math question help??

Question

there are three different numbers all less then 50. when added two at a time the totals are 41, 49, and 56. Whats the largest of these numbers?

Answer 1

x+y=41
y+z=49
z+x=56
adding
2(x+y+z)=146
x+y+z=73
z=32
x=24
y=17

Answer 2

Let the first number be x
the second y and
the third z

x + y = 41 ... (1)
y + z = 49 ... (2)
z + x = 56 ... (3)

(3) - (2)
x - y = 56 - 49 (cos z-z = 0)
x - y = 7 ... (4)

Now we are left with (1) and (4)
x + y = 41 ... (1)
x - y = 7 ... (4)

From (1) x = 41 - y .... (5)
Plug this value in (4)
41 - y - y = 7
41 - 2y = 7
- 2y = 7 - 41
-2y = -34
y = -34/-2
y = 17

Plug this value of y in (5)
x = 41 - y
x = 41 - 17
x = 24

To find z, plug the values of x in (3) or the value of y in (2)
y + z = 49 ... (2)
z + x = 56 ... (3)

From (2),
17 + z = 49
z = 49 - 17
z = 32

or from (3)
z + x = 56
z + 24 = 56
z = 56 - 24
z = 32

So the numbers are 17, 24 and 32 with 32 being the largest

Answer 3

a + b = 41 (1)

b+c=49 (2)

a+c = 56 (3)

(3) - (2) => a-b=56-49=7 (4)
(4) + (1)=> 2a=48 => a=24
b=17 and c= 32
so c=32 is the largest number

Answer 4

The numbers are 17, 24 and 32.

32 + 17 = 49...one of the numbers given.

17 + 24 = 41...also one of the given numbers.

24 + 32 = 56...the last given number.

Out of 17, 24 and 32, which is the biggest?

Yes, 32 is the biggest.

Guido

Answer 5

n1 + n2 = 41
n2 + n3 = 49
n3 + n1 = 56 so.... n3=56-n1 sub. into eq2

n2 + (56-n1) = 49
n2 - n1 = 49 - 56 = -7

now...
n2 + n1 = 41 (reversed ... you'll see)
n2 - n1 = -7 add these two equations together
2(n2) = 34 n2 = 17
if n2 = 17 ... 17 - ? = -7 n1 = 24
n3 must be 17 + ? = 49 n3 = 32

(24 , 17 , 32) 1+2= 41 2+3= 39 3+1= 56

largest is 32

Answer 6

they are 17, 24, and 32, so 32 is the largest number

Answer 7

32

Answer 8

Sorry, I can't figure that out.