller number.
mathmate
  • mathmate
If you want the smaller number, Calculate (110-28)/2
anonymous
  • anonymous
I calculated that and got 96.
anonymous
  • anonymous
@alain relax you have it!
anonymous
  • anonymous
you have the answer. use it :P
anonymous
  • anonymous
Why does everyone call me Alain? Lol.. my name is Alaina. xD Oh well, it can be short for it. :P And thanks everyone! :)
anonymous
  • anonymous
there only is one solution.
anonymous
  • anonymous
MMMKKKKK!!!!
anonymous
  • anonymous
So 41 right?
anonymous
  • anonymous
oh and btw. (110-28) /2 = 41 :P
anonymous
  • anonymous
check your subtraction ^^
anonymous
  • anonymous
RONNCC has a good answer, basically it is best to use substitution in order to solve. He said: "x + y = 110 x - y = 28. what do you think? assume y 0 meaning x>y. y=x-28 from the second equation, so just substitute this into the first to give: x+x-28 = 110 ---> 2x=138 --> x=69. To find y simply use your first equation: 69+y=110 ---> y=41.
anonymous
  • anonymous
substitution is nice for harder equations, but generally elimination is a good method when the coefficient of x is 1 and they are in a simple form like this.

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