X +Y + Z = 200 <--- the sum of the three numbers is 200; This is the first equation!!!
X = Y + 15 <-- first number is 15 more than the second number; This is the second equation
X + Y = 2z + 25 <--- the sum of the first two numbers is 25 more than twice the third number; this is the third equation
We have 3 equations with three unknown quantities.
Substitutes the second equation into the first and third,
the system becomes:
(y+15)+y + z = 200
(y+15) + y = 2z + 25
which, after combining like terms, simplifies to:
2y + z + 15 = 200 <--- please label this equation ALPHA, we will need it later
2y + 15 = 2z + 25
Subtracts second equation FROM the first equation:
(2y + z + 15) - (2y + 15) = 200 - (2z + 25)
2y + z + 15 - 2y - 15 = 200 - 2z - 25
z = 175 - 2z
3z = 175
z = 58 and 1/3
Per equation ALPHA above in bold
2y + z + 15 = 200
2y + z = 200 - 15
2y + z = 185
2y + 58 and 1/3 = 185
2y = 185 - (58 and 1/3)
2y = 126 and 2/3 = 380/3
y = 380/3 * 1/2 = 190/3 = 63 and 1/3
Per the original first equation:
X + Y + Z = 200
X + (63 and 1/3) + (58 and 1/3) = 200
X + 121 and 2/3 = 200
X = 200 - (121 and 2/3)
= 78 and 1/3
IN summary, X = 78 and 1/3
Y = 63 and 1/3
Z = 58 and 1/3
The first number is in fact 15 more than the second as 63+15 = 78 with 1/3 not being a factor.
The sum of the first two numbers is 141 and 2/3. 25 less than that is, 116 and 2/3 = 2 x ( 58 + 1/3).
So, despite the fractions, the solutions check out and the answers in bold are correct.
