the sum of three numbers is 12 The sum of twice the first nu

the sum of three numbers is 12. The sum of twice the first number, 5 times the second number, and 6 times the third number is 39. The difference between 4 times the first number and the second is 31. Find the three numbers.

Solution

Given the sum of three numbers is 12. The sum of twice the first number, 5 times the second number, and 6 times the third number is 39. The difference between 4 times the first number and the second is 31.

Let the required three numbers be a,b,c.

sum of three numbers is 12.

a+b+c=12

The sum of twice the first number, 5 times the second number, and 6 times the third number is 39.

2a+5b+6c=39

The difference between 4 times the first number and the second is 31.

4a-b=31

From the given data, we got three equations.

Solve first and two equations to eliminate \'c\'.

a+b+c=12 and 2a+5b+6c=39

Multiply the first equatipn by 6 and then subtract second equation from it.

6(a+b+c)-(2a+5b+6c)=6(12)-39

6a+6b+6c-2a-5b-6c = 72-39

4a+b = 33

Solve the above equation 4a+b = 33 and 4a-b=31 for a and b.

Add 4a+b = 33 and 4a-b=31

4a+b +4a-b=33+31

8a = 64

a=8

Substitute a=8 in third equation 4a-b=31

32-b=31

b=1

Substitute a=8, b=1 in first equation a+b+c=12

8+1+c = 12

c=12-9

c=3

Therefore, the required three numbers are a=8, b=1 and c=3