can you help me with some math problems for the math final?

Question

1. If a and b are intergers, which equation is always true?
1. a/b = b/a
2. a+2b=b+2a
3. a-b=b-a
4. a+b=b+a

2. The sum of the ages of the three anderson brothers is 63. If their ages can be represented as consecutive intergers, what is the age of the middle brother?
explain please.

Answer 1

The answer to the first question is 4 - I think it's the reflexive property.

The answer the second question... 21. You could work it this way:

X + X+1 + X+2 = 63
3X+3 = 63
3X = 60
X = 20
The first brother is 20, the second brother is 21.

Answer 2

#1. Is answer 4. Adding numbers together is not dependent on any sequencing. For example 7 + 3 = 10 and 3 + 7 = 10 the sequence doesn't matter a bit.

#2. Works like this. 3 people with consecutive ages with a total age of 63. If you divide the total age by the number of people it will will give 3 equal divisions. That is the center brother's age. 63/3 = 21. If all the brothers were the same age that would be the answer across the board. But one is a year younger and the other is a year older. So if you subtract a year from the younger brother and add it to the old brother you get 20 + 21 + 22 = 63.

I have to tell you if there had been three I wouldn't have helped you. I don't normally do other peoples' homework. I will be watching for you name on other questions.

Good luck on the final.

Answer 3

1. Doesn't really matter if a and b are integers or not, #4 is the only one that is always true. It's easy to check, just let a = 1, b = 2 and try them.
a/b = 1/2, b/a = 2 - not equal
a+2b = 5, b+2a = 4 - not equal
a-b = -1, b-a = 1 - not equal
a+b = 3, b+a = 3 - equal.

In general a/b = 1 / (b/a) and a-b = - (b-a).

2. If they are consecutive integers and the middle one is x, the other two must be x-1 and x+1. So the total is x + (x+1) + (x-1) = 3x = 63, so x = 21. So the middle brother is 21.

Answer 4

1. The 4th answer is correct. The sum of any numbers is always the same no matter how the numbers are arranged in the addition. I think this is called the commutative process.

2. Divide the age of the brothers by the number of brothers (63/3=21) Subtract 1 for the youngest brother (21-1=20). Add 1 for the oldest brother (21+1=22). Leave the middle brother the same age (21). 20+21+22=63. The middle brother is 21.

Answer 5

For #1, the answer is choice 4.

For #2, it will probably be easiest to solve by setting up a system of equations.
the youngest brother could be represented by x, the next y, and the oldest z.
therefore, x+y+z=63
you also know that: x+1=y and x+2=z because as stated in the problem, their ages can be represented by consecutive integers.
so if you substitute those values into the original equation you get: x + (x+1) + (x+2) = 63
Solve for x, you get x=20
add 1, y=21, add 2, z=22

:)

Answer 6

The first answer is 4.
a+b and b+a is the same thing, so it's always true...

second one is explained below:

(n)+(n+1)+(n+2)=63
combine!
3n+3=63
-3 -3
3n=60
divide both sides by three
n=20
that makes the middle integer 21
CHECKING YOUR WORK:
20+21+22=63!

SO YOUR ANSWER IS:
21

Answer 7

1. a+b=b+a

2. idk....sry! =l