Multivariate mystery

x+y+z=67 x-y=7 y-z=3 (x=28)

x < y < z | x + y + z = 67 | z - y = 7 | y - x = 3 | z - x = 10 | x = smallest number, so take the 3 and 10 | 67 - ( 3 + 10 ) = 67 - 13 = 54 | there are three numbers, so | 54/3 = 18 | 18 = smallest number = x | x = 18 | y = x + 3 = 18 + 3 = 21 | z = largest number = x + 10 = 18 + 10 = 28

a>b>c

a=b+7

b=c+3 => a = c+10

a +b +c=67

c+10+c+3+c=67

3c+13=67

3c=54

c=18

a=c+10

a=18+10

a=28

$(x-3) + x + (x+7) = 68$ $x = 21 \Rightarrow x+7 = 28$

First remove the 3s and 7s: 67-7-3-3=54(3 of C)

1C=54 divided by 3=18

18+3+7=28 (A)

Let the greatest number be x

Let the middle number be y

Let the smallest number be z

x + y + z = 67

x - y = 7 ----------------------------(i)

x = 7 + y-------------------------------(ii)

y - z = 3 ------------------------------(iii)

y = z + 3-------------------------------(iv)

7 + y = x

7 + z + 3 = x

x = z + 10

z = x - 10 --------------------------------------------(v)

x + y + z = 67

x + x - 7 + x - 10 = 67

3x - 17 = 67

3x = 84

x = 84 / 3 = 28

Thus,

largest number is 28

let the numbers be x+7,x,x-3 then x+7+x+x-3=67 solving this we get, 3x=63 from this x=21 then we get numbers 28,21,18 from this the highest number is 28

x+y+z=67 x-y=7 y-z=3 y+7+y+y-3=67 3y+4=67 3y+4-67=0 3y-63=0 y=63/3=21 x=28 z=18

28+21+18

Start by finding any random number between 20 and 30, ie: 23. Now we want this to have a difference with 7. 23- 7 = 16. So we now think the first two numbers largest to smallest is 23 and 16. 16-3 = 13. Now it's 23, 16, 13. Add them up, which equals 52. But we want 67. The difference between 67 and 52 is 15, and there are 3 numbers. 15÷3 = 5. Therefore we add each of the numbers (23, 16, 13) by 5; 28, 21, 18. To make sure we have 67, add them up, which equals 67!!!! The largest number in the equation is 28, so the answer is 28!!!!!

let a,b,c are three different numbers & a>b>c now according to problem we get z-y=7 &y-x=3 & x+y+z=67 by solving we get x=18,y=21,z=28 largest nmber is 28

Let x,y,z be the three distinct numbers with z as largest, y as second largest and x as smaller Lets form the algebraic equations with given conditions: x+y+z=67.........(1) z-y=7..................(2) y-x=3...................(3) (1)+(2): x+2z=74.......(4) (2)+(3): z-x=10..........(5) (4)+(5): 3z=84: z=28 x=18 y=21