Answer:
The numbers are 7, 4 and 1
Step-by-step explanation:
Let the numbers =x, y, z.
The sum of all three numbers is 12, x+y+z=12--------i
The sum of twice the first number, 4 times the second number, and 5 times the third number is 35, 2x+4y+5z=35-------ii
The difference between 8 times the first number and the second number is 52, 8x-y=52---------iii
Now, from (i),z =1-x-y--------iv
substitute (iv) in (ii),
2x+4y+5(12-x-y)=35
2x+4y+60-5x-5y=35
collect like terms
-3x-y=-25
that is, 3x-y=25--------v
solving (iii) and (v) simultaneously (add iii and v), we have
11x=77
divide both sides by 11 to obtain x=7
Put x=7 in (v)
3(7)+y=25
21+y=25
implies y=25-21=4
put x=7 and y=4 in (iv)
z=12-7-4=1
Hence, the numbers are 7, 4 and 1.
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