The sum of three numbers is 86. The third number is 4 times the first. The first number is 10 more than the second. What are the numbers?

The first number is 16

The second number is 6

The third number is 64

Step-by-step explanation:

The information from the word problem are;

The count of the numbers in the sum = 3 numbers

The value of the sum of the three numbers = 86

The value of the third number = 4 × The value of the first number

The value of the first number = 10 + The value of the second

Let 'x' represent the first number, let 'y' represent the second number and let 'z' represent the third number, we have;

x + y + z = 86...(1)

z = 4 × x...(2)

x = y + 10...(3)

From equation (3), we get;

x = y + 10

∴ y = x - 10...(4)

Substituting the value of z from equation (2) and the value of y from equation (4) in equation (1) gives;

x + y + z = 86

z = 4 × x = 4·x

y = x - 10

∴ x + y + z = x + (x - 10) + 4·x = 86

x + (x - 10) + 4·x = x + x - 10 + 4·x = 6·x - 10 = 86

6·x = 86 + 10 = 96

∴ x = 96/6 = 16

x = 16

From equation (2), we get;

z = 4 × x

∴ z = 4 × 16 = 64

z = 64

From equation (4), we get

y = x - 10

∴ y = 16 - 10 = 6

y = 6

Therefore; x = 16, y = 6, and z = 64

The numbers are 16, 6, and 64